1. Preliminary
We now expand \(\mathbb{R}\) to \(\mathbb{R} \cup \{-\infty, \infty\}\). Note that we do not define \((\infty)-(\infty)\), \((-\infty)-(-\infty)\), \((\infty)+(-\infty)\), \((-\infty)+(\infty)\), \(\displaystyle \frac{a}{0}\), and \(\displaystyle \frac{\pm\infty}{0}\).
We define \(f: E \to \mathbb{R}\) as a finite function, which is different from bounded function. For example, \(f(x)=x^{-1}\), where \(x \in (0, \infty)\), is finite but not bounded.
2. Definition
Definition 2.1 Suppose \(E\) is a measurable set. \(f: E \to \mathbb{R} \cup \{-\infty, \infty\}\) is a measurable function if \[E[f>a]:=\{x \in E: f(x)>a\}\] is measurable for all \(a \in \mathbb{R}\).
\(f(x)>a \Leftrightarrow f(x) \in (a, \infty]\).
Theorem 2.1 \(f\) is a measurable function if and only if \(E[f \leq a]\) is measurable for all \(a \in \mathbb{R}\).
Theorem 2.2 \(f\) is a measurable function if and only if \(E[f \geq a]\) is measurable for all \(a \in \mathbb{R}\).
Proof. \((\Rightarrow)\) We have \[\begin{aligned} E[f \geq a]&=f^{-1}([a, \infty])=f^{-1}\left(\bigcap_{n=1}^\infty \left(a-\frac{1}{n}, \infty\right]\right) \\&=\bigcap_{n=1}^\infty f^{-1}\left(\left(a-\frac{1}{n}, \infty\right]\right)=\bigcap_{n=1}^\infty E\left[f>a-\frac{1}{n}\right]. \end{aligned}\] Since \(f\) is measurable, then \(\displaystyle E\left[f>a-\frac{1}{n}\right]\) is measurable and thus \(E[f \geq a]\) is measurable.
\((\Leftarrow)\) We have \(\displaystyle E[f>a]=\bigcup_{n=1}^\infty E\left[f \geq a+\frac{1}{n}\right]\), and thus \(f\) is measurable.\(\square\)
Theorem 2.3 \(f\) is a measurable function if and only if \(E[f<a]\) is measurable for all \(a \in \mathbb{R}\).
Theorem 2.4 \(f\) is a measurable function if and only if when \(|f(x)|<\infty\), \(E[a \leq f<b]\) is measurable for all \(a, b \in \mathbb{R}\) (\(a<b\)).
Proof. \((\Rightarrow)\) Since \(f\) is measurable function and \[E[a \leq f<b]=E[f<b]-E[f<a],\] then \(E[a \leq f<b]\)is measurable. Note that we do not need \(|f(x)|<\infty\) here.
\((\Leftarrow)\) Suppose \(|f(x)|<\infty\) and \(E[a \leq f<b]\) is measurable. Since \[E[f \geq a]=E[f \in [a, \infty)]=\bigcup_{n=1}^\infty E[a \leq f<a+n]\] and \(E[a \leq f<a+n]\) is measurable, then \(E[f \geq a]\) is measurable, and thus \(f\) is measurable.
Corollary 2.5 If \(f: E \to \mathbb{R} \cup \{-\infty, \infty\}\) is measurable function, then \(E[f=a]\) is measurable.
Proof. When \(a \in \mathbb{R}\), we have \[E[f=a]=E[f \geq a]-E[f>a].\] Since \(E[f \geq a]\) and \(E[f>a]\) are measurable, then \(E[f=a]\) is measurable.
When \(a=\infty\), we have \[E[f=\infty]=\bigcap_{n=1}^\infty E[n<f \leq \infty].\] Since \(E[n<f \leq \infty]\) is measurable, then \(E[f=\infty]\) is measurable.
When \(a=-\infty\), we have \[E[f=-\infty]=\bigcap_{n=1}^\infty E[-\infty \leq f<-n].\] Since \(E[-\infty \leq f<-n]\) is measurable, then \(E[f=-\infty]\) is measurable.\(\square\)
3. Operation
Theorem 3.1 Suppose \(E\) is measurable, and \(g: E \to \mathbb{R} \cup \{-\infty, \infty\}\) is a constant function, then \(g\) is a measurable function.
