Def: For \(a,b \in \mathbb{R}, a < b\), a partition \(p\) of \([a,b]\) is a finite list of the form \(x_0, x_1, ..., x_n\) where \(a = x_0 < x_1 < \cdots < x_n = b\). If the gaps between the elements of \(p\) are equal, \(p\) is called a regular partition.
\[p = \{x_0, x_1, ..., x_n\}\]
Def: Let \(f: [a,b] \rightarrow \mathbb{R}\) be a bounded function (the range of the function is a bounded interval, so there exists \(M > 0: |f(x)| ≤ M \hspace{2mm} \forall \hspace{2mm}x \in [a,b]\)). This is a disadvantage of Riemann integrals because it places the boundedness restriction on the functions.
For a subset \(A \subset [a,b]\), we use the notation
\[\inf\limits_A f = \inf \{ f(x): x \in A \} \hspace{1cm} \sup\limits_A f = \sup \{f(x) : x \in A\}\]
With these, we can define the lower and upper Riemann sums.
Let \(f: [a,b] \rightarrow \mathbb{R}\) be a bounded function, and let \(p\) be a partition of \([a,b]\).
The lower Riemann sum of \(f\) is defined by
\[L(f,p,[a,b]) = \sum\limits_{i=1}^n \left(\inf\limits_{[x_{i-1}, x_i]}f\right)(x_i - x_{i-1})\]
The upper Riemann sum of \(f\) is defined by
\[U(f,p,[a,b]) = \sum\limits_{i=1}^n \left(\sup\limits_{[x_{i-1}, x_i]}f\right)(x_i - x_{i-1})\]
Denote \(M_i = \sup\limits_{[x_{i-1}, x_i]}f\), \(m_i = \inf\limits_{[x_{i-1}, x_i]}f\), and \(\Delta x_i = x_i - x_{i-1}\), then we can rewrite
\[L(f,p) = \sum\limits_{i=1}^n m_i \Delta x_i \hspace{1cm} U(f,p) = \sum\limits_{i=1}^n M_i \Delta x_i\]
Example: consider \(f(x) = x^2\) with \(x \in [0,1]\). For this interval, we can make a regular partition \(p\) of \(n\) parts such that \(x_0 = 0, x_1 = 1/n, x_2 = 2/n, ..., x_n = 1\). Then \(p = \{0, 1/n, ..., 1\}\).
The upper Riemann sum can be calculated as
\[U(f, p) = \frac{1}{n^3}\sum\limits_{i=1}^n i^2 = \frac{n(n+1)(2n+1)}{6n^3} = \frac{(n+1)(2n+1)}{6n^2}\]
The lower Riemann sum can be calculated as
\[L(f,p) = \frac{1}{n^3}\sum\limits_{i=1}^n (i-1)^2 = \frac{n(n-1)(2n-1)}{6n^3} = \frac{(n-1)(2n-1)}{6n^2} \]
Def: Let \(p\) and \(p'\) be partitions of \([a,b]\). We say that \(p'\) is a refinement of \(p\) if \(p \subset p'\). We can describe \(p'\) as being ‘finer’ than \(p\).
Let \(p\) and \(p'\) be partitions of \([a,b]\) such that \(p'\) is a refinement of \(p\). Let \(f: [a,b] \rightarrow \mathbb{R}\) be a bounded function. Then
\[L(f,p) ≤ L(f,p') ≤ U(f,p') ≤ U(f,p) \tag{1}\]
Theorem: If we have two arbitrary partitions \(p\) and \(q\) of \([a,b]\), then
\[L(f,p) ≤ U(f,q)\]
Let \(R = p \cup q\) be the partition of \([a,b]\) obtained by including all points from \(p\) and \(q\). Then \(R\) is a refinement of both \(p\) and \(q\). Then using the refinement property and (1) above,
\[L(f,p) ≤ L(f,R) ≤ U(f,R) ≤ U(f,q) \]
The upper and lower Riemann sums of a function are
\[U(f,p) = \sum\limits_{i=1}^n M_i \Delta x_i \hspace{1cm} L(f,p) = \sum\limits_{i=1}^n m_i \Delta x_i\]
where \(M_i = \sup\limits_{[x_{i-1},x_i]} f\), \(m_i = \inf\limits_{[x_{i-1},x_i]} f\) and \(\Delta x_i = x_i - x_{i-1}\).
Previously, we also saw that if we have two arbitrary partitions, \(P\) and \(Q\), of an interval, then \(L(f,P) ≤ U(f,Q)\). In other words, every upper Riemann sum is greater than every lower Riemann sum.
The lower integral of the function is \(\inf L(f,P) = \underline\int_a^b f\), and the upper integral is \(\sup U(f,P) = \overline\int_a^bf\).
Def: Let \(f: [a,b] \rightarrow \mathbb{R}\) be a bounded function. Define the upper and lower Riemann integral of \(f\) on \([a,b]\), respectively, by
\[\overline\int_a^b f = \inf\{U(f,p)\} \hspace{1cm} \underline\int_a^b f = \sup\{L(f,p)\}\]
where \(p\) is a partition of \([a,b]\). The set of all lower Riemann sums is bounded above by the lower Riemann integral, while the set of all upper Riemann sums is bounded below by the upper Riemann integral.
In the case where the upper integral of the function is equal to the lower integral of the function on \([a,b]\) (ie. \(\overline\int_a^b f =\underline\int_a^b f\)), then we say that \(f\) is Riemann integrable on \([a,b]\), and the common value is
\[\overline\int_a^b f =\underline\int_a^b f = \int_a^b f = \int_a^b f(x) dx\]
Aside: why do we always assume that \(f\) is a bounded function? The supremum and infimum of the function within an interval are only defined if the function is bounded (respectively) above or below.
Example: Let \(f(x) = \left\{\begin{array}{lr} \frac{1}{x} & 0 ≤ x ≤ 1 \\ 1 & x = 1 \end{array}\right.\)
Let \(p = \{x_0, x_1, ..., x_n\}\) be an arbitrary partition of the interval \([0,1]\). Then \(x_0 = 0 < x_1 < \cdots < x_n = 1\).
\(U(f,p) = \sum\limits_{i = 1}^n M_i \Delta x_i\) and \(L(f,p) = \sum\limits_{i = 1}^n m_i \Delta x_i\), where \(M_i, m_i\) have the same definitions as previously. In the first interval \([0,x_1]\), the infimum is \(\min\{1/x_1, 1\}\), but the supremum is not defined because the function is not bounded from above on that interval.
Example: Let \(f(x) = x^2, x \in [0,1]\). Find \(\int_0^1 f\) using the definition of the Riemann integral.
Let \(P_n = \{0, \frac{1}{n}, \frac{2}{n}, ..., 1\}\). Then
\[U(f,P_n) = \frac{2n^2 + 3n + 1}{6n^2} \hspace{1cm} L(f,P_n) = \frac{2n^2 - 3n + 1}{6n^2} \]
From the definition of the Riemann integral, we have that every lower Riemann sum is less than every upper Riemann sum. And the lower integral is the upper bound of the lower Riemann sums. The upper integral is the lower bound of all upper Riemann sums. In particular,
\[\underline\int_a^b f ≥ L(f,P_n) \hspace{2mm}\forall\hspace{2mm} n \in \mathbb{N} \hspace{1cm} \overline\int_a^b f ≤ U(f,P_n) \hspace{2mm}\forall\hspace{2mm} n \in \mathbb{N} \]
Combining these for our example,
\[\frac{2n^2 - 3n + 1}{6n^2} = L(f,P_n) ≤ \underline\int_0^1 f ≤ \overline\int_0^1 f ≤ U(f,P_n) = \frac{2n^2 + 3n + 1}{6n^2} \hspace{2mm}\forall\hspace{2mm} n \in \mathbb{N}\]
With this inequality, we can apply the squeeze theorem. Focus on the lower integral first. We can imagine it as a constant sequence:
\[\frac{2n^2 - 3n + 1}{6n^2} ≤ \underline\int_0^1 f ≤ \frac{2n^2 + 3n + 1}{6n^2}\]
The left and right terms approach \(\frac{1}{3}\) as \(n \rightarrow ∞\), so \(\underline\int_a^b f = \frac{1}{3}\).
Similarly, \(\overline\int_0^1 f = \frac{1}{3}\). Because the upper and lower integrals are equal, \(f\) is Riemann integrable on \([0,1]\), and \(\int_0^1f = \frac{1}{3}\).
Let \(f: [a,b] \rightarrow \mathbb{R}\) be a bounded function. Then \(f\) is Riemann integrable on \([a,b]\) iff for any \(\epsilon > 0\), there exists a partition \(p_\epsilon\) of \([a,b]\) such that \(U(f,p) - L(f,p) ≤ \epsilon\).
Theorem: any continuous function on \([a,b]\) is Riemann integrable. Any monotone function on \([a,b]\) is Riemann integrable.
Proof: any continuous functions on a closed bounded interval \([a,b]\) is uniformly continuous on that interval. A function is uniformly continuous if whenever two input values are close, the output is close.
\[U(f,p) - L(f,p) = \sum\limits_{i=1}^n(M_i - m_i)\Delta x_i \]
Because the function is uniformly continuous, \(M_i\) and \(m_i\) will be obtained at functional values in any interval.
Take any \(\epsilon > 0\). Our goal is to show that there exists a partition \(p_\epsilon\) of \([a,b]\) such that \(U(f,p) - L(f,p) < \epsilon\). If we can show that, then by the Riemann integrability criterion, we will be done.
Since \(f\) is continuous on \([a,b]\), it is also uniformly continuous on that interval because \([a,b]\) is a compact set. So there exists \(\delta > 0\) such that \(|f(u) - f(v)| < \frac{\epsilon}{b-a}\) whenever \(|u-v| < \delta\) for \(u,v \in [a,b]\). In other words, whenever the inputs are close, the outputs are close.
Pick a partition \(p\) of \([a,b]\) such that \(\Vert p\Vert = \max\limits_{1 ≤ i≤n} \Delta x_i < \delta\). That is a partition where the width of the widest subinterval is less than \(\delta\).
Then \(U(f,p) - L(f,p) = \sum\limits_{i=1}^n (M_i - m_i)\Delta x_i\).
Now we can break out the extreme value theorem. Because the function is continuous, \(M_i = f(u_i)\) and \(m_i = f(v_i)\) for some \(u_i,v_i \in [x_{i-1},x_i]\).
So \(U(f,p) - L(f,p) = \sum\limits_{i=1}^n (f(u_i) - f(v_i))\Delta x_i\). Because \(u_i,v_i \in [x_{i-1},x_i]\), \(|v_i - u_i| < \delta\)
Then because the function is continuous,
\[U(f,p) - L(f,p) < \sum\limits_{i=1}^n \frac{\epsilon}{b-a}\Delta x_i = \frac{\epsilon}{b-a}\sum\limits_{i=1}^n\Delta x_i = \frac{\epsilon}{b-a}(b-a) = \epsilon\]
So the function is Riemann integrable.
The Riemann integral can’t handle functions with many discontinuities. For example, the Dirichlet function:
\[f(x) = \left\{\begin{array}{lr}1 & x \in \mathbb{Q}\cap [0,1] \\ 0 & x \in \mathbb{I} \cap [0,1]\end{array}\right. \]
This function is discontinuous at every point.
Let \(p = \{x_0, x_1, ..., x_n\}\) be a partition of \([0,1]\).
Then \(L(f,p) = \sum\limits_{i=1}^n m_i \Delta x_i\). Becuase both rational and irrational numbers are dense in \(\mathbb{R}\), the infimum value in any interval \([x_{i-1},x_i]\) will be zero. And the supremum will be 1. And \(U(f,p) = \sum\limits_{i=1}^n M_i \Delta x_i\).
This means that every lower Riemann sum equals zero, every upper Riemann sum equals one, so they never equal one another.
\[\underline\int_0^1f = 0 \hspace{1cm} \overline\int_0^1f = 1 \]
In \([0,1]\), there are more irrational numbers than rational numbers. While the rational numbers are countable, the irrational numbers are not. The Lebesgue measure of the rational and irrational numbers:
\[\lambda(\mathbb{Q} \cap [0,1]) = 0 \hspace{1cm} \lambda(\mathbb{I} \cap [0,1]) = 1\]
So Riemann integrals can’t handle functions with uncountably infinitely many discontinuities. What about a finite number of discontinuities?
\[f(x) = \left\{\begin{array}{lr}1 & x = \frac{1}{2} \\ 0 & x \in \left[0,\frac{1}{2}\right) \cup \left(\frac{1}{2}, 1\right]\end{array}\right. \]
This function is Riemann integratable.
Proposition: \(f\) is Riemann integrable on \([a,b]\) iff \(\lambda(D) = 0\) where \(D\) is the set of discontinuity points in \([a,b]\).
Example:
\[f(x) = \left\{\begin{array}{lr}\frac{1}{\sqrt x} & 0 < x ≤ 1 \\ 0 & x=0\end{array}\right. \]
Although there is only one discontinuity point, \(f\) is not bounded in \([0,1]\).
Let \(p = \{x_0, x_1, ..., x_n\}\) be a partition of \([0,1]\).
Then \(U(f,p) = \sum\limits_{i=1}^n M_i \Delta x_i = M_1\Delta x_1 + M_2 \Delta x_2 + \cdots + M_n \Delta x_n\)
\(M_1 = \sup\limits_{[x_0,x_1]} f = ∞\), so the upper Riemann sum does not exist no matter what partition we select.
Say we have a sequence of Riemann integrable functions \(f_n\) that converges pointwise to \(f\) on \([0,1]\). It is not necessarily the case that the Riemann integral of \(f_n\) converges to the Riemann integral of \(f\).
\(\{f_n\}\) converges pointwise to \(f\) on \([0,1]\) if for any \(x \in [0,1]\), we have \(\lim\limits_{n \rightarrow ∞} f_n(x) = f(x)\).
For example, take the sequence \(f_n(x) = x^n, x \in [0,1]\).
For \(0 ≤ x < 1, \lim\limits_{n \rightarrow ∞} f_n(x) = 0\).
For \(x = 1, \lim\limits_{n \rightarrow ∞} f_n(x) = 1\). Therefore, \(\{f_n\}\) converges pointwise to
\[f(x) = \left\{\begin{array}{lr}0 & 0 ≤ x < 1 \\ 1 & x = 1\end{array}\right. \]
But let’s compare the Riemann integrals. We will build a sequence of functions that converges pointwise to the zero function on \([0,1]\), but for which the integral does not converge to zero.
Example where pointwise convergence results in Riemann integral convergence:
\(f_n(x) = \frac{x}{n}, 0 ≤ x ≤ 1\). It is easy to see that \(f_n(x)\) converges pointwise to \(f(x) = 0\).
Then the Riemann integral is \(\int_0^1 \frac{x}{n} dx = \frac{1}{2n}\), which converges to zero.
Example:
\[f_n(x) = \left\{\begin{array}{lr} nx & 0 ≤ x < 1/2^n \\ 8n - nx & x = 1 \\ 0 & \frac{1}{n} ≤ x ≤ 1\end{array}\right.\]
So we have an example where a sequence of Riemann integrable functions converges pointwise to a function that is not Riemann integrable
Suppose \(\mathbb{Q} \cap [0,1] = \{r_n: n \in \mathbb{N}\}\) is the set of all rational numbers in \([0,1]\).
Let
\[f_1(x) = \left\{\begin{array}{lr}1 & x = r_1\\ 0 & x = [0,1] \backslash \{r_1\}\end{array}\right. \]
\[f_2(x) = \left\{\begin{array}{lr}1 & x \in \{r_1,r_2\}\\ 0 & x = [0,1] \backslash \{r_1,r_2\}\end{array}\right. \]
\[f_n(x) = \left\{\begin{array}{lr}1 & x \in \{r_1,..., r_n\}\\ 0 & x = [0,1] \backslash \{r_1,..., r_n\}\end{array}\right. \]
For this sequence, the set of all discontinuities for a function \(f_n(x)\) is \(D = \{r_1, ..., r_n\}\). Because this is a finite set, any \(f_n(x)\) will be Riemann integrable.
We can show that \(\{f_n\}\) converges pointwise to the Dirichlet function, which we know is not integrable.
Let \(\mathbb{Q} \cap [0,1] = \{r_1, r_2, ...\}\). And define
\[f_n(x) = \left\{\begin{array}{lr}1 & x \in \{r_1, r_2, .., r_n\} \\ 0 & o/w\end{array}\right. \]
where \(f_n: [0,1] \rightarrow \mathbb{R}\). Then \(\{f_n\}\) converges pointwise to the Dirichlet function:
\[f(x) = \left\{\begin{array}{lr}1 & x \in \mathbb{Q}\cap [0,1] \\ 0 & x \in \mathbb{I} \cap [0,1]\end{array}\right. \]
Def: \(\{f_n\}\) converges pointwise to \(f\) on \([0,1]\) if for any fixed \(x \in [0,1]\) we have \(\lim\limits_{n \rightarrow ∞} f_n(x) = f(x)\). For a fixed \(x\), \(\{f_n(x)\}\) is a sequence of real numbers.
Fix any \(x \in [0,1]\). Our goal is to show that \(\lim\limits_{n \rightarrow ∞} f_n(x) = f(x)\).
Case 1: \(x\) is irrational; \(x \in \mathbb{I}\). Then \(f_n(x) = 0\) for all \(n \in \mathbb{N}\). So \(\{f_n(x)\} = \{0,0,...\}\)
Case 2: \(x \in \mathbb{Q}\). Recall we denoted \(\mathbb{Q} \cap [0,1] = \{r_1, r_2, ...\}\). This is a countably infinite set. Then there exists some \(\hat n \in \mathbb{N}\) such that \(x = r_{\hat n}\). Then
\[\begin{align*} f_1(x) &= 0 \\[1em] f_2(x) &= 0 \\[1em] &\cdots \\[1em] f_{\hat n - 1} &= 0 \\[1em] f_{\hat n} &= 1 \\[1em] f_{\hat n + 1} &= 1 \end{align*}\]
So no matter what \(x \in \mathbb{Q} \cap [0,1]\) we pick, the limit of \(f_n(x) = 1\). And so \(\{f_n\}\) converges pointwise to \(f\).
Prove that \(\{f_n\}\) is Riemann integrable on \([0,1]\), and prove that \(\int_0^1f_n(x)dx = 0\) for all \(n \in \mathbb{N}\).
We saw last week that Riemann integration can handle a finite number of discontinuties.
Consider the case where \(n = 1\) and \(r_1 \in (0,1)\). (The proof for \(r_1 = 0\) or \(r_1 = 1\) proceeds similarly.) Take any \(\epsilon > 0\), where \(\epsilon\) is small enough that \(0 < r_1 - \frac{\epsilon}{2} < r_1 + \frac{\epsilon}{2} < 1\). And define a partition \(p = \{0, r_1 - \frac{\epsilon}{2}, r_1 + \frac{\epsilon}{2}, 1\}\). Then we can compute the lower and upper Riemann sums.
\[\begin{align*} U(f,p) &= M_1\Delta x_1 + M_2\Delta x_2 + M_3\Delta x_3 \\[1em] &= 0 + (\epsilon/2) + 0 &= \epsilon / 2 < \epsilon \end{align*}\]
By the Riemann criterion for integrability, \(f\) is Riemann integrable on \([0,1]\). This means that \(\int_0^1f = \underline\int_0^1f = \overline\int_0^1f\).
We know from last week that every lower Riemann sum is less than every upper Riemann sum. In any case, \(\underline\int_0^1f ≤ \overline\int_0^1f\).
Every lower Riemann sum of this function is zero, and every upper Riemann sum is less than \(\epsilon\), so we can write
\[0 = L(f,p) ≤ \underline\int_0^1f ≤ \overline\int_0^1f < \epsilon / 2 \]
So by the squeeze theorem, letting \(\epsilon \rightarrow 0\),
\[\int_0^1f = \underline\int_0^1f = \overline\int_0^1f = 0\]
We can extend this to \(f_n\). For simplicity, assume that \(0 < r_1 < r_2 < \cdots < r_n < 1\). Again, take any \(\epsilon > 0\) and let \(p = \{0, r_1 - \epsilon/4n, r_1 + \epsilon/4n, r_2 -\epsilon/4n, r_2 + \epsilon/4n, \cdots, r_n - \epsilon/4n, r_n+ \epsilon/4n, 1\}\), where \(\epsilon\) is sufficiently small that \(p\) forms a partition of \([0,1]\).
We have a concept of length already. For instance we have \(U(f,p) = \sum M_i\Delta x_i\), where \(\Delta x_i = x_i - x_{i-1}\). So \(\Delta x_i\) is a measure, where the distance between the two points is the absolute value of their difference.
Our question in this chapter is, how can we measure the ‘length’ of other objects like sets.
\(I = (a,b), a,b \in \mathbb{R}, a < b\). This is an open interval in \(\mathbb{R}\).
\(I = (a, ∞), a \in \mathbb{R}\)
\(I = (-∞,b), b in \mathbb{R}\)
\(I = (-∞,∞)\)
\(\emptyset\)
These are all open intervals in \(\mathbb{R}\). So how can we define the length of these intervals? Let \(I\) be an open interval in \(\mathbb{R}\). Then define
\[|I| = \left\{\begin{array}{lcr} b-a & \mbox{if} & I = (a,b), a,b \in \mathbb{R}, a < b \\ ∞ & \mbox{if} & I = (a,∞), a \in \mathbb{R}\\ ∞ & \mbox{if} & I = (-∞,b), b \in \mathbb{R}\\ ∞ & \mbox{if} & I = (-∞,∞) \\ 0 & \mbox{if} & I = \emptyset \end{array}\right.\]
Definition: the Lebesgue outer measure on \(\mathbb{R}\).
First, for example, let \(A = (0,1) \cup (2,3)\). Then \(|A| = 2\). If we have a more complicated set, we can write it as the union of a collection of open intervals, and the Lebesgue outer measure is the sums of the lengths of the open intervals.
Let \(A \subset \mathbb{R}\). The Lebesgue outer measure of \(A\) is defined by
\[|A| = \inf\left\{\sum\limits_{n=1}^∞ l(I_n): A \subset \bigcup\limits_{n=1}^∞ I_n\right\} \tag{1}\]
where \(I_n\) are open intervals. Call \(S = \left\{\sum\limits_{i=1}^n l(I_n): A \subset \bigcup\limits_{i=1}^n I_n\right\}\). So the Lebesgue outer measure of \(A\) is the greatest lower bound of the sum of the lengths of a sequence of open intervals whose union contains A.
Example: Let \(A = \{a_1, a_2, ..., a_m\} \subset \mathbb{R}\). Then the Lebesgue outer measure of \(A\) is as in (1). Then \(|A| ≤\) any element of \(S\).
Then for any \(\epsilon > 0\), there exists \(a \in S: a < \inf(S) + \epsilon\).
Suppose that \(|A| < ∞\). If \(|A|\) is finite, then for any \(\epsilon > 0\), there exists a collection of open intervals \(\{I_n\}_{n=1}^∞\) such that \(A \subset \cup I_n\) and
\[\sum\limits_{i=1}^n l(I_n) < |A| + \epsilon \]
Now we can apply these properties to see that if we have a finite set, then the Lebesgue outer measure is zero.
Let \(A = \{a_1, a_2, ..., a_m\} \subset \mathbb{R}\).
Let
\[I_i = \left\{\begin{array}{lcr} (a_i - \frac{\epsilon}{2m}, a_i + \frac{\epsilon}{2m}) & \mbox{if} & i = 1, .., m \\ \emptyset & \mbox{if} & i > m \end{array}\right. \]
Then each \(I_i\) is an open interval. And because \(a_i \subset (a_i - \frac{\epsilon}{2m}, a_i + \frac{\epsilon}{2m})\), \(A \subset \cup_{n=1}^∞ I_n\).
Then
\[\sum\limits_{n=1}^∞ l(I_n) = \sum\limits_{n=1}^m l(I_n) = \sum\limits_{n=1}^m l(I_n) = m\left(\frac{\epsilon}{m}\right) = \epsilon\]
The infimum of the positive real numbers is zero, so \(|A| = 0\).
Every countable subset of \(\mathbb{R}\) has Lebesgue outer measure zero.
We say that a subset of \(A\subset \mathbb{R}\) is infinitely countable if there exists a bijection of \(A\) to the natural numbers:
\(f: \mathbb{N} \rightarrow A\), meaning that \(1 \mapsto f(1) = a_1, 2 \mapsto f(2) = a_2, ...\)
\(A\) is said to be countable if it is finite or infinitely countable. We already proved that the Lebesgue measure of a finite set is zero, so we can turn to the infinitely countable case.
Let \(A = \{a_n: n \in \mathbb{N}\}\). We have to modify our small open interval strategy to ensure it still sums to something that will go to zero. We can do this by using open intervals with lengths that get smaller and smaller.
Fix any \(\epsilon > 0\) and let $I_n =
\[I_n = \left\{\begin{array}{lcr} (a_n - \frac{\epsilon}{2^n}, a_i + \frac{\epsilon}{2^n}) & \mbox{if} & i \in \mathbb{N} \end{array}\right. \]
Then the length of the intervals is \(\epsilon, \frac{\epsilon}{2}, \frac{\epsilon}{4}, \frac{\epsilon}{8}\) and so on.
Then \(A \subset \bigcup\limits_{n=1}^∞ I_n\), and so \(|A| ≤ \sum\limits_{n=1}^∞ l(I_n)\) for the same reason as in the finite case.
\[\begin{align*} \sum\limits_{n=1}^∞ l(I_n) &= \sum\limits_{n=1}^∞ \frac{\epsilon}{2^{n-1}} \\[1em] &= \epsilon\sum\limits_{n=1}^∞ \frac{1}{2^{n-1}} \\[1em] &= \epsilon\left(1 + \frac{1}{2} + \frac{1}{2^2} + \cdots \right) \\[1em] &= 2\epsilon \end{align*}\]
So for instance, \(|\mathbb{Q}| = 0\).
(Missing last 5min of lecture video.)
A nice property of the Lebesgue outer measure is the order-preserving property. Let \(A\) and \(B\) be two subsets of \(\mathbb{R}\). If \(A \subset B\), then \(|A| ≤ |B|\). For instance,
\[|A| = \inf\left\{\sum\limits_{n=1}^∞ l(I_n): A \subset \bigcup\limits_{n=1}^∞ I_n\right\} \hspace{1cm} |B| = \inf\left\{\sum\limits_{n=1}^∞ l(I_n): B \subset \bigcup\limits_{n=1}^∞ I_n\right\} \tag{1}\]
where \(I_n\) are open intervals. Because \(A \subset B\),
\[A \subset B \subset \bigcup\limits_{n=1}^∞ I_n \]
Now we should define two sets.
\[S_1 = \left\{\sum\limits_{n=1}^∞ l(I_n): A \subset \bigcup\limits_{n=1}^∞ I_n\right\} \hspace{1cm} S_2= \left\{\sum\limits_{n=1}^∞ l(I_n): B \subset \bigcup\limits_{n=1}^∞ I_n\right\}\]
These are infinite sets of real numbers. Each number is the length of a collection of sets that contains either \(A\) or \(B\). If a collection of sets contains \(B\), that is, \(B \subset S_{2i}\). Any collection of sets corresponding to an element of \(S_2\) will also contain \(A\), so it also corresponds to an element of \(S_1\).
But there are elements of \(S_1\) corresponding to collections of sets that do not contain \(B\).
Take \(x \in S_2\). Then \(x = \sum\limits_{n=1}^∞ l(I_n): B \subset \bigcup\limits_{n=1}^∞ I_n\). Then \(A \subset B \subset \bigcup\limits_{n=1}^∞ I_n\), so \(x \in S_1\). Therefore \(S_2 \subset S_1\).
Proposition: translation-invariant property
If we have a set \(A \subset \mathbb{R}\) and \(t \in \mathbb{R}\), then we can define \(t + A = \{t+a:a \in A\}\). For example, if \(A = [6,7], t = 5, t + A = [11,12]\). So \(l(A) = 1\) and \(l(A + t) = 1\).
If \(A\subset \mathbb{R}, t \in \mathbb{R}\), then \(|t+A| = |A|\).
Proof: observe that if \(I\) is an open interval, eg. \(I = (5,6)\),
Let \(\{I_n\}_{n=1}^∞\) be an arbitrary collection of open intervals such that \(A \subset \cup_{n=1}^∞ I_n\). Then
\[A+t \subset \bigcup_{n=1}^∞(t + I_n)\]
Then, by the definition of the Lebesgue outer measure,
\[|A+t| ≤ \sum_{n=1}^∞ l(t + I_n) = \sum_{n=1}^∞ l(I_n) \]
So \(|t + A|\) is a lower bound for \(\sum_{n=1}^∞ l(I_n)\), which is the greatest lower bound for \(|A|\). Therefore,
\[|A+t| ≤ |A| \]
To prove the reverse inequality, we can note that \(|A| = |(t+A)+(-t)|\). Then if we add \(t\), we obtain from the previous result that
\[|A| ≤ |t+A|\]
Theorem: subadditivity. Let \(\{A_n\}_n \subset \mathbb{R}\). Then
\[\left|\bigcup_{n=1}^∞ A_n\right| ≤ \sum_{n=1}^∞ |A_n| \]
Last week: countable additivity of the Lebesgue outer measure
Theorem (countable subadditivity): Let \(\{A_n\}_{n=1}^∞\) be a collection of subsets of \(\mathbb{R}\). Then
\[\left|\bigcup_{n=1}^∞ A_n\right| ≤ \sum_{n=1}^∞ |A_n| \tag{1}\]
Proof: If we have a set \(S \subset \mathbb{R}\), then if \(\alpha = \inf S\), then for all \(\epsilon > 0\), there exists \(a \in S\) such that \(\alpha ≤ a < \alpha + \epsilon\).
If there exists \(n \in \mathbb N\) such that \(|A_n| = ∞\), then the inequality in (1) is satisfied. But if all \(|A_n| < ∞\), then we need to consider this case.
Recall that we defined
\[|A_n| = \inf \left\{\sum_{k=1}^∞l(I_k)\right\} \]
where \(I_1, I_2, ...\) are open intervals such that \(A \subset \cup I_k\). We can use the \(\epsilon\)-characterization of this set to prove our theorem.
Take any \(\epsilon > 0\). Then for any \(n \in \mathbb N\), \(\frac{\epsilon}{2^n} > 0\). Then there exist open intervals \(I_{n,1}, I_{n,2}, ...\) such that \(A_n \subset \cup I_{n,k}\) and
\[\sum_{k=1}^∞l(I_k) < |A_n| + \frac{\epsilon}{2^n} \tag{2}\]
Then,
\[\bigcup_{n=1}^∞ A_n \subset \bigcup_{n=1}^∞\bigcup_{k=1}^∞ I_{n,k} = \bigcup_{(n,k) \in \mathbb N \times \mathbb N} I_{n,k}\]
which is a countable union of open intervals. So because the left-hand collection is a subset of the right-hand collection, it follows that
\[\left|\bigcup_{n=1}^∞ A_n\right| ≤ \sum_{(n,k) \in \mathbb N \times \mathbb N} l(I_{n,k}) = \sum_{k=1}^∞\sum_{n=1}^∞ l(I_{n,k}) = \sum_{n=1}^∞\sum_{k=1}^∞ l(I_{n,k})\]
The Lebesgue outer measure of the union of \(A_n\) is less than the sum of the lengths of all \(I_{n,k}\). Then, invoking (2),
\[≤ \sum_{n=1}^∞\left(|A_n| + \frac{\epsilon}{2^n}\right) = \sum_{n=1}^∞ |A_n| + \sum_{n=1}^∞ \frac{\epsilon}{2^n} = \left(\sum_{n=1}^∞ |A_n|\right) + \epsilon\]
So
\[\left|\bigcup_{n=1}^∞ A_n\right| ≤ \left(\sum_{n=1}^∞ |A_n|\right) + \epsilon\]
And letting \(\epsilon \rightarrow 0^+\),
\[\left|\bigcup_{n=1}^∞ A_n\right| ≤ \sum_{n=1}^∞ |A_n|\]
If \(A \subset \bigcup_{n=1}^∞ J_n\), where each \(J_n\) is an open interval, then
\[|A| ≤ \sum_{n=1}^∞ l(J_n) \]
because the lhs is simply the infimum of all elements of the type on the rhs. Then
\[\bigcup_{n=1}^∞ A_n \subset \bigcup_{(n,k) \in \mathbb N \times \mathbb N} l(I_{n,k}) \]
So
\[\left|\bigcup_{n=1}^∞ A_n\right| ≤ \sum_{(n,k) \in \mathbb N \times \mathbb N} l(I_{n,k}) \]
Finite additivity: If \(A_1, A_2, ..., A_m \subset \mathbb{R}\), then
\[\left|\bigcup_{n=1}^m A_n\right| ≤ \sum_{n=1}^m |A_n|\]
This follows by letting \(A_{m+1} = \emptyset\) and so on for all \(> m\).
Monotonicity. If \(A \subset B \subset \mathbb{R}\), then \(|A| ≤ |B|\).
Proof: based on subadditivity. If \(S_1 \subset S_2 \subset \mathbb{R}\), then \(\inf S_1 ≥ \inf S_2\).
Theorem: the Lebesgue outer measure of a closed bounded interval \([a,b]\) is \(b-a\).
If \([a,b] \subset \bigcup_{n=1}^∞ I_n\), then because the RHS is an open cover of \([a,b]\), \(\bigcup_{n=1}^∞ I_n\) has a finite subcover.
1B.1 - Popcorn function. First , for any \(\epsilon > 0\), take \(A = \{x \in [0,1]: f(x) ≥ \epsilon\}\).
First, this is a finite set. If \(x \in A\), then \(x\) is rational. So \(x = m/n\) where \(m,n\) are relatively prime integers. And \(1/n ≥ \epsilon\) means that \(n ≤ 1/\epsilon\). There are only finitely many natural numbers such that this is true. So there are also finitely many \(m \in \mathbb N\) such that \(f(x) = m/n ≥ \epsilon\).
We can define small intervals around each of the elements \(a_1, a_2, ..., a_m \in A\) such that their rectangles with the function values are small. If \(h_1 = f(a_1)\), then we can make the intervals of the form \((a_1 - \epsilon'/h_1, a_1 + \epsilon'/h_1)\).
Then \(L(p,f) = 0\) for any partition because the irrationals are dense.
And
\[U(p,f) ≤ \epsilon(\Delta x_1) + \epsilon(\Delta x_1) + \cdots + \epsilon(\Delta x_m) \sum_{i=1}^m(2\epsilon')\]
2A.1 - Just two steps.
\(A \subset A \cup B\) implies \(|A| ≤ |A\cup B|\).
\(|A \cup B| ≤ |A| + |B| = |A|\).
2A.6 - \(|(a,b)| = b-a\).
\((a,b) \subset (a,b) \cup \emptyset \cup \emptyset \cup \cdots\).
So \(|(a,b)| ≤ \sum_{n=1}^∞ l(I_n) = b-a\).
Take any \(\epsilon > 0\), then \([a + \epsilon, b - \epsilon] \subset (a,b)\). We already know the length of a closed interval in \(\mathbb{R}\) is \(|[a,b]| = b-a\).
So \(|[a + \epsilon, b - \epsilon]| ≤ |(a,b)|\)
\(b - \epsilon - a - \epsilon ≤ |(a,b)|\)
Letting \(\epsilon \rightarrow 0^+\), we have that \(b-a < |(a,b)|\).
2A.10 - show that \(|[0,1] \setminus \mathbb Q| = 1\). This is the set of irrational numbers in \([0,1]\). Use the fact that \(\mathbb Q\) is countable, so \(|\mathbb Q| = 0\).
\([0,1] \setminus \mathbb Q \subset [0,1]\) so \(|[0,1] \setminus \mathbb Q| ≤ |[0,1]| = 1\).
We can write our set as \([0,1] = A \cup B\) where \(B\) is the set of all rational numbers in \([0,1]\). So \(|B| = 0\). Then use the subadditivity property to show that \(1 ≤ |A|\).
2A.2 - use the definition of the Lebesgue outer measure
1B.5 -
First inequality: For any \(\epsilon > 0\), \([a,b] \subset (a - \epsilon, b + \epsilon)\). We could rewrite this as \([a,b] \subset (a - \epsilon, b + \epsilon) \cup \emptyset \cup \emptyset \cup \cdots\).
Then by the definition of the Lebesgue outer measure,
\[|[a,b] ≤ l((a - \epsilon, a + \epsilon)) + l(\emptyset) + \cdots = b-a + 2\epsilon\]
So \(|[a,b]| ≤ b-a\).
Second ineqality: show that \(b - a ≤ |[a,b]|\). We start with a collection of open intervals that cover \([a,b]\).
Let \(\{I_n\}_{n=1}^∞\) be an arbitrary collection of open intervals such that
\[[a,b] \subset \bigcup_{n=1}^∞ I_n \]
Then if we can show that \(b-a ≤ \sum_{n=1}^∞ l(I_n)\), then this implies that \(b-a ≤ |[a,b]|\) becuase the Lebesgue outer measure is the greatest lower bound (infimum) of that sum.
Finite example: \(n = 2\),
We have intervals \([a,b] = [2,5] \subset (0,4) \cup (2,6)\). Then \(5-2 ≤ (4-0) + (6-2)\). We want to prove the general case of this.
Lemma: If \([a,b] \subset \bigcup_{k=1}^n I_k\), then \(b-a ≤ \sum_{k=1}^n l(I_k)\).