Proof. For all \(a \in \mathbb{R}\), we have \[E[f>a]=\begin{cases} E, &a<c \\ \varnothing, &a \geq c \end{cases},\] which is measurable, then \(g\) is a measurable function.\(\square\)
Theorem 3.2 Suppose \(f\) and \(g\) are constant functions, then \(f \pm g\), \(f \cdot g\), and \(\displaystyle \frac{f}{g}\) are measurable functions.
Theorem 3.3 Suppose \(f\) is a measurable function, and \(g=c\) is a constant function, then \(f \pm g\), \(f \cdot g\), and \(\displaystyle \frac{f}{g}\) are measurable functions.
Proof. For all \(a \in \mathbb{R}\), we have \[E[f+g>a]=E[f+c>a]=E[f>a-c],\] which is measurable, then \(f+c\) is a measurable function. Besides, \[E[f \cdot c>a]=\begin{cases} \displaystyle E\left[f>\frac{a}{c}\right], &c>0 \\ \displaystyle E\left[f<\frac{a}{c}\right], &c<0 \\ \end{cases},\] which is measurable, and when \(c=0\), \(f \cdot c=0\) is measurable. Therefore, \(f \cdot c\) is a measurable function.
As a consequence, \(f-g\) and \(\displaystyle \frac{f}{g}\) are measurable functions.\(\square\)
Theorem 3.4 If \(f\) and \(g\) are measurable functions on \(E\), then \(E[f>g]\) and \(E[f \geq g]\) are measurable.
Proof. Take an arbitrary \(x \in E[f>g]\). There exists an \(r \in \mathbb{Q}\) such that \(f(x)>r>g(x)\), i.e., \[x \in E[f>r] \cap E[g<r].\] Hence, \[E[f>g] \subset \bigcup_{r \in \mathbb{Q}} (E[f>r] \cap E[g<r]).\] For any \(r \in \mathbb{Q}\), take an arbitrary \(x \in E[f>r] \cap E[g<r]\), \(f(x)>g(x)\), i.e., \(x \in E[f>g]\). Hence, \[E[f>g] \supset \bigcup_{r \in \mathbb{Q}} (E[f>r] \cap E[g<r]).\] Therefore, \[E[f>g]=\bigcup_{r \in \mathbb{Q}} (E[f>r] \cap E[g<r]).\] Since \(f\) and \(g\) are measurable functions, then \(E[f>r]\) and \(E[g<r]\) are measurable, and thus \[\bigcup_{r \in \mathbb{Q}} (E[f>r] \cap E[g<r])\] is measurable. Therefore, \(E[f>g]\) is measurable.
In addition, since \(E[f \geq g]=(E[g>f])^c\), then \(E[f \geq g]\) is measurable.\(\square\)
Theorem 3.5 If \(f\) and \(g\) are measurable functions, then \(f \pm g\) are measurable functions.
Proof. We have \(E[f+g>a]=E[f>a-g]\). Since \(a-g\) is a measurable function, then \(E[f>a-g]\) is measurable, i.e., \(E[f+g>a]\) is measurable. Similarly, \(E[f-g>a]\) is measurable.\(\square\)
Theorem 3.6 If \(f\) is a measurable function, then \(f^2\) is a measurable function.
Proof. We have \[E[f^2>a]=\begin{cases} E, &a<0 \\ E[f>\sqrt{a}] \cup E[f<-\sqrt{a}], & a \geq 0 \end{cases}.\] Therefore, \(f^2\) is a measurable function.\(\square\)
Theorem 3.7 If \(f\) and \(g\) are measurable functions, then \(f \cdot g\) is a measurable function.
Proof. Since \[f \cdot g=\frac{1}{4}((f+g)^2-(f-g)^2),\] then it is obvious that \(f \cdot g\) is a measurable function.
Theorem 3.8 If \(f\) and \(g\) are measurable functions, then \(\displaystyle \frac{1}{g}\) and \(\displaystyle \frac{f}{g}\) are measurable functions.
Proof. We have \[E\left[\frac{1}{g}>a\right]=\begin{cases} \displaystyle E[g>0] \cap E\left[g<\frac{1}{a}\right], &a>0 \\ E[g>0] \cap E[g \neq \infty], &a=0 \\ \displaystyle E[g>0] \cup E\left[g<\frac{1}{a}\right], &a<0 \end{cases},\] i.e., \(\displaystyle E\left[\frac{1}{g}>a\right]\) is measurable, and thus \(\displaystyle \frac{1}{g}\) and \(\displaystyle \frac{f}{g}\) are measurable functions.\(\square\)
Theorem 3.9 If \(f\) is a measurable function, then \(|f|\) is a measurable function.