Proof (induction): the conclusion is obvious when \(n = 1\).
Suppose that \([a,b] \subset \bigcup_{k=1}^{n+1}I_k\). Our goal is to show that \(b-a ≤ \sum_{k=1}^{n+1} l(I_k)\).
Assume that \([a,b] \subset \bigcup_{k=1}^{n+1}I_k\). Then \(b \in \bigcup_{k=1}^{n+1}I_k\). So \(b \in I_k\) for some \(k = 1, 2, ... n + 1\). However, by re-labeling the intervals, we can assume WLOG that \(b \in I_{n+1}\). Here, we can assume that \(I_k\) is a bounded open interval for all \(k = 1, 2, ..., n + 1\). We can denote \(I_{n + 1} = (c,d)\).
If \(a ≥ c\), then \(b-a ≤ d - c = l(I_{n+1})\), so the conclusion is true in this case.
If \(a<c\), then we can recall that \([a,b] \subset \bigcup_{k=1}^{n+1}I_k\), so also, \([a,c] \subset \bigcup_{k=1}^{n+1}I_k\). This implies that \([a,c] \subset \bigcup_{k=1}^{n}I_k\). Not \(n+1\), but \(n\).
Take \(x \in [a,c]\). Then it is in the union of the \(n+1\) open intervals. But it is not in \(I_{n+1}\), so it must be in one of the first \(n\) open intervals.
By the induction hypothesis, the conclusion holds for \(n\): \(c - a ≤ \sum_{k=1}^n l(I_k)\).
Then
\[\begin{align*} c - a + l(I_{n+1}) &≤ \sum_{k=1}^n l(I_k) + l(I_{n+1}) \\[1em] c - a + d - c &≤ \sum_{k=1}^{n+1} l(I_k) \\[1em] d - a &≤ \sum_{k=1}^{n+1} l(I_k) \\[1em] b - a &≤ \sum_{k=1}^{n+1} l(I_k) \\[1em] \end{align*}\]
So for the finite case, it is proved. To extend this to the countable case, we can use the compactness of closed intervals.
Thm: \(|[a,b]| = b-A\).
Proof: From above, \(|[a,b]| ≤ b-a\). Let \(\{I_k\}_{k=1}^∞\) be an arbitrary collection of open intervals such that \([a,b] \subset \bigcup_{k=1}^∞ I_k\).
Then because \([a,b]\) is compact, there exists \(n \in \mathbb N\) such that \([a,b] \subset \bigcup_{k=1}^n I_k\).
By the previous lemma, whenever we have \([a,b] \subset \bigcup_{k=1}^n I_k\), it is true that \(b-a ≤ \sum_{k=1}^n l(I_k) ≤ \sum_{k=1}^∞ l(I_k)\).
Then \(b-a\) is a lower bound for the summation on the RHS, which is simply \(|[a,b]|\).
For \((a,b), a < b\), there are three cases:
\(t > 0: t(a,b) = (ta,tb)\)
\(t < 0: t(a,b)= (tb, ta)\).
\(t = 0: t(a,b)= \{0\}\)
In the general case,
\(t = 0: tA = \{0\}\), and the Lebesgue outer measure of a finite set is zero. Then use \(0\cdot ∞ = 0\) in the case that \(|A| = ∞\).
\(t > 0: Let \{I_n\}_{n=1}^∞\) be an arbitrary collection of open intervals such that \(tA \subset \bigcup_{n=1}^∞ I_n\). Then \(A \subset \frac{1}{t} \bigcup_{n=1}^∞ I_n = \bigcup_{n=1}^∞\left(\frac{1}{t}I_n\right)\).
It follows that because \(I_n\) are open intervals that cover \(A\), \(|A| ≤ \sum_{n=1}^∞ l(I_n/t)\).
So \(t|A| ≤ \sum_{n=1}^∞l(I_n) = |tA|\).
Then, a trick to show the opposite inequality is to let \(t' = \frac{1}{t}\) and perform the same proof to show that \(\frac{1}{t}|tA| ≤ \left|\frac{1}{t}(tA)\right| = |A|\).
The Lebesgue outer measure is not additive. Let \(S\) be a nonempty set and let \(\sim\) be a binary relation on \(S\). We say that \(\sim\) is an equivalence relation if the following properties are satisfied:
Reflexivity. \(x \sim x\) for all \(x \in S\)
Symmetry. If \(x \sim y\), then \(y \sim x\)
Transitivity. If \(x \sim y\) and \(y\sim z\), then \(x \sim z\).
Simple example is \(S\) is the set of all people, and \(\sim\) is ‘has the same last name.’ It is easy to see why the three properties hold.
Fix any \(a \in S\). Define \(\tilde a = \{x \in S: x \sim a\}\). The equivalence class of \(a\) is all elements \(x\) that are related to \(a\).
\(\bigcup_{a \in S}\tilde a = S\)
If \(\tilde a \cap \tilde b ≠ \emptyset\), then \(\tilde a = \tilde b\).
Lemma: Let \(S = [-1,1]\). For any \(x,y \in S\), define \(x \sim y\) iff \(x - y \in \mathbb Q\).
Then \(\sim\) is an equivalence relation on \(S\). Can easily check the three properties.
For \(x \in S\), \(x - x = 0 \in \mathbb Q\).
For \(x,y \in S\) suppose \(x \sim y\). Then \(y - x = r \in \mathbb Q\). Then \(x - y = -r \in \mathbb Q\), so \(y \sim x\).
For \(x,y,z \in S\), suppose \(x \sim y\) and \(y \sim z\). Then \(y - x = r_1 \in \mathbb Q\) and \(z - y = r_2 \in \mathbb Q\). Then \(z - x = r_1 + r_2 \in \mathbb Q\), so \(x \sim z\).
Therefore, \(\sim\) is an equivalence relation on \(S\).
For any \(a \in S\), denote $
\[\begin{align*} \tilde a &= \{x \in S: x \sim a\} \\[1em] &= \{x \in S: x - a \in \mathbb Q\}\\[1em] \end{align*}\]
Eg. \(\overset{\sim}{1/\sqrt 2}\) would be all numbers in \([-1,1]\) whose difference with \(1/\sqrt 2\) is rational.
The set of equivalence classes of elements of \(S\) form a partition of \(S\). The equivalence classes are pairwise disjoint.
Let us select one element from each equivalence class. Let \(V\) be a set that contains one element from each of the distinct equivalence classes from the equivalence relation \(\sim\).
Let \(\mathbb Q \cap [-2,2] = \{r_1, r_2, ...\}\). And we still have \(S = [-1,1]\). Then
\([-1,1] \subset \bigcup_{k=1}^∞(r_k + V) \subset [-3,3]\)
\((r_i + V) \cap (r_j + V) = \emptyset\) if \(i ≠j\). This is beacuse \(\mathbb Q \cap [-2,2] = \{r_1, r_2, ...\}\) contains distinct elements, so no two elements will be equal to each other.
(Recall that if \(A \subset \mathbb{R}\) and \(x \in \mathbb{R}\), then \(x + A = \{x + a: a \in A\}\)).
Proof (a):
Take any \(x \in [-1,1]\). Then \(x \in \tilde a\) for some \(a \in S\). We have our set \(V\) that has one representative from each equivalence class. Let \(v \in V\) such that \(x \sim v\). Then because \(x \in \tilde a\), \(v \in \tilde a\).
Then \(x - v \in \mathbb Q\). Note that \(x \in [-1,1]\) and \(v \in [-1,1]\) because \(V\) draws its elements from \(S\).
So \(x - v \in [-2,2] \cap \mathbb Q\). So \(x - v = r_k\) for some \(k \in \mathbb N\). Then \(x = r_k + v \in r_k + V \subset \bigcup_{k=1}^∞(r_k + V)\).
\(r_k \in [-2,2]\) and elements of \(V\) are in \([-1,1]\).
So \([-1,1] \subset \bigcup_{k=1}^∞(r_k + V) \subset [-3,3]\)
Proof (b): Try proving by contradiction. If \(x \in (r_i + V) \cap (r_j + V)\), then we will have \(x \sim v_1\) and \(x \sim v_2\), so \(v_1, v_2 \in \tilde x\). But we only took one element of each equivalence class, so the only way this could be true is if \(v_1 = v_2\), so \(r_i = r_j\) and \(i = j\) which is a contradiction.
There exist two sets \(A\) and \(B\) in \(\mathbb{R}\) such that \(A \cap B = \emptyset\) and \(|A \cup B| ≠ |A| + |B|\). They are disjoint, but the outer meausre of their union is different from the sum of their outer measures.
Prove by contradiction. Suppose that for \(A,B \subset \mathbb{R}\), \(|A \cup B| = |A| + |B|\) whenever \(A \cap B = \emptyset\).
By induction, we can show that it is true for finitely many sets.
\(|A_1 \cup A_2 \cup \cdots \cup A_n| = |A_1| + \cdots |A_n|\) whenever \(A_1, ..., A_n \subset \mathbb{R}\) such that (see notes)
\([-1,1] \subset \bigcup_{k=1}^∞ (r_k + V)\). Then
$|[-1,1]| &≤ |{k=1}^∞ (r_k + V)| \[1em] &≤ {k=1}^∞
So we know that \(|V| > 0\).
Then we have that \(\bigcup_{k=1}^∞ (r_k + V) \subset [-3,3]\). Then for any \(n \in \mathbb N\), \(\bigcup_{k=1}^n (r_k + V) \subset [-3,3]\). Just switched index limit to \(n\).
\[\begin{align*} \left|\bigcup_{k=1}^∞ (r_k + V)\right| ≤ \left|[-3,3]\right| = 6 \\[1em] &≤ \sum_{k=1}^∞ |r_k + V| \\[1em] \end{align*}\]
So \(\sum_{k=1}^n|r_k + V| ≤ 6\)
\[\begin{align*} \sum_{k=1}^n|r_k + V| &≤ 6 \\[1em] \sum_{k=1}^n |V| &≤ 6 \\[1em] n|V| &≤ 6 \end{align*}\]
Here we have our contradiction, since as \(n \rightarrow ∞\), this is not possible.
1.1: \(\mathcal P(\mathbb{R}) \rightarrow [0,∞]\). Previously, we saw that the Lebesgue outer measure \(A \mapsto |A| = \inf\{\sum_{n=1}^∞ l(I_n)\}\) where \(I_n\) are open intervals and \(A\) is a subset of the union of the \(I_n\). So we saw for instance that \(|(1,∞)| = ∞\)
We hope that the Lebesgue measure can serve as an extension of the concept of length. But we want to examine its properties. A few were that
\(|t + A| = |A|\), translation invariance
Lebesgue outer measure is countably subadditive,
But not countably additive. There exist \(A,B \subset \mathbb{R}\) such that \(A \cap B = \emptyset\), but \(|A \cup B| ≠ |A| + |B|\). This is a problem for extending the concept of length to all subsets of \(\mathbb{R}\).
Theorem: there does not exist a function \(\mu\) with the following properties:
\(\mu: \mathcal P(\mathbb{R}) \rightarrow [0,∞]\). \(\mu\) is a function on the power set of \(\mathbb{R}\) that maps to a real nonnegative number.
\(\mu(I) = l(I)\) for every open interval \(I\) of \(\mathbb{R}\).
\(\mu(\cup_{n=1}^∞ A_n) = \sum_{n=1}^∞ \mu(A_n)\) for every disjoint sequence \(A_1, A_2, ...\). (Countable additivity)
\(\mu(t + A) = \mu(A)\) for all \(A \subset \mathbb{R}\) and \(t \in \mathbb{R}\). (Translation invariance)
“That’s not good news. Wow.” -MNN
Proof:
\(\mu(\emptyset) = 0\). The empty set is considered as an open interval. And \(\mu(I)\) is equal to the length of \(I\) for an open interval \(I\).
If \(A \subset B\), then \(\mu(A) ≤ \mu(B)\). We know this because
\[\begin{align*} B &= A \cup (B \setminus A) \cup \emptyset \cup \cdots \\[1em] \mu(B) &= \mu\left(\bigcup_{n=1}^∞ A_n\right) \end{align*}\]
where \(A_1 = A\), \(A_2 = B \setminus A\), \(A_i = \emptyset\) for all \(i ≥ 3\). So the \(A_i\) are mutually disjoint sets, so
\[\begin{align*} \mu(B) &= \mu\left(\bigcup_{n=1}^∞ A_n\right) \\[1em] &= \mu(A) + \mu(B\setminus A) ≥ \mu(A) \end{align*}\]
Next, \(\mu([a,b]) = b - a\). We know that \((a,b) \subset [a,b]\), so
\[b-a = l((a,b)) = \mu((a,b)) ≤ \mu([a,b])\]
And we can also show that \(\mu([a,b]) ≤ b-a\) by considering the open interval \((a-\epsilon, b + \epsilon)\) for \(\epsilon > 0\), using the comparison theorem to obtain the desired inequality.
Countable subadditivity: Let \(A_1, A_2, ...\) be a sequence of subsets of \(\mathbb{R}\) Let us show that
\[\mu\left(\bigcup_{n=1}^∞ A_n\right) ≤ \sum_{n=1}^∞ \mu(A_n)\]
In this case, the \(A_n\) are not assumed to be mutually disjoint, so we cannot apply property (c) from above.
Say we have \(A_1, A_2, A_3 \subset \mathbb{R}\), not necessarily mutually disjoint. Let \(B_1 = A_1, B_2 = A_2 \setminus A_1, B_3 = A_3 \setminus (A_1 \cup A_2)\). We could continue this with more than three subsets of \(\mathbb{R}\) such that \(B_n = A_n \setminus(\cup_{n=1}^{n-1}A_i)\)
If we do this, then \(B_1, B_2, ...\) is a mutually disjoint sequence of subsets of \(\mathbb{R}\) with \(B_n \subset A_n\), and
\[ \bigcup_{n=1}^∞ B_n = \bigcup_{n=1}^∞ A_n\]
It follows that
\[\mu\left(\bigcup_{n=1}^∞ A_n\right) = \mu\left(\bigcup_{n=1}^∞ B_n\right) = \sum_{n=1}^∞ \mu(B_n) ≤ \sum_{n=1}^∞ \mu(A_n)\]
Now, we only need to repeat the procedure of proving that there exist two sets \(A,B \subset \mathbb{R}\) such that \(|A \cup B| ≠ |A| + |B|\).
We derived this property on Monday. Then we used the subadditivity property.
\[\begin{align*} \mu([-1,1]) &≤ \mu(\cup_{k=1}^∞(r_k + V)) \\[1em] &≤ \sum_{k=1}^∞ \mu(r_k + v) \\[1em] &= \lim_{n \rightarrow ∞} \sum_{k=1}^n \mu(r_k + V) \\[1em] &= \lim_{n \rightarrow ∞} \sum_{k=1}^n \mu(V) \\[1em] &= \lim_{n \rightarrow ∞} n \mu(V) \end{align*}\]
Then we use the property that \(\cup_{k=1}^∞(r_k + V) \subset [-3,3]\) to show that
\[\begin{align*} \mu(\cup_{k=1}^∞(r_k + V)) &≤ \mu([-3,3]) = 6 \\[1em] \sum_{k=1}^∞ \mu(r_k + V) &≤ 6 \\[1em] n\mu(V) &≤ 6 \end{align*}\]
which is a contradiction. And so there exist two sets \(A, B\) such that \(A \cap B = \emptyset\) but \(|A\cup B| ≠ |A| + |B|\).
Therefore, we cannot find a set function \(\mu\) on the power set of \(\mathbb{R}\) such that the four properties above are satisfied. So if we want to extend the concept of length, we need a different approach.
We still want the property that the measure of an open interval is equal to its length (b), and we would like our measure to satisfy countable additivity (c), as well as translation invariance (d). That leaves only property (a), that \(\mu\) is a function on the power set of \(\mathbb{R}\). Instead of defining \(\mu\) on all subsets of \(\mathbb{R}\), we will define it on a subset of the power set that satisfies certain properties.
Let \(\mathcal A \subset \mathcal P(\Omega)\) where \(\Omega\) is a given set. We say that \(\mathcal A\) is a \(\sigma\)-algebra if the following properties are satisfied:
\(\emptyset \in \mathcal A\).
If \(B \in \mathcal A\), then \(B^C \in \mathcal A\).
If \(A_1, A_2, ... \in \mathcal A\), then \(\cup_{n=1}^∞ A_n \in \mathcal A\)
2A.5 - If a set of closed subsets of \(\mathbb{R}\) has an empty intersection, then there exists a finite subset with an empty intersection. We can use the connectedness property to prove this. Assume \(B \in A\) is bounded. Then \(B\) is compact because it is closed and bounded.
We are given that \(\cap_{F \in \mathcal A}F = \emptyset\). So \(\left(\cap_{F \in \mathcal A}F\right)^C = \mathbb{R}\), and \(\cup_{F \in \mathcal A}F^C = \mathbb{R}\), and \(B \subset \mathbb{R}\).
Then we can find a sequence such that \(B \subset F_1^C \cup F_2^C \cup \cdots \cup F_m^C\)
Then it wouldbe the case that \(B \cap \left(\cap){i=1}^n F_i\right) = \emptyset\).
Then MNN mentioned something about \(n = m+1\).
2A.6 (or was is 2A.7?) - Start with the union of two closed bounded intervals in \(\mathbb{R}\):
\[|[a,b] \cup [c,d]| = b-a + d - c\] when the intervals are disjoint.
Prove by induction that if \(\left([a,b] \cup [c,d]\right) \subset \bigcup_{k=1}^n I_k\) where \(I_k\) are open boudned intervals, then
\[b-a + d - c ≤ \sum_{k=1}^n l(I_k) \]
Previously, we defined a \(\sigma\)-algebra.
Def: suppose \(X\) is a nonempty set and \(S\) is a set of subsets of X. Then S is called a \(σ\)-algebra on \(X\) if the following three conditions are satisfied:
Contains the empty set. \(\emptyset \in S\)
Closed under complimentation If \(A\in S\), then \(A^C \in S\).
Closed under countable union. If \(A_1, A_2, ...\) is a sequence of sets in \(S\), then \(\cup_{n=1}^∞ A_n \in S\).
Each element of \(S\) is also a subset of \(\mathbb{R}\), so \(S \subset \mathcal P(X)\).
\(\emptyset \in S\)
\(\emptyset^C = \mathbb{R}\) and \((-∞,0]^C = (0,∞)\)
\(\emptyset \cup \mathbb{R}\cup (-∞, 0] \cup (0, ∞) \cup \emptyset \cup \emptyset \cup \cdots = \mathbb{R}\in S\)
So we can conclude that \(S\) is a \(\sigma\)-algebra on \(\mathbb{R}\).
Let \(X = \mathbb{R}\), and let \(S = \mathcal P(\mathbb{R})\). Is \(S\) a \(\sigma\)-algebra on \(\mathbb{R}\)? Easy to verify the three criteria.
Let \(X = \mathbb{R}\), and let \(S = \{A \subset \mathbb{R}: A\) is countable or \(A^C\) is countable \(\}\).
For example, \(\mathbb Q\) and \(\mathbb I\) could be in \(S\), \(\{0\}\) and \((-∞,0) \cup (0,∞)\) could be in \(S\).
For counterexample, \((0,1)\) can’t be in \(S\), because neither \((0,1)\) nor \((0,1)^C = (-∞,0]\cup[1,∞)\) are countable. The Lebesgue outer measure of \((0,1) = 1 > 0\), so it cannot be countable.
The first and second properties come easily. For the third, we need to show that a countable union of elements of \(S\) is also in \(S\).
Let \(A_1, A_2, ...\) be a sequence of sets in \(S\). Then for each \(A_n\), either \(A_n\) or \(A_n^C\) is countable.
Case 1: \(A_n\) is countable for every \(n\), then the union is also countable.
Case 2: \(A_n\) is not countable for some \(n_0 \in \mathbb N\). Then \(A_{n_0}^C\) is countable. So we have by de Morgan’s laws that
\[\left(\bigcup_{n=1}^∞ A_n\right)^C = \bigcap_{n=1}^∞ A_n^C \subset A_n^C\]
So \(\left(\bigcup_{n=1}^∞ A_n\right)^C\) is countable, therefore it is in \(S\).
2A.12
The compliment of an open set is closed.
Prove by contradiction. Suppose it contains two elements, then use the density property: in between two real numbers, there exists a rational number.
Prove by contradiction. Suppose outer measure is finite. \(\mathbb{R}= F \cup\) with the RHS, then use the subadditivity property to show that the Lebesgue outer measure of \(\mathbb{R}\) is the LOM of the union of those two sets, and because the outer measure of \(\mathbb{R}\) is infinite, then one of the two RHS must have LOM = \(∞\). And it turns out it has to be \(F\).
2A.11
Prove both inequalitites. LHS ≤ RHS by subadditivity. RHS ≤ LHS, prove for the finite consultant. Show that LOM of finite union of disjoint I_k is sum of lengths. Then use monotonicity and let $n –> ∞. Can also use #8 to solve this problem.
\[\left|\bigcup_{k=1}^n I_k\right| = \sum_{k=1}^n l(I_k)\]
2A.8
Hint: LHS RHS using de Morgan’s law:
\[A = (A \cap(-t,t)) \cup (A \cap (-t,t)^C) \]
so
\[|A| ≤ |(A \cap(-t,t))| + |(A \cap (-t,t)^C)| \]
Then apply the LOM property of unions.
For the opposite inclusion, we can prove it for when \(A\) is an interval
Previously, we defined \(\sigma\) algebras and measurable spaces.
A \(\sigma\) algebra contains the empty set, is closed under complimentation, and is closed under countable union. We showed that a \(\sigma\) algebra is closed under the intersection of two sets, and under countable intersection. A measurable space is simply an ordered pair of a set and an associated \(\sigma\) algebra. Then we defined the Borel \(\sigma\) algebra as the smallest \(\sigma\) algebra containing all open sets in \(\mathbb{R}\).
Proposition: Every closed set in \(\mathbb{R}\) is a Borel set.
Let \(\mathcal B\) be the Borel \(\sigma\) algebra on \(\mathbb{R}\). Then
\[\mathcal B = \sigma(O) \]
where \(O\) is the collection of all open sets in \(\mathbb{R}\). That means that \(\mathcal B\) is the smallest \(\sigma\) algebra containing all open sets in \(\mathbb{R}\). Then clearly, every open set is a Borel set (it belongs to the Borel \(\sigma\) algebra). Further we can see that \(O \subset \mathcal B\). Second, if \(O \subset S\), where \(S\) is a \(\sigma\) algebra, then \(\mathcal B \subset S\), because \(\mathcal B\) is the smallest \(\sigma\) algebra containing \(O\).
We have that \(\mathcal B\) is a \(\sigma\) algebra containing all open sets. We can observe that every closed set in \(\mathbb{R}\) is the compliment of an open set in \(\mathbb{R}\). Let \(F\) be a closed set in \(\mathbb{R}\). Then \(F^C\) is open. Then \(F^C \in \mathcal B\). And because \(\mathcal B\) is closed under complimentation, \(F \in \mathcal B\).
If \(S\) is a \(\sigma\) algebra on \(X\), then \((X,S)\) is called a measurable space.
Let \((X,S)\) be a measurable space. Then \(X\) is a universal set. And \(S\) is a collection of subsets of \(X\) that meet the empty set, copmlimentation, and countable union criteria. Consider a function \(f: X \rightarrow \mathbb{R}\). (We can characterize the continuity of \(f\) using open sets: the preimage of any open set is open.)
We say that \(f\) is an \(S\)-measurable function if \(f^{-1}(B) \in S\) for every Borel set \(B\).
Lemma: Let \((X,S)\) be a measurable space. Consider a function \(f: X\rightarrow \mathbb{R}\). Let \(S_1 = \{A \in \mathbb{R}: f^{-1}(A) \in S\}\). (And recall that each element of \(S\) is a subset of \(X\).) We take a subset of \(\mathbb{R}\), and the inverse function of \(A\) is in this set \(S_1\). Then \(S_1\) is a \(\sigma\) algebra on \(\mathbb{R}\). So any set
\[A \in S_1 \hspace{2mm}\mathrm{iff}\hspace{2mm}f^{-1}(A) \in S \]
Proof:
First, the empty set. The inverse of the empty set in \(\mathbb{R}\) is the empty set in \(X\). And \(\emptyset \in S\) always, so \(f^{-1}(\emptyset) = \emptyset \in S\), so \(\emptyset \in S_1\).
Second, complimentation. Take any set \(A \in S_1\). Then \(f^{-1}(A) \in S\). \(f^{-1}(A^C) = \left(f^{-1}(A)\right)^C \in S\), because \(S\) is closed under complimentation.
Third, countable union. Take \(A_1, A_2, ...\) be a sequence of sets in \(S_1\). Then \(f^{-1}(A_n) \in S\) for all \(n \in \mathbb N\). Then \(f^{-1}(\cup_{n=1}^∞ A_n) = \cup_{n=1}^∞f^{-1}(A_n) \in S\).
With this lemma in hand, we can prove a useful theorem
Theorem: Let \((X,S)\) be a measurable space, and let \(f: X \rightarrow \mathbb{R}\) be a function. Then the following properties are equivalent.
“There are so many Borel sets. It is really difficult to find a set that is not a Borel set.”
\(f\) is \(S\)-measurable
\(f^{-1}(O) \in S\) for every open set in \(\mathbb{R}\).
\(f^{-1}((a,∞)) \in S\) for every \(a \in \mathbb{R}\).
First, \((a) \Rightarrow (b)\). Suppose (a) is satisfied, so \(f^{-1}(B) \in S\) for every \(B \in \mathcal B\). Then because every open set is a Borel set, \(f^{-1}(O) \in S\) for every open set \(O\).
Next, \((b) \Rightarrow (a)\). Suppose (b) is satisfied, so \(f^{-1}(O) \in S\) for every open set \(O \in \mathbb{R}\). So the collection of all open sets in \(\mathbb{R}\) is a subset of \(S_1\): \(\mathcal O \subset S_1\). And we know that \(S_1\) is a \(\sigma\) algebra. Therefore, \(\mathcal B \subset S_1\). So any element \(B \in \mathcal B\) is in \(S_1\), so \(f^{-1}(B) \in S_1\).
Example: Let \(X\) be a set and let \(F \subset X\). Define \(\chi_F: X \rightarrow \mathbb{R}\), where
\[\chi_F = \left\{\begin{array}{lcr} 1 & \mbox{if} & x \in F \\ 0 & \mbox{if} & x \notin F \end{array}\right.\]
2B.1.
2B.2 - Done in class
2B.9 - Find a function that is either -1 or 1. Then abs = 1.
2B.11 - Write definition of \(\mathcal S\), write element of each side and show it is in the other side. 2B.13 - If all \(c_i\chi_{E_i}\) is measurable, and we know that measurability is preserved under scalar multiplication. And the sum of measurable functions is measurable.
Midterm Weds, 2/10
All day take-home format
Practice will be posted on Sat
Measurable space: an ordered pair \((X,S)\) where \(X\) is a set and \(S\) is a \(\sigma\)-algbra on \(X\).
Measurable function: Let \((X,S)\) be a measurable space, and let \(f: X \rightarrow \mathbb{R}\). We say that \(f\) is an \(S\)-measurable function if the pre-image of every open set is in \(S\).
Example: let \(X = \mathbb{R}\). Let \(S\) be \(P(\mathbb{R})\), the power set of \(\mathbb{R}\). Then \((X,S)\) is a measurable space because the power set of \(\mathbb{R}\) is a \(\sigma\)-algebra.
Consider an arbitrary function \(f: X \rightarrow \mathbb{R}\). Then \(f\) is \(S\)-measurable. If we take any open set or Borel set in \(\mathbb{R}\), and take the preimage of that set, then it should be in \(S\).
Example: Let \(X = \mathbb{R}\) and let \(S = \{\emptyset, X\}\). We can see that this is a \(\sigma\)-algebra. Consider \(f: X \rightarrow \mathbb{R}\). What conditions are necessary for \(f\) to be an \(S\)-measurable function? We can show that \(f\) has to be constant.
Backwards: suppose \(f\) is a constant function. We should show that it is \(S\)-measurable. Suppose \(f(x) = c\), and let \(O\) be an open set in \(\mathbb{R}\). Then
\[f^{-1}(O) = \left\{\begin{array}{lcr} \mathbb{R}& \mbox{if} & c \in O \\ \emptyset &\mbox{if} & c \notin O \end{array}\right.\]
So \(f^{-1}(O) \in S\) for every open set \(O\). Therefore, \(f\) is \(S\)-measurable.
Forwards: suppose that \(f\) is \(S\)-measurable. Let us show that \(f\) must be constant. Suppose \(f\) is not constant. Then there exist \(x_1\) and \(x_2 \in X\) such that \(c_1 = f(x_1) ≠ f(x_2) = c_2\). Then since \(\{c_1\}\) and \(\{c_2\}\) are Borel sets (any closed set is a Borel set), \(f^{-1}(c_1) \in S\) and \(f^{-1}(c_2) \in S\) are both nonempty, because they must at least contain \(x_1\) and \(x_2\). Therefore, because the inverse images are not empty but are in \(S\), they must equal \(X\).
\(f^{-1}(c_1) = f^{-1}(c_2) = X\). Here is our contradiction. If we take \(x_0 \in X\), then we have \(f(x_0) = c_1 = c_2\) which is not a functional relationship. Therefore \(f\) must be constant.
Example: Let \(X = \mathbb{R}\) and let \(S = \mathcal B\), the Borel \(\sigma\)-algebra on \(\mathbb{R}\). The intersection of all \(\sigma\)-algebras that contain all open sets from \(\mathbb{R}\).
Let \(f: \mathbb{R}\rightarrow \mathbb{R}\). We say that \(f\) is a Borel-measurable function if \(f^{-1}(B) \in \mathcal B\) for all Borel sets \(B\).
And let \(f: \mathbb{R}\rightarrow \mathbb{R}\) be a continuous function. Is \(f\) Borel-measurable? (Proof 2.41). Let \(O\) be an open set in \(\mathbb{R}\). Then \(f^{-1}(O)\) is open. Any open set in \(\mathbb{R}\) is a Borel set, so \(f^{-1}(O) \in \mathcal B\).
Theorem (Proof 2.44): Let \((X,S)\) be a measurable space. And let \(f: X \rightarrow \mathbb{R}\) be an \(S\)-measurable function. Then
If \(\phi: \mathbb{R}\rightarrow \mathbb{R}\) is Borel-measurable, then \(\phi \circ f\) is \(S\)-measurable.
\(\phi \circ f: X \rightarrow \mathbb{R}\rightarrow \mathbb{R}\). So \((\phi \circ f)^{-1}(B) = f^{-1}(\phi^{-1}(B)) \in S\). Because \(g\) is Borel measurable, the preimage of a Borel set is a Borel set
Eg. let \(O\) be an open set in \(\mathbb{R}\). Then \(\phi^{-1}(O)\) is also an open set in \(\mathbb{R}\), so the preimage of this under \(f\) is in \(S\).
Review Thms 2.41, 2.43, Ex 2.45, Thm 2.46
\(A = \{x \in X: f(x) + g(x) < a\} = \{x\in X: f(x) < a - g(x)\}\).
\(Q = \{r_n: n \in \mathbb N\}\)
\(x \in A\) iff \(f(x) < a - g(x)\), which is true iff there exists some \(n \in \mathbb N\) such that \(f(x) < r_n < a - g(x)\).
Which we can rewrite as requiring that there exists \(n \in \mathbb N\) such that \(x \in f^{-1}((-∞,r_n))\) and \(x \in g^{-1}((-∞, a - r_n))\)
\[x \in \bigcup_{n=1}^∞\left[f^{-1}((-∞,r_n) \cap g^{-1}((-∞,a-r_n))\right] \]
therefore \(A \in S\). Then because \(-g\) is \(S\)-measurable.
HW 2B.13:
One direction, See Def 2.37 and Ex 2.38. So if we assume each \(E_i \in S\), then each \(c_i \chi_{E_i}\), then use sum of \(n\) S-measurable functions is S-measurable.
Other direction, suppose the sum is \(S\)-measurable and show that \(E_i\) are in \(S\). Write the function in terms of indicator functions.
The motivation is to extend the concept of length of an open interval in \(\mathbb{R}\) to other contexts. Eg. how do we measure the size of the rational numbers.
Definition 2.54: Let \(X\) be a nonempty set and let \(S\) be a \(\sigma\) algebra on \(X\). Then a measure on \((X,S)\) is a function \(\mu: S \rightarrow [0,∞]\) such that \(\mu(\emptyset) = 0\) and
\[\mu\left(\bigcup_{n=1}^∞ A_n\right) = \sum_{n=1}^∞ \mu(A_n) \]
Whenever \(\{A_n\}_{n=1}^∞\subset S\), and whenever the sets \(A_i \cap A_j = \emptyset\) for all \(i ≠ j\).
That is, \(\mu\) is countably additive. This will allow us to prove good limit theorems regarding measures.
Remark 1: Let \((X,S)\) be a measurable space and let \(\mu: S \rightarrow [0,∞]\) be a measure. Then \(\mu\) is also finitely additive. That is,
\[\mu\left(\bigcup_{k=1}^n A_k\right) = \sum_{k=1}^n \mu(A_k) \]
Remark 2: \((\mathbb{R}, \mathcal P(\mathbb{R}))\) is a measurable space. That means that the power set of \(\mathbb{R}\) is a \(\sigma\) algebra on \(\mathbb{R}\). Then consider the Lebesgue outer measure function
\[|\cdot |: \mathcal P(\mathbb{R}) \rightarrow [0,∞] \]
\[A \mapsto |A| = \inf\left\{\sum_{n=1}^∞ l(I_n)\right\}\]
where \(I_1, I_2, ...\) are open intervals such that \(A \subset \cup_{n=1}^∞ I_n\). Then this is not a measure. Recall that we proved that there exist two sets \(A,B \subset \mathbb{R}\) with \(A \cap B = \emptyset\), such that \(|A \cup B | ≠ |A| + |B|\) (See Jan 27 notes for proof of this).
Part of the reason we spent time definint \(\sigma\) algebras is that if we define a Lebesgue \(\sigma\) algebra, then we will not have this problem.
Definition: Let \(X\) be a nonempty set, let \(S\) be a \(\sigma\) algebra on \(X\), and let \(\mu: S \rightarrow [0,∞]\) be a measure. Then the ordered triplet \((X,S,\mu)\) is called a measure space. Important to distinguish between a measure space and a measurable space.
Let \((X,S,\mu)\) be a measure space. If \(\mu(X) = 1\), then \(\mu\) is called a probability measure.
Example: Let \(X\) be a nonempty set, and let \(S\) be \(\mathcal P(X)\). Define \(\mu: S \rightarrow [0,∞]\):
\[A \mapsto \left\{\begin{array}{lcr} \mathrm{card}(A) & \mbox{if} & \mathrm{card}(A) < ∞ \\ 0 & \mbox{if} & \mathrm{otherwise} \end{array}\right. \]
**Example: Let \(X\) be a nonempty set, and let \(x_0 \in X\) be a fixed element in \(X\). \(S\) be \(\mathcal P(X)\). Define \(\delta_{x_0}: \mathcal P(X) \rightarrow [0,∞]\):
\[A \mapsto \delta_{x_0}(A) = \left\{\begin{array}{lcr} 0 & \mbox{if} & x_0 \notin A \\ 1 & \mbox{if} & x_0 \in A \end{array}\right.\]
Then \(\delta_{x_0}\) is a measure on \(\mathcal P(X)\) and is called the direct measure.
eg, if \(X = \mathbb{R}\) and \(x_0 = 3\), then \(\delta_{x_0}([0,1]) = 0\). And because \(\delta_{x_0}(\mathbb{R}) = 1\), this is a probability measure.
\[ \frac{f(y) - f(x)}{y-x} = f'(c)\]
for some \(c \in [x,y]\), then
\[\left|\frac{f(y) - f(x)}{y-x}\right| = |f'(c)| ≤ 1\]
So \(|f(y) - f(x)| ≤ |y-x|\).
\[|y_1 - y_2| = |f(x_1) - f(x_2)| ≤ |x_1 - x_2| ≤ b-a \]
Finally because \((y_1, y_2)\) is an open interval in \(\mathbb{R}\), \(|f(I)| ≤ |I| = b-a\).
\[\sum_{n=1}^∞ l(I_n) < |A| + \epsilon \]
and because \(A\) is a subset of the above union,
\[f(A) \subset f\left(\cup_{n=1}^∞ I_n\right) \subset \cup_{n=1}^∞ f(I_n)\]
Then
\[\begin{align*} |f(A)| &≤ |\cup_{n=1}^∞ f(I_n)| \\ &≤ \sum_{n=1}^∞ |f(I_n)| \\ &≤ \sum_{n=1}^∞ |I_n| \\ &= \sum_{n=1}^∞ l(I_n) < |A| + \epsilon \end{align*}\]
Then letting \(\epsilon \rightarrow 0^+\), we obtain the result.
Take any \(a \in \mathbb{R}\) and consider
\[A = \{x \in X: |f(x)| < a\} \]
Then by the three equivalent properties, we can say that this is equivalent to
\[A = \{x \in X: -a < f(x) < a\} = f^{-1}((-a,a)) \in S\]
when \(a > 0\). If \(a < 0\), then \(A = \emptyset\). The inverse image is in \(S\) because \((-a,a)\) is an open set, and thus measurable.
\[\lim_{n \rightarrow ∞} f_n(x) = \lim_{n \rightarrow ∞} \frac{f(x + 1/n) - f(x)}{1/n} = f'(x)\]
If we fix \(n\), then \(f_n(x)\) is continuous, so it is Borel Measurable. And it converges pointwise to \(f'(x)\). We proved earlier that the pointwise limit of a sequence of Borel measurable functions is also Borel measurable.