Theorem 3.10 Suppose \(\{f_n\}\) includes at most countable measurable functions on \(E\). Let \[\mu=\inf_n f_n, \lambda=\sup_n f_n.\] Then \(\mu\) and \(\lambda\) are measurable on \(E\).
Proof. For all \(a \in \mathbb{R}\), we have \[E[\mu \geq a]=\bigcap_n E[f_n \geq a]\] and \[E[\lambda \leq a]=\bigcap_n E[f_n \leq a].\] Therefore, \(\mu, \lambda\) are measurable functions.\(\square\)
关于\(\displaystyle E[\mu \geq a]=\bigcap_n E[f_n \geq a]\)的证明如下.
任取\(x \in E[\mu \geq a]\)，我们知道\(E[\mu \geq a]=\{x \in E: \inf_n f_n(x) \geq a\}\). 因此对于所有的\(n \in \mathbb{N}\)，我们有\(f_n(x) \geq a\). 若不然，存在\(n \in \mathbb{N}\)使得\(f_n(x)<a\)，与下确界定义矛盾. 因此\(\displaystyle x \in \bigcap_n E[f_n \geq a]\)，也即\(\displaystyle E[\mu \geq a] \subset \bigcap_n E[f_n \geq a]\).
反之，任取\(\displaystyle x \in \bigcap_n E[f_n \geq a]\)，则对于所有的\(n \in \mathbb{N}\)，我们有\(f_n(x) \geq a\). 因此\(\inf_n f_n(x) \geq a\)，即\(x \in E[\mu \geq a]\). 所以，\(\displaystyle E[\mu \geq a] \supset \bigcap_n E[f_n \geq a]\).
所以，\(\displaystyle E[\mu \geq a]=\bigcap_n E[f_n \geq a]\). 同理我们可证\(\displaystyle E[\lambda \leq a]=\bigcap_n E[f_n \leq a]\).
连续函数不具备定理3.10的性质. 假设\[f_n(x)=\begin{cases}0, &x \leq 0 \\nx, &\displaystyle 0<x \leq \frac{1}{n} \\1, &\displaystyle x>\frac{1}{n}\end{cases}.\] 显然，\(f_n\)在\(\mathbb{R}\)上是连续的. 不难发现，\[\lambda(x)=\sup_n f_n(x)=\begin{cases}0, &x \leq 0 \\1, &x>0\end{cases}\]并非连续函数. 因此至多可数个连续函数的上下确界函数未必连续.
Theorem 3.11 Suppose \(\{f_n\}\) includes at most countable measurable functions on \(E\). Then \(\displaystyle \varliminf_{n \to \infty} f_n\) and \(\displaystyle \varlimsup_{n \to \infty} f_n\) are measurable. Specifically, if \(\displaystyle \lim_{n \to \infty} f_n\) exists, then \(\displaystyle \lim_{n \to \infty} f_n\) is measurable.
Theorem 3.12 We define \(f^+=\max\{f, 0\}\) and \(f^-=-\min\{f, 0\}\). If \(f\) is measurable, then \(f^+\) and \(f^-\) are measurable. Besides, \(f=f^+-f^-\) and \(|f|=f^++f^-\).
4. Example
Example 4.1 Suppose \(E \subset \mathbb{R}^n\) is measurable, and \(f: E \to \mathbb{R}\) is a constant function, then \(f\) is a measurable function.
Example 4.2 Suppose \(f: E \to \mathbb{R}\) is a continuous function, then \(f\) is a measurable function.
Proof. For all \(a \in \mathbb{R}\), we have \[E[f>a]=\{x \in E: f(x)>a\}=\{x \in E: f^{-1}((a, \infty])=U \cap E,\] where \(U \subset \mathbb{R}^n\) is an open set. Therefore, \(E[f>a]\) is measurable, and thus \(f\) is a measurable function.\(\square\)
Example 4.3 Suppose \(f: [a, b] \to \mathbb{R}\) is a monotonic function, then \(f\) is a measurable function.
Proof. Assume without loss of generality that \(f\) is an increasing function.