Definition: a measure is defined on a \(\sigma\) algabra. Let \(S\) be a \(\sigma\) algebra. Then a function \(\mu: S \rightarrow [0,∞]\), such that \(A \mapsto \mu(A)\) is called a measure if \(\mu(\emptyset) = 0\) and \(\mu(\cup_{n=1}^∞ A_n) = \sum_{n=1}^∞ \mu(A_n)\) whenever \(A_1, A_2, ... \in S\) and \(A_i \cap A_j = \emptyset\) whenever \(i ≠ j\).
Why is it required that \(S\) be a \(\sigma\) algebra? Because if \(S\) is not a \(\sigma\) algebra, it might not contain the empty set, and it might not contain a countable union of its component sets.
Previously we mentioned the Lebesgue outer measure, \(|\cdot|: \mathcal P(\mathbb{R}) \rightarrow [0,∞]\) such that \(A \mapsto |A|\), but this is not countably additive, so it is not a measure under our definition. We can define a new measure.
Define \(\mathcal L \subset \mathcal P (\mathbb{R})\) such that \(\mathcal L\) is a \(\sigma\) algebra. It i the collection of all Lebesgue measurable sets. We call it the Lebesgue \(\sigma\) algebra. This will yield some desirable properties, because when we restrict our lebesgue outer measure to this \(\sigma\) algebra, we get \(\lambda : \mathcal L \rightarrow [0,∞]\) such that \(A \mapsto |A|\), where this is just the Lebesgue outer measure function with new notation \(\lambda\), restricted to components of \(\mathcal L\) instead of on the power series of \(\mathbb{R}\).
This is a different approach than that used by the textbook.
Definition: A subset \(M \subset \mathbb{R}\) is called a Lebesgue measurable set if \(|A| = |A \cap M| + |A \cap M^C|\) for all subsets \(A \subset \mathbb{R}\). The collection of all Lebesgue measurable sets is denoted by \(\mathcal L\).
Another way of thinking of the sets \(M\) is that \(M\) is a Lebesgue measurable set if it can be used to cut any other subset of \(\mathbb{R}\) into two pieces, the sum of whose Lebesgue outer measures is equal to the Lebesgue outer measure of the whole.
Example: \(\mathbb{R}\in \mathcal L\). We can see that \(|A| = |A \cap \mathbb{R}| + |A \cap \mathbb{R}^C| = |A| + |\emptyset|\) for any subset \(A \subset \mathbb{R}\).
Example: Similarly, \(\emptyset \mathcal L\).
Example: \((t, ∞) \in \mathcal L\). Like on the midterm.
Example: \((-t,t) \in \mathcal L\). \(|A| = |A \cap (-t,t)| + |A \cap (-t,t)^C|\). Like on the second homework.
Remark: \((A \cap M) \cup (A \cap M^C) =A\) always, by de Morgan’s laws. Thus, \(|A| = |(A \cap M) \cup (A \cap M^C)|\). Then, the Lebesgue outer measure is always countably subadditive, and so \(|A| ≤ |A \cap M| + |A \cap M^C|\).
Therefore, \(M \in \mathcal L \Leftrightarrow |A| ≥ |A\cap M| + |A \cap M^C|\) for all subsets \(A\subset \mathbb{R}\). This will be helpful, because to show that a set \(M\) is in \(\mathcal L\), we only need to prove the inequality, rather than the equality.
Example: Let \(M \subset \mathbb{R}\). If \(|M| = 0\), then \(M \in \mathcal L\). In particular, the LOM of the rational numbers is zero, so \(\mathbb Q \in \mathcal L\).
To prove this, we need to show that \(|A| ≥ |A\cap M| + |A \cap M^C|\). \(A \cap M \subset M\), so \(|A \cap M | = 0\) by monotonicity. Then \(A \cap M^C \subset A\), so \(|A \cap M^C| ≤ |A|\), again by monotonicity. So no matter what set \(A \subset \mathbb{R}\) we take, \(|A\cap M| + |A \cap M^C| ≤ |A|\), so \(M \in \mathcal L\).
Our goal is to show that \(\mathcal L\) is a \(\sigma\) algebra on \(\mathbb{R}\). It already contains the empty set and is closed under complimentation, so we need to show that it is closed under countable union.
Theorem: First we will show that \(\mathcal L\) is closed under finite union.
If \(M_1, M_2, ..., M_n \in \mathcal L\), then \(\cup_{i=1}^n M_i \in \mathcal L\). By induction, it suffices to prove for the case where \(n = 2\). Let \(M_1, M_2 \in \mathcal L\). We should show that \(|A| ≥ |A \cap (M_1 \cup M_2)| + |A \cap (M_1 \cup M_2)^C|\). By de Morgan’s laws, this is equivalent to \(|A| ≥ |A \cap (M_1 \cup M_2)| + |A \cap M_1^C \cap M_2^C|\).
First, on the lefthand side. We should remember that \(M_1, M_2 \in \mathcal L\). So
\[\begin{align*} |A| &= |A \cap M_1| + |A \cap M_1^C| \\[1em] &= |A \cap M_1| + |A \cap M_1^C \cap M_2| + |A \cap M_1^C \cap M_2^C| \\[1em] &= |(A\cap M_1) \cup (A \cap M_1^C \cap M_2)| + |A \cap M_1^C \cap M_2^C| \tag{Subadditivity} \\[1em] &= |A \cap (M_1 \cup M_2)| + |A \cap M_1^C \cap M_2^C| \end{align*}\]
for all \(A \subset \mathbb{R}\). Therefore \(M_1 \cup M_2 \in \mathcal L\).
Definition: The set \(\mathcal L \subset \mathcal P(\mathbb{R})\). \(M \in \mathcal L\) iff \(|A| = |A \cap M| + |A \cap M^C|\) for all \(A \subset \mathbb{R}\). This set is going to allow us to deal with the major problem with the Lebesgue outer measure, which was that there exist disjoint sets in \(\mathbb{R}\) such that the LOM of their union is not equal to the sum of their LOMs.
Proposition: \(\mathcal L\) is an algebra.
\(\emptyset in \mathcal L\).
Closed under complimentation
Closed under finite union. If \(M_1, M_2, ..., M_n \in \mathcal L\), then \(\cup_{i=1}^n M_i \in \mathcal L\).
In the definition of \(\sigma\) algebra, we require that the algebra be closed under countably infinite union. Here we only require closure under finite union. We saw an intuitive representation of this using Venn diagrams during 6.3, but today we can prove it.
Proof:
\[(A \cap M_1) \cup (A \cap M_2 \cap M_1^C) = A \cap (M_1 \cup M_2)\]
\[\begin{align*} LHS &= [(A \cap M_1) \cup (A \cap M_2)] \cap [A \cap M_1) \cup M_1^C] \\[1em] &= [(A \cap M_1) \cup (A \cap M_2)] \cap [(A \cup M_1^C) \cap (M_1 \cup M_1^C)] \\[1em] &= [(A \cap M_1) \cup (A \cap M_2)] \cap (A \cup M_1^C) \\[1em] &= (A \cap M_1) \cup (A \cap M_2) \\[1em] &= A \cap (M_1 \cup M_2) \tag{which is the RHS} \end{align*}\]
Now we have that \(\mathcal L\) is an algebra. To show that it is also a \(\sigma\) algebra, we need to use a Lemma we have used before.
If \(A \subset \mathbb{R}\), and we have four disjoint sets \(M_1, ..., M_4 \in \mathcal L\), then \(|A \cap (\cup_{i=1}^4 M_i)| = \sum_{i=1}^4 |A \cap M_i|\).
Lemma: Let \(M_1, ..., M_n \in \mathcal L\) such that \(M_i \cap M_j =\emptyset\) whenever \(i ≠ j\) and \(i, j = 1, ..., n\). Then
\[|A \cap (\cup_{i=1}^n M_i)| = \sum_{i=1}^n |A \cap M_i| \tag{*}\]
for all \(A \subset \mathbb{R}\). This is the big distinction from the power set of \(\mathbb{R}\). On \(\mathcal L\), it is countably additive. In particular, if we choose as our subset \(A = \mathbb{R}\), then \(|\cup_{i=1}^n M_i| = \sum_{i=1}^n |M_i|\).
Proof: We can look at the base case when \(n = 1\). The LHS is \(|A \cap M_1|\), and the RHS is \(|A \cap M_1|\). So (*) is satisfied for \(n = 1\).
Suppose that \(|A \cap (\cup_{i=1}^n M_i)| = \sum_{i=1}^n |A \cap M_i|\) for all subsets \(A \in r\), mutually disjoint sets \(M_i \in \mathcal L\), and \(n \in \mathbb N\). Our goal is to show that it is true for \(n + 1\):
\(|A \cap (\cup_{i=1}^{n+1} M_i)| = \sum_{i=1}^{n+1} |A \cap M_i|\) for all \(A \subset \mathbb{R}\).
We know that a set \(M \in \mathcal L\) iff The LOM of any subset of \(\mathbb{R}\) can be split into the LOM of its intersection with \(M\) and the LOM of its intersection with \(M^C\). So in our target equation,
\[\begin{align*} LHS &= |A \cap (\cup_{i=1}^{n+1} M_i)| \\[1em] &= |A \cap (\cup_{i=1}^{n+1} M_i) \cap M_{n+1}| + |A \cap (\cup_{i=1}^{n+1} M_i) \cap M_{n+1}^C| \\[1em] &= |A \cap M_{n+1}| + |A \cap(\cup_{i=1}^{n} M_i)| \\[1em] &= |A \cap M_{n+1}| + \sum_{i=1}^n |A \cap M_i| \\[1em] &= \sum_{i=1}^{n+1} |A \cap M_i| \end{align*}\]
which is the inductive hypothesis. So for sets in \(\mathcal L\), the Lebesgue outer measure is finitely additive.
Theorem: \(\mathcal L\) is a \(\sigma\) algebra on \(\mathbb{R}\) and it is called the Lebesgue \(\sigma\) algebra on \(\mathbb{R}\).
Proof: We need to show that \(\mathcal L\) meets the three criteria. We already showed #1 and #2, so we should show that if \(\{M_k\}_{k=1}^∞ \in \mathcal L\), then \(\cup_{k=1}^∞ M_k \in \mathcal L\). We will show that
\[|A| = |A \cap (\cup_{k=1}^∞ M_k )| + \left|A \cap (\cup_{k=1}^∞ M_k )^C\right| \tag{*}\]
Note that the way we set up this equation is exactly the definition of an element of \(\mathcal L\). For any \(n \in \mathbb N\) we know that for sets \(M_i \in \mathcal L, \cup_{i=1}^{n} M_i \in \mathcal L\).
Without loss of generality, assume that \(M_i \cap M_j = \emptyset\) whenever \(i ≠ j\). If we have non-disjoint sets \(M_i\), then we can define a new set of disjoint sets
\[\begin{align*} D_1 &= M_1 \\ D_2 &= M_2 \setminus M_1 \\ \cdots \\ D_n &= M_n \setminus (\cup_{i=1}^{n-1}M_i) \end{align*}\]
Previously, we proved that \(\mathcal L\) is closed under finite union, meaning that for any \(n \in \mathbb N\), $
\[\begin{align*} |A| &= |A \cap (\cup_{k=1}^n M_k )| + \left|A \cap (\cup_{k=1}^n M_k )^C\right| \\[1em] &= \sum_{i=1}^n|A \cap M_i| + \left|A \cap (\cup_{k=1}^n M_k )^C\right| \\[1em] &≥ \sum_{i=1}^n|A \cap M_i| + \left|A \cap (\cup_{k=1}^∞ M_k )^C\right| \tag{monotonicity, note the index change} \\[1em] \end{align*}\]
So letting \(n \rightarrow ∞\),
\[\begin{align*} |A| &≥ \sum_{i=1}^∞ |A \cap M_i| + \left|A \cap (\cup_{k=1}^∞ M_k )^C\right| \\[1em] &≥ | \cup_{i=1}^∞(A \cap M_k)| + |A \cap (\cup_{i=1}^∞ M_k)^C| \\[1em] &= |A \cap (\cup_{i=1}^∞ M_k)| + |A \cap (\cup_{i=1}^∞ M_k)^C| \end{align*}\]
Therefore, \(\cup_{i=1}^∞ M_k \in \mathcal L\).
Theorem: Consider the function \(\lambda: \mathcal L \rightarrow [0,∞]; A \mapsto \lambda(A) = |A|\). Then \(\lambda\) is a measure on \(\mathcal L\) and it is called the Lebesgue measure, “wow.”
After this theorem, we will have a solid definition of the Lebesgue measure. To define a measure, we need a \(\sigma\) algebra. We’ve done this with \(\mathcal L\). But lastly we need to show that \(\lambda\) is countably additive to confirm that it is a measure.
Proof: First, it is clear that \(\lambda(\emptyset) = |\emptyset| = 0\).
Let \(\{M_k\}_{k=1}^∞ \subset \mathcal L\) satisfy \(M_i \cap M_j = \emptyset\) whenever \(i ≠ j\). We should show that
\[\lambda\left(\cup_{k=1}^∞ M_k\right) = \sum_{k=1}^∞ \lambda(M_k)\]
or equivalently,
\[|\cup_{k=1}^∞ M_k| = \sum_{k=1}^∞ |M_k| \]
The fact that \(|\cup_{k=1}^∞ M_k| ≤ \sum_{k=1}^∞ |M_k|\) is obvious from the subadditivity of the LOM. So we need to prove the opposite direction.
For any \(n \in \mathbb N\), \(\sum_{k=1}^n |M_k| = |\cup_{k=1}^n M_k| ≤ |\cup_{k=1}^∞ M_k|\) (from the previous lemma).
Letting \(n \rightarrow ∞\) gives
\[\sum_{k=1}^∞ |M_k| ≤ |\cup_{k=1}^∞ M_k| \]
So the theorem is proved.
Exercise 2C.3: the notation is referring to all outputs \(\mu(E)\) such that the input is a positive integer. MNN suggests looking at binary expansion.
Exercise 2A.9: hint is to use the lemma we proved today.
Previously, we defined the Lebesgue \(\sigma\) algebra on \(\mathbb{R}\), \(\mathcal L\). A subset \(M \subset \mathbb{R}\) belongs to \(\mathcal L\) iff \(|A| = |A \cap M| + |A \cap M^C|\) for all subsets \(A \in \mathbb{R}\).
We also defined the function \(\lambda\) that maps elements of \(\mathcal L\) to their Lebesgue outer measure in \([0,∞]\). \(\lambda\) is called the Lebesgue measure on \(\mathbb{R}\) rather than the Lebesgue outer measure because its domain is restricted to elements of \(\mathcal L\).
Our eventual motivation is to explore the Lebesgue Integral. But today we will look at an equivalent definition of the Lebesgue \(\sigma\) algebra from the textbook.
First, we can note that the Borel \(\sigma\) algebra is a subset of \(\mathcal L\).
Proof:
Step 1. Any open interval in \(\mathbb{R}\) of the form \(I = (a,b)\) belongs to \(\mathcal L\). This is because \(|A| = |A \cap I| + |A \cap I^C|\) for all \(A \subset \mathbb{R}\).
Step 2. Let \(G\) be an open set in \(\mathbb{R}\). Then \(G = \bigcup_{n=1}^∞(a_n, b_n)\). So any open set in \(\mathbb{R}\) can be represented as a countable union of open intervals. And the individual intervals can be disjoint. Because \(\mathcal L\) is a \(\sigma\) algebra, it is closed under countable union, meaning that \(G = \bigcup_{n=1}^∞(a_n, b_n) \in \mathcal L\).
Therefore, every Borel set is in \(\mathcal L\), so \(\mathcal B \subset \mathcal L\). \(\mathcal B\) is the smallest \(\sigma\) algebra containing all open sets in \(\mathbb{R}\), so it should be a subset of \(\mathcal L\).
How can we find a subset of \(\mathbb{R}\) that is not in \(\mathcal L\)? First, there are so many sets in \(\mathcal L\). \(\emptyset, \mathbb Q, [1,2]\). \(mathbb Q\) can be represented as the countable union of singleton sets, each of which is a Borel set, so each is in \(\mathcal L\) and the countable union is in \(\mathcal L\) as well.
Theorem 2.71: Suppose \(A \subset \mathbb{R}\). Then the following are equivalent (below, \(M\) stands for a measurable set):
\(A\) is Lebesgue measurable.
For each \(\epsilon > 0\), there exists an open set \(G \supset M\) such that \(|G \setminus M| < \epsilon\).
There exists a sequence of open sets \(G_1, G_2, ...\), each \(G_n\) containing \(M\), such that \(\left|\bigcap_{k=1}^∞ G_k \setminus M\right| = 0\).
There exsists a Borel set \(B \subset M\) such that \(|B \setminus M| = 0\).
Proof \(A \Rightarrow B\): Suppose that \(M \in \mathcal L\). We should show that no matter what \(\epsilon > 0\) we select, we can find an open set \(G\) such that \(M \subset G\), and \(|G \setminus M| < \epsilon\). So we can approximate the Lebesgue measure of a measurable set using the Lebesgue measure of an open set. Another way of writing this is that \(\lambda(G \setminus M) < \epsilon\). We can write it this way because \(G, M,\) and \(G \setminus M\) are all in \(\mathcal L\).
Case 1: \(|M| < ∞\). Take any \(\epsilon > 0\). Then by the definition of the LOM, there exists a collection of open intervals \(\{I_n\}_{n=1}^∞\) in \(\mathbb{R}\) such that \(M \subset \cup_{n=1}^∞ I_n\) and \(\sum_{n=1}^∞ l(I_n) < |M| + \epsilon\). So we can denote \(G = \cup_{n=1}^∞ I_n\). Then \(G\) is a union of open sets, which means that it is also open. Also, \(|G| = |\cup_{n=1}^∞ I_n| ≤ \sum_{n=1}^∞ |I_n|\) using the subadditivity property. The LOM of each interval is just its length, so \(|G| ≤ \sum_{n=1}^∞ l(I_n) < |M| + \epsilon\). Then, becuase \(M \subset G\), \(\lambda(G) < \lambda(M) + \epsilon\), and \(\lambda(G) - \lambda M) < \epsilon)\), so \(\lambda(G \setminus M) < \epsilon\).
Case 2: \(|M| = ∞\). \(M\) can be represented as the countable union of sets with finite Lebesgue measures. MNN’s hint is to use \(\epsilon/2^n\) (see screenshots).
Proof \(B \Rightarrow C\): from (b), for every \(\epsilon > )\) there exists an open set \(G \supset M\) such that \(|G \setminus M| < \epsilon\). Part (c) says that there exists a sequence of open sets \(G_1, G_2, ...\), each \(G_n\) containing \(M\), such that \(\left|\bigcap_{k=1}^∞ G_k \setminus M\right| = 0\).