For all \(c \in \mathbb{R}\), assume \(E[f \geq c] \neq \varnothing\). Let \(x_0=\inf E[f \geq c]\). Take an arbitrary \(x>x_0\), if \(f(x)<c\), then for all \(z \in E[f \geq c]\), \(z>x\), i.e., \(x\) is a lower bound, then \(x_0 \geq x\), which is a contradiction. Hence, \(f(x) \geq c\). Take an arbitrary \(y<x_0\). Since \(x_0\) is an infimum, then \(y \notin E[f \geq c]\).
Therefore, when \(x_0 \in E[f \geq c]\), \(E[f \geq c]=[x_0, b]\); when \(x_0 \notin E[f \geq c]\), \(E[f \geq c]=(x_0, b]\). Therefore, \(E[f \geq c]\) is measurable, and thus \(f\) is a measurable function.\(\square\)
Example 4.4 We divide \(E\) into a finite number of disjoint measurable subsets, \(E_1, \ldots, E_s\). A simple function \(f: E \to \mathbb{R} \cup \{-\infty, \infty\}\) is a constant function on \(E_i\). Then \(f\) is a measurable function.
Lemma 4.4.1 Suppose \(f\) is a measurable function on a measurable set \(E\), and \(E'\) is a measurable subset of \(E\), then \(f|_{E'}\) is measurable on \(E'\).
Proof. For all \(a \in \mathbb{R}\), \(E'[f>a]=E' \cap E[f>a]\), which is measurable. Hence, \(f|_{E'}\) is measurable on \(E'\).\(\square\)
Lemma 4.4.2 Suppose \(f\) is a function on \(\displaystyle E=\bigcup_{i=1}^s E_i\), and \(f\) is measurable on \(E_i\), then \(f\) is measurable on \(E\).
Proof. For all \(a \in \mathbb{R}\), \[E[f>a]=\bigcup_{i=1}^s E_i[f>a],\] which is measurable, then \(f\) is measurable on \(E\).\(\square\)
\(\square\)
Example 4.5 Dirichlet function on \([0, 1]\) is a simple function.
5. Approximating Measurable Function with Simple Functions
Theorem 5.1 Suppose \(E \subset \mathbb{R}^n\) is measurable. If \(f\) is non-negative and measurable on \(E\), then there exists a sequence of non-negative and simple functions \(\{\varphi_k\}\) such that for all \(x \in E\), \(\varphi_k(x) \leq \varphi_{k+1}(x)\), and \(\displaystyle \lim_{k \to \infty} \varphi_k(x)=f(x)\).
Proof. Let \[E_{k, j}=E\left[\frac{j-1}{2^k} \leq f<\frac{j}{2^k}\right], j=1, \ldots, k \cdot 2^k,\] and \(E_k=E[f \geq k]\). Note that \(E_{k, j}\) and \(E_k\) are measurable.
We define \[\varphi_k(x)=\begin{cases} \displaystyle \frac{j-1}{2^k}, &x \in E_{k, j} \\ k, &x \in E_k \end{cases},\] which is a simple function and satisfies \(\varphi_k(x) \leq \varphi_{k+1}(x) \leq f(x)\).
When \(f(x) \neq \infty\), take an arbitrary \(\varepsilon>0\), there exists a \(k\) such that \(f(x)<k\) and \(\displaystyle \frac{1}{2^k}<\varepsilon\). Hence, \(x\) must lie in one of \(E_{k, j}\), and thus \[0 \leq f(x)-\varphi_k(x) \leq \frac{1}{2^k}.\] For all \(n \geq k\), \[0 \leq f(x)-\varphi_n(x) \leq f(x)-\varphi_k(x) \leq \frac{1}{2^k}<\varepsilon.\] Therefore, \(|f(x)-\varphi_n(x)|<\varepsilon\), i.e., \(\displaystyle \lim_{k \to \infty} \varphi_k(x)=f(x)\).
When \(f(x)=\infty\), for all \(k \in \mathbb{Z}^+\), \(\varphi_k(x)=k\). Therefore, \(\displaystyle \infty=f(x)=\lim_{k \to \infty} \varphi_k(x)\).\(\square\)
Theorem 5.2 If \(f\) is measurable on \(E\), then there exists a sequence of simple functions \(\{\varphi_k\}\) such that for all \(x \in E\), \(\displaystyle \lim_{k \to \infty} \varphi_k(x)=f(x)\). Moreover, if \(f\) is bounded, then \(\{\varphi_k\}\) is uniformly convergent.