Take \(\epsilon = 1\). Then there exists an open set \(G_1 \supset M\) such that \(|G_1 \setminus M| < 1\).
Take \(\epsilon = 1/2\). Then there exists an open set \(G_2 \supset M\) such that \(|G_2 \setminus M| < 1/2\).
Generally, for \(\epsilon = 1/n\), there exists an open set \(G_n \supset M\) such that \(|G_n \setminus M| < 1/n\).
So consider the sequence \(G_1, G_2, ...\). Then \(M \subset G_n\) for all \(n \in \mathbb N\). And let \(G = \bigcap_{n=1}^∞ G_n\). Then \(M \subset G\). And \(\left|\left(\bigcap_{n=1}^∞ G_n\right) \setminus A\right| ≤ |G_n \setminus A| < 1/n\) for all \(n \in \mathbb N\) because the intersection will be a subset of any one \(G_n\).
So \(0 ≤ |G \setminus A| < 1/n\), so the limit is zero, which is the criterion for (c).
In the homework, MNN is mentioning coming up with a sequnce that meets the nestedness criterion. Where \(G_1 = O_1, G_2 = O_1 \cup O_2, ...\).
Proof \(C \Rightarrow D\): Suppose (c) is satisfied. So there exists a sequence of open sets \(G_1, G_2, ...\), each \(G_n\) containing \(M\), such that \(\left|\bigcap_{k=1}^∞ G_k \setminus M\right| = 0\).
We want to prove that there exsists a Borel set \(B \subset M\) such that \(|B \setminus M| = 0\).
Let \(B = \bigcap_{n=1}^∞ G_n\). Then \(B\) is a Borel set because it is the intersection of Borel sets \(G_n\), and the Borel \(\sigma\) algebra is closed under countable intersection.
Also, \(|B \setminus M| = 0\). So \(C \Rightarrow D\).
Proof \(D \Rightarrow A\): We have a set \(B \in \mathcal B\) such that \(M \subset B\) and \(|B \setminus M | = 0\). Now we need to prove that under these conditions, \(M \in \mathcal L\).
\(|B \setminus M| = 0\). And previously we showed that any set with LOM = 0 is a member of \(\mathcal L\). Therefore \(B \setminus M \in \mathcal L\).
And \(\mathcal B \subset \mathcal L\) so \(B \in \mathcal L\).
Finally, \(M = B \setminus (B \setminus M)\). Both elements on the RHS are in \(\mathcal L\), and \(\sigma\) algebras are closed under set operations, so \(M \in \mathcal L\).
In the case where we want to look at closed sets, we can modify the proofs.
Hint: \(M^C \in \mathcal L\). Then \(M^C \subset G\) and \(G^C \subset M\).
Constructing the Cantor Set:
Step 1: Remove the middle third of the interval \([0,1]\), so remove \((1/3, 2/3)\). Denote
\[P_1 = [0,1/3]\cup [2/3,1]\hspace{1cm}G_1 = (1/3, 2/3)\]
Step 2: Remove the middle third of each of the components of \(P_1\), that is remove \((1/9,2/9)\) and \((7/9,8/9)\). Denote \(P_2\) as the remaining closed intervals, and \(G_2\) as the union of the two open intervals we just removed.
\[P_2 = \left[0,\frac{1}{9}\right]\cup \left[\frac{2}{9},\frac{3}{9}\right]\cup \left[\frac{6}{9},\frac{7}{9}\right]\cup\left[\frac{8}{9},1\right] \hspace{1cm} G_2 =\left(\frac{1}{9},\frac{2}{9}\right)\cup \left(\frac{7}{9},\frac{8}{9}\right)\]
After the \(n-1\)th step, there remain \(2^{n-1}\) closed intervals of length \(\frac{1}{3^{n-1}}\)
Step \(n\): remove the middle third open interval from every closed interval in \(P_{n-1}\). The union of the remaining closed intervals after the \(n\)th step is \(P_n\), while \(G_n\) is the union of all open intervals removed during step \(n\).
We can observe that \(P_1 \supset P_2 \supset \cdots\), and \(P_n\) is a compact set for every \(n\) because it is a union of closed and bounded (compact) sets.
\(P = \bigcap_{n=1}^∞ P_n ≠ \emptyset\) is called the Cantor Set. We can denote another set \(G = \bigcup_{n=1}^∞ G_n\).
Observations about \(P\):
Obs 1: \(P\) is nonempty and closed.
Obs 2: \(G\) is nonempty and open.
Obs 3: \(P_n\) is the set remaining after the \(n\)th step, and is the union of \(2^n\) closed intervals, each of length \(\frac{1}{3^n}\).
Obs 4: \(G_n\) is the set removed at the \(n\)th step. It is composed of \(2^{n-1}\) open intervals of length \(\frac{1}{3^n}\).
Obs 5: \(|G_n| = \lambda(G_n) = \sum_{i=1}^{2^{n-1}}\frac{1}{3^n} = \frac{2^{n-1}}{3^n}\).
Obs 6: \(|P_n| = \lambda(P_n) = \sum_{i=1}^{2^n}\frac{1}{3^n} = \left(\frac{2}{3}\right)^n\).
Obs 7: \(\lambda(P) = |P| = 0\).
\[P = \bigcap_{n=1}^∞ P_n \subset P_n\]
Then \(\lambda(P) = |P| ≤ |P_n| = \lambda(P_n)\) for all \(n\). Letting \(n \rightarrow ∞\) gives \(\lambda(P) = 0\).
Obs 8: \(P\) contains no nonempty open intervals. We can easily prove this by contradiction, as the Lebesgue measure of a nonempty open interval is greater than zero. This means that \(P\) is nowhere dense. \(\mathrm{int}(\overline P) = \emptyset = \mathrm{int}(P)\).
Let \(p \in \mathbb Z\), \(p > 1\). Then for any \(x \in [0,1]\) we can create a base\(-p\) expansion of \(x\). There is a sequence \(\{a_n\}_{n=1}^∞\) with \(0 ≤ a_n ≤ p-1\) such that \(x = \frac{a_1}{p} + \frac{a_2}{p^2} + \frac{a_3}{p^3} + \cdots\). The notation for this is \(x = 0.a_1a_2a_3...\).
The base-\(p\) expansion of \(x\) is unique unless \(x ≠ 1\) and \(x = \frac{q}{p^m}\) where \(q,m\) are positive integers.
Example: In base-3, a ternary expansion of \(0 = 0.0000...\), and two expansions of \(1 = 1 = 0.222...\). The expansion of \(1/3 = 0.1\).
\[\begin{align*} x = \frac{a_1}{p} + \frac{a_2}{p^2} + \frac{a_3}{p^3} + \cdots \\[1em] px = a_1 + \frac{a_2}{p} + \frac{a_3}{p^2} + \cdots \\[1em] \end{align*}\]
And we know that the sum of the 2nd term onward on the RHS, so denoting \(x_0 = x\), \(a_1 = \lfloor px_0 \rfloor\). \(x_1 = p x_0 - a_1\), \(a_2 = \lfloor p x_1 \rfloor\), and so on.
Example: Converting 1/4 to base 3.
\[\begin{align*} x_0 &= \frac{1}{4} \\[1em] px_0 &= \frac{3}{4} \\[1em] a_1 &= \lfloor px_0\rfloor = 0 \\[1em] x_1 &= px_0 - a_1 = \frac{3}{4} \\[1em] px_1 &= \frac{9}{4} \\[1em] a_2 &= \lfloor px_1 \rfloor = 2 \\[1em] x_2 &= px_1 - a_2 = \frac{1}{4} \end{align*}\]
We can see the pattern from here, so the base three expansion of \(\frac{1}{4} = 0.\overline{02}\)
Previously we looked at an algorithm for converting base-10 numbers to a different base using the flooring function.
\[x = a_1/p + a_2/p^2 + \cdots \hspace{1cm} px = a_1 + a_2/p + \cdots\]
where the sum of the second term onward on the RHS is between 0 and 1. Then \(px - a_1 \{p_x\}\) is the decimal part of the RHS. Then we iterate this. Previously we saw that for \(x = 1/4\),
\[a_1 = \lfloor 3(1/4)\rfloor = 0 \hspace{1cm} a_2 = \lfloor 3(3/4)\rfloor = 2 \hspace{1cm} a_3 = \lfloor 3(1/4)\rfloor = 0\hspace{1cm} \cdots\]
so that \(x = 0.\overline{02}\) in base 3.
At the first step, we take out \(\left(\frac{1}{3},\frac{2}{3}\right)\), or in base 3, \((0.1, 0.2)\).
At the second step, we take out \(\left(\frac{1}{9},\frac{2}{9}\right)\) and \(\left(\frac{7}{9},\frac{8}{9}\right)\), or in base-3, \((0.01, 0.02)\) and \((0.21, 0.22)\).
Theorem: an interval \((a,b)\) is one of \(2^{n-1}\) open intervals removed from the interval \([0,1]\) at the \(n\)th step in the construction of the cantor set iff \(a\) and \(b\) are of the following form:
\[a = 0.(2c_1)(2c_2)\cdots(2c_{n-1})100\cdots \hspace{1cm} b = 0.(2c_1)(2c_2)\cdots(2c_{n-1})200\cdots \tag{1}\]
Proof: Induction. If we look at the left endpoint of an interval removed at the \(n\)th step is the right endpoint of an interval removed at the \(n-1\)th step, adding \(1/3^n\).
Theorem: Let \(x \in [0,1]\). Then \(x\) belongs to the Cantor set iff \(x\) has a ternary expansion without the digit 1.
Example: \(1/4\) is in the Cantor set. \(0\) is in the Cantor set. \(1/3 = 0.1\) OR \(0.0\overline 2\), so \(1/3\) is in the Cantor set. \(2/3 = 0.2\) is in the Cantor set.
\([0,1] = G \cup P\) and \(G \cap P = \emptyset\). \(x \in G\) iff \(x \in (a,b)\) where \((a,b)\) is removed at some step. if \((a,b)\) is removed, then \(a,b\) will have the form in (1). And
\[x = 0.(2c_1)(2c_2)\cdots(2c_{n-1})1d_{n+1}d_{n+2}\cdots \]
where not all the \(d\) values are equal to zero, and not all equal to two (otherwise \(x\) would equal \(a\) or \(b\), so it wouldn’t be in the open interval).
\(f: P \rightarrow \mathbb{R}\). Let \(x \in P\). Then \(x\) has a ternary expansion without the digit 1. Then \(x \mapsto 0.c_1c_2\cdots\) in base 2.
Example: \(x = 1/3 = 0.0\overline 2\) in base 3. Then \(f(x) = 0.0\overline 1\) in base 2 is equal to \(1/2\) in base 10.
If \(x = 2/3 = 0.2\), then \(f(x) = 0.1\) which equals \(1/2\) in base 10.
\(\psi: [0,1] \rightarrow \mathbb{R}\). If \(x \in P\), then \(\psi(x) = f(x)\). If \(x \notin P\), then \(x \in G\), where \(G\) is the set of intervals removed from \([0,1]\).
Then \(x \in (a,b)\) where \(a,b\) have the form in equation (1). We can rewrite \(a= 0.(2c_1)(2c_2)\cdots(2c_{n-1})0\overline 2\). Then when we convert to base 2, we see that \(f(a) = f(b)\). So for \(x \notin G\), \(\psi(x) = f(a) = f(b)\). These correspond to the flat portions of the Cantor function.
Properties of the Cantor function: \(\psi\) is nondecreasing, onto, and continuous.
Find a measure \(\mu: \mathbb Z^+ \rightarrow [0,∞]\) such that it maps to \([0,1]\). Binary expansion. Use \(\sum_{i=1}^∞ \frac{1}{2^{\alpha_i}}\) if \(\alpha\) is an infinite set. If it is a finite set, then use a finite sum.
Take \(A = \{2,3\}\). then \(\mu(A) = \frac{1}{2^2} + \frac{1}{2^3}\).
Then \(\mu(\mathbb Z^+) = \sum \frac{1}{2^i} = \frac{1/2}{1-1/2} = 1\) and \(\mu(\emptyset) = 0\)
\(A_1 \supset A_2 \supset \cdots\), hint, \(\mu = \lambda\). Think about \(A_n = [n,∞)\). Find \(G_1 \supset G_2\) following the procedure from class on Monday.
Recall some definitions:
\(\mathcal L\), the collection of all Lebesgue measurable sets.
A set is Lebesuge measurable, ie. \(M \in \mathcal L\) if and only if \(|A| = |A \cap M| + |A \cap M^C|\) for all subsets \(A \subset \mathbb{R}\).
\(\mathcal L\) is a \(\sigma-\) algebra on \(\mathbb{R}\) and is called the Lebesgue \(\sigma\) algebra on \(\mathbb{R}\).
The Lebesgue measure is the function \(\lambda: \mathcal L \rightarrow [0,∞]; M \mapsto \lambda(M) = |M|\).
If \(\mathcal S\) is a \(\sigma\) algebra on \(X\), then \((X,\mathcal S)\) is a measurable space.
A measurable function: If \((X,\mathcal S)\) is a measurable space, then a function \(f: X \rightarrow \mathbb{R}\) is called a measurable function iff \(f^{-1}(O) \in \mathcal S\) for every open set \(O \in \mathbb{R}\). Equivalently, the preimage of any Borel set is in \(\mathcal S\). We can also use the preimage of an interval \((-∞,a)\) or \((a,∞)\).
Because \(\mathcal L\) is a \(\sigma\) algebra on \(\mathbb{R}\), then \((\mathbb{R}, \mathcal L)\) is a measurable space. A function \(f: \mathbb{R}\rightarrow \mathbb{R}\) is called a Lebesgue measurable function if and only if \(f^{-1}(O) \in \mathcal L\) for every open set \(O \in \mathbb{R}\). So a Lebesgue measurable function is simply an \(\mathcal L\) measurable function.
(Good distinction from Eric, it is not necessarily the case that the preimage of a Lebesgue measurable set is in \(\mathcal L\), because there are Lebesgue measurable sets that are not Borel sets. \(\mathcal B \varsubsetneq \mathcal L\).)
Say you have a collection of coins. Pennies, nickels, dimes, quarters. If we want to know how much money we have total, we count up each individual coin. This is analogous to Riemann integration. Under Lebesgue integration, we group the coins and count the value of the groups.
**Definition*: Simple functions. A function \(f: \mathbb{R}\rightarrow \mathbb{R}\) is called a simple function if the range of \(x\) is a finite subset of \(\mathbb{R}\).
Example: Let \(f:\mathbb{R}\rightarrow \mathbb{R}\) be given by
\[f(x) = \left\{\begin{array}{lcr}1 & \mbox{if} & x \in [-1,0) \\ 2 &\mbox{if} & x \in [0,1] \\0 & \mbox{if} & o/w \end{array}\right.\tag{1}\]
Example: Let \(f:\mathbb{R}\rightarrow \mathbb{R}\) be given by
\[f(x) = \left\{\begin{array}{lcr}1 & \mbox{if} & x \in \mathbb Q \\ 0 & \mbox{if} & x \notin \mathbb Q \end{array}\right.\]
Definition: Characteristic functions. Let \(A \subset \mathbb{R}\). Define \(\chi_A: \mathbb{R}\rightarrow \mathbb{R}\) by
\[\chi_A(x) = \left\{\begin{array}{lcr}1 & \mbox{if} & x \in A \\ 0 & \mbox{if} & x \notin A \end{array}\right.\]
Now suppose that \((X,\mathcal S)\) is a measurable space, and that \(f: X \rightarrow \mathbb{R}\) is a simple function, and \(c_1, ..., c_n\) are the distinct nonzero values of \(f\). Then
\[f = c_1\chi_{E_1} + \cdots + c_n \chi_{E_n} \]
where \(E_k = f^{-1}(\{c_k\})\). Or equivalently, \(E_k = \{x \in \mathbb{R}: f(x) = c_k\}\). That is, any simple function can be written as a linear combination of characteristic functions.
For example, the function in (1) could be written as \(f = \chi_{[-1,0)} + 2\chi_{[0,1]}\). To show that these are equal, take
Case 1: \(x\in [-1,0)\): LHS = 1, RHS = 1.
Case 2: \(x\in [0,1]\): LHS = 2, RHS = 2.
Case 3: \(x\in (-∞,-1) \cup (1,∞)\): LHS = 0, RHS = 0
So generally, \(f = \sum_{i=1}^n c_i\chi_{A_i}\), and this is called the canonical representation of a simple function \(f\).
Example: find the canonical representation of
\[f(x) = \left\{\begin{array}{lcr}2 &\mbox{if} & x ≤ 0 \\ 6 & \mbox{if} & x > 0 \end{array}\right.\]
\(f = 2\chi_{(-∞,0]} + 6\chi_{(0,∞)}\). So \(c_1 = 2, c_2 = 6\), \(A_1 = (-∞,0], A_2 = (0,∞)\).
\[f(x) = \left\{\begin{array}{lcr}0 &\mbox{if} & x ≤ -1 \\ 2 & \mbox{if} & -1 < x < 1 \\6 & \mbox{if} & x ≥ 1 \end{array}\right.\]
Definition: \(f\) is called a Lebesgue measurable simple function if it is simple and Lebesgue measurable. We need the simple function to have the characteristic that the preimage of any open set is Lebesgue measurable.
We can show that in the canonical representation of \(f = \sum_{i=1}^n c_i\chi_{A_i}\), that if \(A_1, ..., A_n \in \mathcal L\), then \(f\) is a Lebesgue measurable simple function.
If a simple function \(f\) is greater than or equal to zero, then
\[\int_\mathbb{R}f d\lambda = \int_\mathbb{R}f(x) d\lambda(x) = \sum_{i=1}^n a_i \lambda(A_i) \]
The Lebesgue integral is the sum of the products of the functional values multiplied by the Lebesgue measures of the sets that map to those values.
Example
\[f(x) = \left\{\begin{array}{lcr}1 & \mbox{if} & x \in \mathbb Q \\ 0 & \mbox{if} & x \notin \mathbb Q \end{array}\right.\]
Recall the definition of Lebesgue integral for a simple, nonnegative function.