Proof. We can write \(f=f^+-f^-\). Because of theorem 5.1, there exist sequences of simple functions \(\{\varphi_k^+\}\) and \(\{\varphi_k^-\}\) such that \(\displaystyle \lim_{k \to \infty} \varphi_k^+(x)=f^+(x)\) and \(\displaystyle \lim_{k \to \infty} \varphi_k^-(x)=f^-(x)\).
Since \(\varphi_k=\varphi_k^+-\varphi_k^-\) is a simple function, then \(\displaystyle \lim_{k \to \infty} \varphi_k(x)=f(x)\).
When \(f\) is bounded, there exists an \(M>0\) such that \(|f(x)| \leq M\). Take an arbitrary \(\varepsilon>0\), there exists a \(k>M\) such that \(\displaystyle \frac{1}{2^{k-1}}<\varepsilon\) and \(E_k=\varnothing\). For all \(x \in E\), \(x\) must lie in one of \(E_{k, j}\).
We have \[0 \leq f^+(x)-\varphi_k^+(x) \leq \frac{1}{2^k}\] and \[0 \leq f^-(x)-\varphi_k^-(x) \leq \frac{1}{2^k}.\] For all \(n \geq k\), \[0 \leq f^+(x)-\varphi_n^+(x) \leq \frac{1}{2^k}\] and \[0 \leq f^-(x)-\varphi_n^-(x) \leq \frac{1}{2^k}.\] Therefore, \[|f(x)-\varphi_n(x)| \leq |f^+(x)-\varphi_n^+(x)|+|f^-(x)-\varphi_n^-(x)| \leq \frac{1}{2^{k-1}}<\varepsilon.\] Therefore, \(\{\varphi_k\}\) is uniformly convergent.\(\square\)
Recall 5.1 For each \(k \in \mathbb{N}\), let \(f_k\) be a function defined on \(E\). The sequence of functions \(\{f_k\}\) converges pointwise on \(E\) to a function \(f\) defined on \(E\) if \[\forall x \in E, \forall \varepsilon>0, \exists N \in \mathbb{N}\ \text{s.t.}\ \forall k \geq N, |f_k(x)-f(x)|<\varepsilon.\] Note that \(N(x, \varepsilon)\) is related to \(X\) and \(\varepsilon\), i.e., for different \(x\), the sequence of functions converges in different speeds.
Recall 5.2 For each \(k \in \mathbb{N}\), let \(f_k\) be a function defined on \(E\). The sequence of functions \(\{f_k\}\) converges uniformly on \(E\) to a function \(f\) defined on \(E\) if \[\forall \varepsilon>0, \exists N \in \mathbb{N}\ \text{s.t.}\ \forall x \in E, \forall k \geq N, |f_k(x)-f(x)|<\varepsilon.\] Note that \(N(\varepsilon)\) is only related to \(\varepsilon\). An equivalent definition is \[\lim_{k \to \infty} \sup_{x \in E} |f_k(x)-f(x)|=0.\]
6. Convergence
6.1. Egorov's Theorem
Definition 6.1 If a property holds after removing a zero measure set, we say that the property holds almost everywhere (a.e.).
Example 6.1 Dirichlet function \(D(x)=0\ \text{a.e.}\ x \in [0, 1]\).
Example 6.2 \(|\tan x|<\infty\ \text{a.e.}\ x \in \mathbb{R}\).
Example 6.3 If \(f(x)=g(x)\ \text{a.e.}\ x \in E\), \(g(x)=h(x)\ \text{a.e.}\ x \in E\), then \(f(x)=h(x)\ \text{a.e.}\ x \in E\).
Theorem 6.1 (Egorov's Theorem) Suppose \(f_k\) and \(f\) are measurable functions defined on \(E\), where \(f_k\) and \(f\) are finite almost everywhere, and \(m(E)<\infty\). If \(f_k(x) \to f(x)\ \text{a.e.}\ x \in E\), when \(k \to \infty\), then for all \(\delta>0\), there exists measurable subset \(E_\delta \subset E\), where \(m(E_\delta) \leq \delta\), such that \(\{f_k\}\) converges uniformly on \(E-E_\delta\) to \(f\).
Lemma 6.1.1 Suppose \(f_k\) and \(f\) are measurable functions defined on \(E\), where \(f_k\) and \(f\) are finite almost everywhere, and \(m(E)<\infty\). If \(f_k(x) \to f(x)\ \text{a.e.}\ x \in E\), when \(k \to \infty\), then for all \(\varepsilon>0\), we have \[\lim_{N \to \infty} m\left(\bigcup_{k=N}^\infty E_k(\varepsilon)\right)=0,\] where \(E_k(\varepsilon)=\{x \in E: |f_k(x)-f(x)| \geq \varepsilon\}\).