Definition: Let \(S:\mathbb{R}\rightarrow \mathbb{R}\) be a Lebesgue measurable simple function such that \(S(x) ≥ 0\) for all \(x \in \mathbb{R}\). Then we can write
\[S = \sum_{k=1}^n a_k\chi_{A_k} \]
as the canonical representation of \(S\). Then
\[\int_{\mathbb{R}}S d\lambda = \int_{\mathbb{R}}S(x) d\lambda (x) = \sum_{k=1}^n a_k \lambda(A_k) \]
Let \(E \in \mathcal L\) be a Lebesgue measurable set. Then
\[\int_E S d\lambda = \int_{\mathbb{R}} \chi_E\cdot S d \lambda \]
Example: \(\int_{[0,1]}S d\lambda = \int_{\mathbb{R}}\chi_{[0,1]}\cdot S d\lambda\). This is because
\[(\chi_{[0,1]}\cdot S)(x) = \left\{\begin{array}{lcr}S(x) & \mbox{if} & x \in [0,1] \\ 0 & \mbox{if} & o/w \end{array}\right.\]
So for a general simple function,
\[\begin{align*} \int_E S d \lambda &= \int_{\mathbb{R}} \chi_E\left(\sum_{k=1}^n a_k \chi_{A_k}d\lambda\right)\\[1em] &= \int_{\mathbb{R}} \sum_{k=1^n} a_k \chi_{E \cap A_k} d \lambda \\[1em] &= \sum_{k=1}^n a_k \lambda(E \cap A_k) \end{align*}\]
Example: step function between 1 and 2: \(S = 0\chi_{(-∞,0]} + 1\chi_{(0,1)} + 2\chi_{[1,2]} + 1\chi_{(2,3)} + 2\chi_{[3,4]} + 0\chi_{(4,∞)}\). This is not the canonical representation because we have duplicate \(a\) values.
The canonical form would involve boiling it down to three terms for the three unique values of \(S(x)\).
If we want to compute the integral of this function,
\[\begin{align*} \int_{\mathbb{R}}S(x) &= 1\lambda((0,1)\cup(2,3)) + 2\lambda([1,2]\cup[3,4]) \end{align*}\]
Remark: Let \(S\) be a simple nonnegative function with disjoint subsets \(A_k\) of \(\mathbb{R}\) given by
\[S = \sum a_k\chi_{E_{A_k}} \]
where \(A_k \in \mathcal L\) and \(a_k ≥ 0\) for all \(k = 1, ..., m\), and \(A_i \cap A_j = \emptyset\) for \(i ≠ j\), \(i,j = 1, ..., m\). Then the Lebesgue integral of this function is
\[\int_{\mathbb{R}}S d\lambda = \sum_{k=1}^m a_k \lambda(A_k) \]
More generally, if \(E \in \mathcal L\), then
\[\int_E S d\lambda = \sum_{k=1}^m a_k \lambda(E \cap A_k) \]
This depends on the additivity of the Lebesuge measure on Lebesgue measurable sets.
Example: \(S = 2\chi_{(-∞,1)} + 3\chi_{(-1,∞)}\). Find \(\int_{\mathbb{R}}Sd\lambda\) and \(\int_{[-0.5, 0.5]} Sd\lambda\). Rewrite as
\[S = \left\{\begin{array}{lcr}2 & \mbox{if} & x ≤ -1 \\ 5 & \mbox{if} & -1 < x < 1 \\ 3 & \mbox{if} & x ≥ 1 \end{array}\right.\]
The first diverges so we say that it is not Lebesgue integrable, the second is 5 because of the overlapping intervals.
\[\begin{align*} \int_E S d\lambda &= \sum_{k=1}^m a_k \lambda(E \cap A_k) \\[1em] &= 2\lambda([-0.5, 0.5] \cap (-∞,-1]) + 5\lambda([-0.5, 0.5] \cap (-1,1)) + 3\lambda([-0.5, 0.5] \cap [1,∞)) \\[1em] &= 0 + 5 + 0 = 5 \end{align*}\]
Theorem: Let \(s, t: \mathbb{R}\rightarrow [0,∞]\) be Lebesgue measurable simple functions. Then
Hint: show that \(\int_{\mathbb{R}} (s + t)d\lambda = \int_{\mathbb{R}} sd\lambda + \int_{\mathbb{R}} td\lambda\). Write \(s\) and \(t\) in their canonical representations and recall that the \(a_k\) and \(b_k\) values are distinct and the \(A_k\) and \(B_k\) are disjoint and their union is the entire real line. We can’t apply our earlier lemma because the intersection of the \(A_k\) and \(B_k\) sets might not be disjoint. Try to rewrite them in a way that makes them disjoint.
\(A_k = A_k \cap \mathbb{R}= A_k \cap (\cup_{j=1}^n B_j) = \cup_{j=1}^n (A_k \cap B_j)\) because the \(B\) sets partition the real line. Then
nd \(\chi_{A\cup B} = \chi_A + \chi_B\) if \(A \cap B = \emptyset\). So
\(S = \sum_{k=1}^m a_k \chi_{A_k} = \sum_{k=1}^m a_k\sum_{j=1}^n \chi_{A_k \cap B_j} = \sum_{k=1}^m\sum_{j=1}^n a_k \chi_{A_k\cap B_j}\).
Similarly, we can write \(t = \sum_{k=1}^m \sum_{j=1}^n b_j \chi_{A_k\cap B_j}\). Combining,
\[\begin{align*} s + t &= \sum_{k=1}^m\sum_{j=1}^n(a_k + b_j)\chi_{A_k \cap B_j} \\[1em] &= \sum_{k=1}^m\sum_{j=1}^n(a_k + b_j)\lambda(A_k \cap B_j) \\[1em] &= \sum_{k=1}^m\sum_{j=1}^n(a_k)\lambda(A_k \cap B_j) + \sum_{k=1}^m\sum_{j=1}^n(b_j)\lambda(A_k \cap B_j)\\[1em] &= \int_{\mathbb{R}} sd\lambda + \int_{\mathbb{R}} t d \lambda \end{align*}\]
Theorem: Let \(f: \mathbb{R}\rightarrow [0,∞]\) be a Lebesgue measurable function. If we have such a function, we can always write it as a pointwise limit of a sequence of simple functions.
There exists a sequence of simple Lebesgue measurable functions \(0 ≤ s_1 ≤ s_2 ≤ ...\) such that \(\{s_n\}\) converges pointwise to \(f\). That is
\[\lim_{n\rightarrow ∞}s_n(x) = f(x)\hspace{5mm} \forall x \in \mathbb{R}\]
Theorem: \(\int_{\mathbb{R}} f d\lambda = \sup\limits_{0≤s≤f} \int_{\mathbb{R}}sd\lambda\).
Example: \(f_n(x) = \frac{x}{n}, x \in [0,∞)\). Does \(\{f_n\}\) converge uniformly to the zero function on \([0,∞)\)?
Prove by contradiction.
Let \(\epsilon = 1\). Then there exists \(\hat n\) such that \(|x/n| < 1\) for all \(x \in [0,∞)\) and for all \(n ≥ \hat n\). If this is true, then let \(n= \hat n\) and let \(x = \hat n + 1\). Then we get \((\hat n + 1)/\hat n < 1\), which is our contradiction.
Definition: Let \(f: \mathbb{R}\rightarrow [0,∞)\) be a Lebesgue measurable function. Then \(\int_\mathbb{R}f d\lambda = \sup_{0≤s≤f} \int_\mathbb{R}s d\lambda\) where \(s\) is simple and Lebesgue measurable. The integral of a Lebesgue measurable function is equal to the supremum of the integrals of simple functions such that the simple function is less than or equal to \(f\).
Define the set \(S_f = \{s: s:\mathbb{R}\rightarrow ∞\) is a simple Lebesgue Measurable function such that \(s ≤ f \}\). So this is the collection of all simple functions that are majorized by \(f\).
If \(E \subset \mathbb{R}\), then
\[\int_Efd\lambda = \int_\mathbb{R}\chi_Efd\lambda \]
Proposition: Let \(f,g:\mathbb{R}\rightarrow [0,∞)\) be Lebesuge measurable functions, and let \(\alpha≥ 0\).
If \(f ≤ g\), then \(\int_\mathbb{R}fd\lambda ≤ \int_\mathbb{R}gd\lambda\). And if \(E\subset \mathcal L\), then \(\int_E fd\lambda ≤ \int_E gd\lambda\).
Proof: Our set \(S_f\) is the collection of simple Lebesgue measurable functions such that elements \(s \in S_f\) are majorized by \(f\). \(S_g\) is defined similarly for \(g\).
So by the definition of the Lebesgue integral,
\[\int_\mathbb{R}fd\lambda = \sup_{s \in S_f}\int_\mathbb{R}sd\lambda ≤ \sup_{s\in S_g}\int_\mathbb{R}sd\lambda\]
If \(E \in \mathcal L\), then \(\chi_E\cdot f ≤ \chi_E\cdot g\). So \(\int_\mathbb{R}\chi_E fd\lambda ≤ \int_\mathbb{R}\chi_E gd\lambda\).
If \(A \subset E\), with \(A, E \in \mathcal L\), then \(\int_A fd\lambda ≤ \int_E fd\lambda\).
We can rewrite \(\int_A fd\lambda = \int_\mathbb{R}\chi_A fd\lambda\) and \(\int_E fd\lambda = \int_\mathbb{R}\chi_E f d\lambda\). \(\chi_Af ≤ \chi_Ef\), so the integral inequality (expressed as a supremum from earlier) follows.
If \(f(x) = 0\) for all \(x \in E\). This is because \((\chi_E \cdot f)(x) = 0\).
If \(\lambda(E) = 0\), then \(\int_E fd\lambda = 0\).
\(\int_E \alpha f d \lambda = \alpha \int_E f d \lambda\).
The Cantor function is continuous and monotone increasing, so the preimage of an open set is open. So it is in the Borel \(\sigma\) algebra, which means that it is Lebesgue measurable.
Theorem: The monotone convergence theorem. Let \(\{f_n\}\) be a sequence of nonnegative Lebesgue measurable functions defined on \(\mathbb{R}\). Suppose that \(f_1 ≤ f_2, \cdots\), and that \(\{f_n\}\) converges pointwise to \(f\). Then \(\int_E fd\lambda = \lim_{n\rightarrow ∞}\int_E f_n d\lambda\) for \(E \in \mathcal L\).
Let \((X,\mathcal S, \mu)\) be a measure space. That is \(X\) is nonempty, \(\mathcal S\) is a \(\sigma\)-algebra on \(X\), and \(\mu: \mathcal S \rightarrow [0,∞)\) where \(\mu(\emptyset) = 0\) and it satisfies additivity. The goal is to define \(\int_X f d\mu\).
Step 1: Let \(s: X \rightarrow [0,∞)\) be an \(\mathcal S\) measurable simple function, so that the preimage of any Borel set is in \(\mathcal S\). Then we can write \(s = \sum_{k=1}^n a_k \chi_{A_k}\) where \(A_k = \{x \in X: f(x) = a_k\} \in \mathcal S\) (because it is the preimage of a closed set).
Then \(\int_X s d\mu = \sum_{k=1}^n a_k\mu(A_k)\).
The integral of the function \(f\) is still defined in terms of supremum:
\[\int_X f d \mu = \sup_{s \in \mathcal S_f} \int_X s d\mu \]
where \(S_f = \{s: s: X \rightarrow [0,∞)\) is a simple \(S\) measurable fucntion with \(s ≤ f\}\).
Example: \((\mathbb Z^+, 2^{\mathbb Z^+}, \mu)\) where \(\mu\) is the counting measure, \(\mu(A)\) is the number of elements in \(A\). Find \(\int_{\{6\}}fd\mu\) where \(f: \mathbb Z^+ \rightarrow [0,∞)\)
\(\int_{\mathbb Z^+} fd\mu = \sum_{n=1}^∞ f(n) = \sum_{n=1}^∞ a_n\) where \(a_n = f(n)\)
Defining Lebesgue integration for a nonnegative measurable function.
Monotone Convergence Theorem: Let \((X, S, \mu)\) be a measure space. Then let \(f: X \rightarrow \mathbb{R}\) be a nonnegative \(S\)-measurable function. This means that the preimage of any Borel set in \(\mathbb{R}\) is in \(S\). Then define
\[\int_X fd\mu = \sup_{s\in A_f} \int_X sd\mu \]
where \(A_f\) is the set of all Lebesgue measurable simple functions that are majorized by \(f\). (\(0≤s ≤ f\) for all \(s \in A_f\).)
We already know how to compute the integral of a simple function. We represent it in its canonical form and use characteristic functions. We also know that if a function \(s≤f\), then \(\int s ≤ \int f\).
If \(E\in S\), meaning that \(E\subset X\), then we define \(\int_E fd\mu = \int_X \chi_E fd\mu\).
Theorem: Let \(\{f_n\}\) be a sequence of nonnegative simple \(S\)-measurable functions such that \(f_1 ≤ f_2 ≤ ...\) and \(\{f_n\}\) converges pointwise to \(f: X \rightarrow \mathbb{R}\). Then
\[\int_X fd\mu = \lim\limits_{n\rightarrow ∞} \int_X f_n d\mu = \int_X\left(\lim_{n\rightarrow ∞}f_n\right)d\mu\]
Sometimes the limit at a certain \(X\) is infinity. For any \(x \in X\), \(f_1(x) ≤ f_2(x) ≤ ...\). Then \(\lim_{n\rightarrow ∞}f_n(x)\) always exists in \([0,∞]\). In the theorem, we assume that the function has real values. But the limit could be infinity.
Remark: The monotone convergence theorem can also be applied when \(f: X\rightarrow [0,∞]\).
Theorem: Let \(\{f_n\}\) be a sequence of nonnegative \(S\) measurable functions defined on \(X\). Then \(\int_X\sum_{n=1}^∞ f_n d\mu = \sum_{n=1}^∞ \int_X f_nd\mu\) (hw exercise?).
What is the sum on the LHS? Let \(f = \sum_{n=1}^∞ f_n\). Then \(f:X\rightarrow [0,∞]\), and so it is also a function. It is the limit of a sequence of partial sums. If we define \(S_n(x) = \sum_{k=1}^n f_k(x)\), with \(x \in X\), then \(S_n X \rightarrow [0,∞)\) and \(f(x) = \lim_{n\rightarrow ∞}S_n(x), x\in X\).
And what is the sum on the RHS? It is again the limit of the partial sum. For each \(n\in \mathbb N\), \(a_n = \int_X f_nd\mu \in [0,∞]\). Then \(\sum_{n=1}^∞\int_X f_nd\mu = \sum_{n=1}^∞ a_n\). If we define \(T_n = \sum_{k=1}^n a_k\), with \(n \in \mathbb N\), then \(\sum_{n=1}^∞ a_n = \lim_{n\rightarrow ∞}T_n \in [0,∞]\).
Now we can prove the infinite sum version using the monotone convergence theorem.
Proof: Let \(f = \sum_{n=1}^∞ f_n\) and let \(S_n = \sum_{k=1}^n f_k\). Then \(0≤S_1≤S_2≤...\) and \(\{S_n\}\) converges pointwise to the function \(f\) on \(X\). That is
\[\lim_{n\rightarrow ∞}S_n(x) = f(x) \]
for every \(x \in X\). Then, by the monotone convergence theorem, $
\[\begin{align*} \int_X f d \mu &= \lim_{n\rightarrow ∞} \int_X S_nd\mu \\[1em] &= \lim_{n\rightarrow ∞} \int_X (\sum_{k=1}^n f_k)d\mu \\[1em] &= \lim_{n\rightarrow ∞}\sum_{k=1}^n\int_Xf_kd\mu \\[1em] &= \sum_{k=1}^n \int_X f_kd\mu \\[1em] &= \sum_{n=1}^∞ \int_X f_n d\mu \end{align*}\]
Corrolary: Let \(f:X\rightarrow \mathbb{R}\) be a nonnegative \(S\) measurable function. Let \(\{E_n\}_{n=1}^∞\) be a sequence of pairwise disjoint \(S\) measurable sets. This means that \(E_n \in S\) for every \(n\). Then
\[\int_{\cup_{n=1}^∞ E_n} fd\mu = \sum_{n=1}^∞ \int_{E_n} fd\mu \]
In particular, if \(A,B \in S\) and \(A\cap B = \emptyset\), then \(\int_{A\cup B} fd\mu = \int_A fd\mu + \int_B fd\mu\).
\(\chi_{\cup_{n=1}^∞ A_n} = \sum_{n=1}^∞\chi_{A_n}\). To prove this, take \(x\in X\). Then either \(x\) is in the union or it isn’t. If it isn’t, then LHS and RHS both equal zero. If it does, then both sides equal 1.
Then
\[\begin{align*} \int_{\cup_{n=1}^∞ E_n} fd\mu &= \int_X \chi_{\cup_{n=1}^∞ A_n}fd\mu \\[1em] &= \int_X \sum_{n=1}^∞ (\chi_{E_n}f)d\mu \\[1em] &= \sum_{n=1}^∞ \int_X \chi_{E_n}fd\mu \\[1em] = \sum_{n=1}^∞ \int_{E_n}fd\mu \end{align*}\]
Example: \((\mathbb Z^+, 2^{\mathbb Z^+}, \mu)\), where \(\mu\) is the counting measure. If \(A \subset \mathbb Z^+\), then \(A \in 2^{\mathbb Z^+}\). The counting measure \(\mu(A)\) is just the number of elements in \(A\).
Now, if we have an infinite series like \(\sum_{n=1}^∞ a_n\), this is a function \(f:\mathbb Z^+ \rightarrow \mathbb{R}\). \(1\mapsto f(1) = a_1\) and so on.
Then this function is always \(S\) measurable, because \(S\) is the power set of \(\mathbb Z^+\). We want to evaluate the claim that \(\int_{\mathbb Z^+} fd\mu = \sum_{n=1}^∞ a_n\), with \(a_n ≥ 0\).
Proof:
\[\begin{align*} \int_{\mathbb Z^+} fd\mu &= \int_{\cup_{n=1}^∞ \{n\}}fd\mu \\[1em] &= \sum_{n=1}^∞\int_{\{n\}}fd\mu \\[1em] &= \sum_{n=1}^∞ \int_{\mathbb Z^+}\chi_{\{n\}}fd\mu \\[1em] &= \sum_{n=1}^∞ \int_{\mathbb Z^+}\chi_{\{n\}}f(n)d\mu \\[1em] &= \sum_{n=1}^∞ \int_{\mathbb Z^+}\chi_{\{n\}}a_nd\mu \\[1em] &= \sum_{n=1}^∞ f(n) \mu(\{n\}) \\[1em] &= \sum_{n=1}^∞ f_n \\[1em] &= \sum_{n=1}^∞ a_n \end{align*}\]
Aside: \(\int_X \sum_{k=1}^n a_k \chi_{A_k} d\mu = \sum_{k=1}^n a_k \mu(A_k)\)
The characteristic function multiplied by \(f\) is
\[\chi_{\{n\}}f(x) = \left\{\begin{array}{lcr}0 & \mbox{if} & x ≠ n \\ f(n) & \mbox{if} & x = n \end{array}\right.\]
Corrolary: Let \(f:\mathbb{R}\rightarrow \mathbb{R}\) be a nonnegative \(\mathcal L\) measurable function. Then
\[\lim_{n\rightarrow ∞} \int_{[-n,n]} fd\lambda = \int_\mathbb{R}fd\lambda \]
Proof:
\[\begin{align*} \int_{[-n,n]} fd\lambda &= \int_\mathbb{R}\chi_{[-n,n]}fd\lambda \\[1em] \end{align*}\]
Then the limit of \(f_n(x)\) is \(f(x)\) and the sequence of functions is monotone increasing.
\(f: X \rightarrow \mathbb{R}\) is an \(S\) measurable function, and \(f ≥ 0\). Then
\[\int_X fd \mu = \sup_{s \in A_f} \int_X sd\mu \]
where \(A_f\) is the collection of nonnegative simple \(S\) measurable functions that are majorized by \(f\).
Let \((X,S,\mu)\) be a measure space. And consider a function \(f\) that is negative for some values of \(x\in X\). Define \(f^+ = \max\{f,0\}\). Then \(f^+\) is nonnegative and \(S\) measurable (the max of two \(S\) measurable functions is also \(S\) measurable).
Define \(f^- = \max\{-f,0\}\).
Proposition: let \(f:X \rightarrow \mathbb{R}\) be \(S\) measurable. Then \(f = f^+ - f^-\) and \(|f| = f^+ + f^-\).
Proof:
Case 1: If \(f ≥ 0\), then \(f^+ = f\) and \(f^- = 0\)
Case 2: If \(f < 0\), then \(f^+ = 0\) and \(f^- = -f\)
Case 3: If \(f\) is neither nonnegative nor strictly negative, then we can similarly show that the proposition is true.
Definition: Let \((X, S, \mu)\) be a measure space and let \(f: X \rightarrow \mathbb{R}\) be an \(S\) measurable function. Then define
\[\int_Xfd\mu = \int_X f^+d\mu - \int_X f^-d\mu \]
provided that the RHS does not result in something like \(∞-∞\).
Example:
\[f(x) = \left\{\begin{array}{lcr}2 & \mbox{if} & 0 < x < 1 \\ -3 & \mbox{if} & x ≥ 1 \\ 0 & \mbox{if} & x ≤ 0 \end{array}\right.\]
Then \(f^+ = 2\chi_{(0,1)}\) and \(f^- = -3\chi_{[1,∞)}\), so \(\int_\mathbb{R}f^+ = 2\lambda((0,1)) = 2\) and \(\int_\mathbb{R}f^- = 3\lambda([1,∞)) = ∞\) and \(\int_\mathbb{R}f = 2 - ∞ = -∞\).
Theorem: Dominated convergence theorem (from 3B).
Let \((X,S,\mu)\) be a measure space and let \(\{f_n\}\) be a sequence of \(S\) measurable functions that converges pointwise on \(X\) to \(f:X\rightarrow \mathbb{R}\).
Suppose that \(|f_n(x)| ≤ g(x)\) for all \(x \in X\) and all \(n \in \mathbb N\), where \(g:X \rightarrow \mathbb{R}\) is a nonnegative \(S\) measurable function such that
\[\int_X gd\lambda = < ∞ \]
Then
\[\lim_{n\rightarrow ∞} \int_X f_nd\mu = \int_X fd\mu \]
Remark: We say that \(f\) is integrable on \(X\) if \(\int_X |f|d\mu < ∞\). This means that both \(\int_Xf^+d\mu < ∞\) and \(\int_Xf^-d\mu < ∞\).
Hint: 3A.2: \(\int_Xfd\delta_c = f(c)\). Need to show that \(\delta_c\) is a measure on \(S\)?
First prove for a characteristic function \(f = \chi_E, E\in S\).
Then prove for a nonnegative simple function that is \(S\) measurable
Then prove for a nonegative \(S\) measurable function.
Final exam, all day Monday 3/15, due end of day. Problems will be similar to homework problems.
If \(\int_X |f|d\mu < \infty\), then we can denote
\[\mathcal L'(X) = \{f:X\rightarrow \mathbb{R}\} \]
Now we can prepare to prove the monotone convergence theorem.
Proposition: Let \((X,S,\mu)\) be a measure space. And suppose that \(\{E_n\}_{n=1}^∞ \subset S\), and \(E_1 \subset E_2 \subset \cdots\). Then
\[\mu(\cup_{n=1}^n E_n) = \lim_{n\rightarrow ∞}\mu(E_n) \]
Example: Using the Lebesgue interval, \(\cup_{n=1}^∞ [-n,n] = \mathbb{R}\). Based on this proposition, we will see that the limit of the measure is
\[\lim_{n\rightarrow ∞}\lambda([-n,n]) = \lambda(\mathbb{R}) \]
LHS: \(\lim_{n\rightarrow ∞}\lambda([-n,n]) = \lim_{n\rightarrow ∞} 2n = ∞\)
RHS: \(\lambda(\mathbb{R}) = ∞\).
Proof: For a general measure, we can only apply the subadditivity property (since it could be something other than the Lebesgue measure). However, we can represent the union as a union of disjoint sets. Let
\[\begin{align*} A_1 &= E_1 \\ A_2 &= E_2 \setminus E_1 \\ A_3 &= E_3 \setminus E_2 \\ \cdots \\ A_n &= E_n \setminus E_{n-1} \end{align*}\]
for all \(n \in \mathbb N\). Then \(A_n \in S\) because \(S\) is a \(\sigma\) algebra, meaning that it is closed under set operations on its members. And \(A_i \cap A_j = \emptyset\) for \(i≠j\). Also, \(\cup_{n=1}^∞ A_n = \cup_{n=1}^∞ E_n\). Then
\[\begin{align*} \mu(\cup_{n=1}^∞ E_n) &= \mu(\cup_{n=1}^∞ A_n \\[1em] &= \sum_{n=1}^∞ \mu(A_n)\\[1em] &= \lim_{n\rightarrow ∞} \sum_{k=1}^n \mu(A_k) \\[1em] &= \lim_{n\rightarrow ∞} \mu(\cup_{k=1}^n A_k) \\[1em] &= \lim_{n\rightarrow ∞} \mu(E_n) \end{align*}\]
Remark: \(\{E_n\}\subset S\) with \(E_1 \supset E_2 \supset \cdots\), and \(\mu(E_1) < ∞\), then
\[\mu(\cap_{n=1}^∞ E_n) = \lim_{n\rightarrow ∞} \mu(E_n) \]
Proposition: Let \(s: X \rightarrow \mathbb{R}\) be a nonnegative \(S\)-measurable simple function. Suppose \(\{E_n\}\subset S\) and \(E_1 \subset E_2 \subset \cdots\). Then
\[\int_{\cup_{n=1}^∞ E_n}sd\mu = \lim_{n\rightarrow ∞} \int_{E_n} sd\mu \]
Proof: Consider an obvious case, then by the linearity of the integral, we can come to the conclusion. Let \(s = \chi_A\), where \(A \in S\). Then the LHS is
\[\int_{\cup_{n=1}^∞ E_n}sd\mu = \mu(A\cap (\cup_{n=1}^∞ E_n))\]
We just need to remember that \(\chi_A\chi_B = \chi_{A\cap B}\).
So we have \((A\cap E_1) \subset (A \cap E_2) \subset \cdots\) with all being in \(S\). So the RHS is
\[= \lim_{n\rightarrow ∞} \mu(A\cap E_n) = \lim_{n\rightarrow ∞} \int_{E_n} \chi_{A} d\mu \]
Now to generalize, let \(s = \sum_{k=1}^m a_k \chi_{A_k}\) with \(A_i \cap A_j = \emptyset\) for \(i≠j\). Then
\[\begin{align*} \int_{\cup_{n=1}^∞ E_n} sd\mu &= \sum_{k=1}^m a_k \int_{\cup_{n=1}^∞ E_n} \chi_{A_k} d\mu \\[1em] &= \sum_{k=1}^m a_k \left(\lim_{n\rightarrow ∞} \int_{E_n} \chi_{A_k} d\mu\right) \\[1em] &= \lim_{n\rightarrow ∞} \int_{E_n} \left(\sum_{k=1}^m a_k \chi_{A_k}\right)d\mu \\[1em] &= \lim_{n\rightarrow ∞} \int_{E_n} sd\mu \end{align*}\]
Now we can prove the monotone convergence theorem.
Theorem: Let \(\{f_n\}_{n=1}^∞\) be a sequence of nonnegative Lebesgue measurable functions defined on \(\mathbb{R}\) such that \(0 ≤ f_1 ≤ f_2 ≤ \cdots\). Suppose that \(\{f_n\}\) converges pointwise to \(f:\mathbb{R}\rightarrow \mathbb{R}\). Then
\[\lim_{n\rightarrow ∞} \int_\mathbb{R}f_nd\lambda = \int_\mathbb{R}fd\lambda \]
More generally, if \(E \in \mathcal L\), then
\[\lim_{n\rightarrow ∞} \int_E f_nd\lambda = \int_E fd\lambda \]
When we say that \(f_n\) converges pointwise to \(f(x)\), this means that \(\lim_{n\rightarrow ∞}f_n(x) = f(x)\) for all \(x \in \mathbb{R}\).
An alternate way of writing the integral is
\[\int_\mathbb{R}fd\lambda = \sup_{s \in A_f}\int_\mathbb{R}sd\lambda \]
where \(A_f = \{s: \mathbb{R}\rightarrow \mathbb{R}:\) s is a nonnegative Lebesgue measurable simple function majorized by \(f \}\).
Proof: Observe that \(f\) is nonnegative and Lebesgue measurable. WE have that the sequence
\[\left\{\int_\mathbb{R}f_nd\lambda\right\}\]
is monotone increasing because the functions are monotone increasing. Eg \(\int_\mathbb{R}f_nd\lambda ≤ \int_\mathbb{R}f_{n+1}d\lambda\).
Then \(\lim_{n\rightarrow ∞} \int_\mathbb{R}f_n d\lambda\) exists in the extended positive real line, and may equal infinity.
And also, \(\lim_{n\rightarrow ∞} f_n(x) = f(x)\) and \(f_n(x) ≤ f(x)\) for all \(x \in \mathbb{R}\). This relies on the monotone convergence theorem from adv calculus. Generally, \(\lim_{n\rightarrow ∞} a_n = \sup\{a_n:n\in\mathbb N\}\). In our case, \(f(x)\) has to be the supremum, so \(f_n(x)\) cannot be greater than the limit.
Thus,
\[\lim_{n\rightarrow ∞} \int_\mathbb{R}f_nd\lambda ≤ \int_\mathbb{R}fd\lambda \]
Now, fix any simple function \(s \in A_f\). Then recall that \(s\) is nonnegative, Lebesgue measurable, and majorized by \(f\).
Recall a property that if \(\gamma = sup A\) and \(\gamma > \alpha \in A\). Then there must exist some element \(a \in A\) such that \(a = \alpha + \epsilon\).
let \(0 < \alpha < 1\). Then \(\alpha s(x) < f(x)\). Now define the following set. \(A_n = \{x\in\mathbb{R}: \alpha s(x) < f_n(x)\}\). Then \(A_1 \subset A_2 \subset \cdots\) because \(f_n\) is increasing. Then \(A_n \in \mathcal L\) for all \(n \in \mathbb N\), and \(\cup_{n=1}^∞ A_n = \mathbb{R}\).
Then
\[\begin{align*} \alpha \int_\mathbb{R}sd\lambda &= \alpha \lim_{n\rightarrow ∞} \int_{A_n} sd\lambda\\[1em] &= \lim_{n\rightarrow ∞} \int_{A_n} \alpha sd\lambda\\[1em] &≤ \lim_{n\rightarrow ∞} \int_{A_n} f_nd\lambda \\[1em] &≤ \lim_{n\rightarrow ∞} \int_\mathbb{R}f_nd\lambda \\[1em] &≤ \lim_{n\rightarrow ∞} = L \end{align*}\]
Now, letting \(\alpha \rightarrow 1^-\) gives us the property that
\[\begin{align*} \int_\mathbb{R}sd\lambda ≤ L \end{align*}\]
for all \(s \in A_f\). So
\[\sup_{s \in A_n}\int_\mathbb{R}s d \lambda ≤ L \hspace{1cm}\mathrm{and}\hspace{1cm} \int_\mathbb{R}f d\lambda ≤ L = \lim_{n\rightarrow ∞} \int_\mathbb{R}f_n d \lambda\]
3B.11: Consider a function \(f \in L'(X)\). Then \(\int_X |f|d\mu < \infty\). Prove that the set \(A = \{x \in X: f(x) ≠ 0 \}\) is a countable union of sets with finite \(\mu\) measure.
Show that \(A = \cup_{n=1}^∞ A_n\), where \(\mu(A_n) < ∞\) for all \(n \in \mathbb N\).
How can we show that \(A\) is \(S\) measurable. Observe that \(A = f^{-1}((-∞,0)\cup(0,∞))\). And because \((-∞,0)\cup(0,∞)\) is an open set, the preimage is in \(S\).
We could rewrite \(A = \{x \in X: |f(x)| > 0\}\). This links our condition on \(A\) to the objective integral.
We can also write \(A\) as a union of a sequence of sets: \(A = \cup_{n=1}^∞\{x \in X: |f(x)| ≥ \frac{1}{n}\} = \cup_{n=1}^∞ A_n\).
This is because if \(|f(x)| > 0\), then there exists \(\hat n \in \mathbb N\) such that \(|f(x)| ≥ 1/\hat n\)
Now we want to show that \(\mu(A_n) < ∞\).
\(a > 0: \mu(\{x\in X: |f(x)| ≥ a\}) < ∞\). Call that set \(A_a\).
Then
\[\int_X |f|d\mu ≥ \int_{A_a} |f|d\mu ≥ \int_{A_a}ad\mu = a\int_{A_a}d\mu = a\int_X \chi_{A_n}d\mu = a\mu(A_a)\]
Then \(\mu(A_a) < 1/a\int_X |f|d\mu < ∞\) and we are done.
3B.5 - Show that \(\lim_{k\rightarrow ∞} \int_{[-k,k]}fd\lambda = \int_\mathbb{R}fd\lambda\), where \(f: \mathbb{R}\rightarrow \mathbb{R}\) and \(\int_\mathbb{R}|f|d\lambda < ∞\).
\[\begin{align*} \int_{[-k,k]} fd\lambda &= \int-\mathbb{R}\chi_{[-k,k]} fd\lambda \\[1em] \end{align*}\]
Let \(f_k = \chi_{[-k,k]}f\), then \(\{f_k\}\) converges pointwise to \(f\). We can show that \(\lim_{k\rightarrow ∞}f_k(x_0) = f(x_0)\) because whenever \(\hat k < |x_0|, f_{\hat k}(x_0) = f(x_0)\). So \(|f_k(x_0) - f(x_0)| = 0\) for all \(k ≥ \hat k\).
Then we have that \(|f_k(x)| = |\chi_k(x)f(x) | ≤ |f(x)|\) for all \(x\in \mathbb{R}\). So \(|f(x)|\) plays the roll of \(g\) in the dominated convergence theorem. And because \(\int_X |f|d\lambda < ∞\), the objective statement \(\lim_{k\rightarrow ∞} \int_{[-k,k]}fd\lambda = \int_\mathbb{R}fd\lambda\) follows immediately.
Theorem: Let \(f:\mathbb{R}\rightarrow \mathbb{R}\) be a nonnegative Lebesgue measurable function. Then there exists a sequence \(\{s_n\}\) of nonnegative Lebesgue measurable simple functions such that \(s_1 ≤ s_n ≤ \cdots\), \(\{s_n\}\) converges pointwise to \(f\) (ie. \(\lim_{n\rightarrow ∞} s_n(x) = f(x)\) for all \(x \in \mathbb{R}\)).
Example: Take the function
\[f(x) = \left\{\begin{array}{lcr}0 & \mbox{if} & x < 0 \\ x & \mbox{if} & x ≥ 0 \end{array}\right.\]
Let
\[f_1(x) = \left\{\begin{array}{lcr}0 & \mbox{if} & x < 1/2 \\ 1/2 & \mbox{if} & 1> x ≥ 1/2 \\ 1 & \mbox{if} & x ≥ 1 \end{array}\right.\]
\[ f_2(x) = \left\{\begin{array}{lcr} 0 & \mbox{if} & x < 1/4 \\ 1/4 & \mbox{if} & 1/2> x ≥ 1/4 \\ 1 & \mbox{if} & x ≥ 1 \\ 2 & \mbox{if} & \end{array}\right.\]
Theorem: If \(f,g\) are nonnegative Lebesgue measurable functions on \(\mathbb{R}\), then we have two properties:
\[\int_\mathbb{R}(f+g)d\lambda = \int_\mathbb{R}fd\lambda + \int_\mathbb{R}gd\lambda \hspace{1cm} \int_E (f+g)d\lambda = \int_E fd\lambda + \int_E gd\lambda\]
Proof: (first property). Take a monotone sequence of nonnegative simple Lebesgue measurable functions \(s_n\) that converges pointwise to \(f\), and a similar sequence \(t_n\) that converges to \(g\). Then \(s_n + t_n\) is also monotone and converges to \(f+g\). Then because of the monotone property of \(s_n+t_n\), by the MCT,
\[\int_\mathbb{R}(f+g)d\lambda = \lim_{n\rightarrow ∞} \int_\mathbb{R}(s_n + t_n)d\lambda\]
We can apply the sum rule for simple functions to the RHS, so that
\[\int_\mathbb{R}(f+g)d\lambda = \lim_{n\rightarrow ∞} \left[\int_\mathbb{R}(s_nd\lambda + \int_\mathbb{R}(s_nd\lambda \right]\]
Then by the MCT,
\[\int_\mathbb{R}(f+g)d\lambda = \lim_{n\rightarrow ∞} \left[\int_\mathbb{R}(s_nd\lambda + \int_\mathbb{R}(s_nd\lambda \right] = \int_\mathbb{R}fd\lambda + \int_\mathbb{R}gd\lambda\]
Important Result: If \(f\) is a bounded function and Riemann integrable, then
\(\lambda(\{x\in [a,b]:f\) is not continuous at x \(\}) = 0\). This means that \(f\) is continuous at every point in \([a,b]\) except for a set of points whose measure is zero.
Also, \(f\) is Lebesgue measurable and the Riemann integral is equal to the Lebesgue integral.
\[\int_a^b f(d)dx = \int_{[a,b]}fd\lambda \]
Example: \(f(x) = 1\) for irrationals, zero for rationals. \(g(x) = 1\). Then \(f = g\) \(\lambda\)-almost everywhere. Any point \(x\) where \(f(x) ≠ g(x)\) has \(x \in \mathbb Q\), and \(\lambda(\mathbb Q) = 0\).
We can use a nice property that if \(\lambda(E) = 0\), then \(\int_Efd\lambda = 0\).
Example: If we want to find
\[\begin{align*} \int_0^1 \frac{1}{\sqrt{x}}dx &= \\[1em] \end{align*}\]
For 2C.12, use the dominated convergence theorem. When \(k \rightarrow ∞\), the function \(f_k\) converges pointwise to zero on \((0,1]\). And \(|f_k(x)| ≤ 1/\sqrt{x}\)