Proof. Since \(f_k(x) \to f(x)\ \text{a.e.}\ x \in E\), then the set of all non-convergent points has measure zero. We can write the set as \[\begin{aligned}&\ \ \ \ \ \{x \in E: \exists \varepsilon>0\ \text{s.t.}\ \forall N \in \mathbb{N}, \exists k \geq N\ \text{s.t.}\ |f_k(x)-f(x)| \geq \varepsilon\}\\&=\bigcup_{\varepsilon>0}\bigcap_{N \geq 1}\bigcup_{k \geq N} \{x \in E: |f_k(x)-f(x)| \geq \varepsilon\}\\&=\bigcup_{\varepsilon>0}\bigcap_{N \geq 1}\bigcup_{k \geq N} E_k(\varepsilon)\\&=\bigcup_{\varepsilon>0}\lim_{N \to \infty}\bigcup_{k \geq N} E_k(\varepsilon).\end{aligned}\] Therefore, all points in \(\displaystyle \bigcap_{N \geq 1}\bigcup_{k \geq N} E_k(\varepsilon)\) are not convergent, for any \(\varepsilon>0\), and thus \[m\left(\lim_{N \to \infty}\bigcup_{k \geq N} E_k(\varepsilon)\right)=0.\]\(\square\)
Proof. Since \(\displaystyle \bigcup_{k \geq N} E_k(\varepsilon)\) is decreasing and \(m(E)<\infty\), then we have \[\lim_{N \to \infty} m\left(\bigcup_{k=N}^\infty E_k(\varepsilon)\right)=m\left(\lim_{N \to \infty}\bigcup_{k=N}^\infty E_k(\varepsilon)\right)=0.\]
Assume without loss of generality that \(f_k\) and \(f\) are finite on \(E\).
By lemma 6.1.1, we know \[\forall \varepsilon>0, \forall \widetilde{\varepsilon}>0, \exists N(\varepsilon, \widetilde{\varepsilon}) \in \mathbb{N}\ \text{s.t.}\ \forall N \geq N(\varepsilon, \widetilde{\varepsilon}), m\left(\bigcup_{k=N}^\infty E_k(\varepsilon)\right)<\widetilde{\varepsilon}.\]
Let \(\displaystyle \varepsilon=\frac{1}{i}\) and \(\displaystyle \widetilde{\varepsilon}=\frac{\delta}{2^i}\), \(i \in \mathbb{N}\). Therefore, \[\forall \delta>0, \forall i \in \mathbb{N}, \exists N(\delta, i) \in \mathbb{N}\ \text{s.t.}\ m\left(\bigcup_{k \geq N(\delta, i)} E_k\left(\frac{1}{i}\right)\right)<\frac{\delta}{2^i}.\] Take \(\displaystyle E_\delta=\bigcup_{i \geq 1}\bigcup_{k \geq N(\delta, i)}E_k\left(\frac{1}{i}\right)\), then we have \[m(E_\delta) \leq \sum_{i \geq 1} m\left(\bigcup_{k \geq N(\delta, i)} E_k\left(\frac{1}{i}\right)\right)=\delta.\] We also have \[E-E_\delta=\bigcap_{i \geq 1}\bigcap_{k \geq N(\delta, i)} E_k\left(\frac{1}{i}\right)^c.\]
Note that for all \(\varepsilon>0\), we can find an \(i\) such that \(\displaystyle \frac{1}{i}<\varepsilon\). Therefore, \[\forall \delta>0, \forall \varepsilon>0, \exists N(\delta, i) \in \mathbb{N}\ \text{s.t.}\ \forall x \in E-E_\delta, \forall k \geq N(\delta, i), |f_k(x)-f(x)|<\frac{1}{i}<\varepsilon.\] Hence, \(\{f_k\}\) converges uniformly on \(E-E_\delta\) to \(f\).\(\square\)
假设\(f_k\)在\(E-N_k\)上是有限函数，其中\(N_k\)为零测集；\(f\)在\(E-N\)上是有限函数，其中\(N\)为零测集. 因此\(\{f_k\}\)和\(f\)在\(\displaystyle E-\left(\bigcup_{k=1}^\infty N_k \cup N\right)\)上是有限的，其中\(\displaystyle \left(\bigcup_{k=1}^\infty N_k \cup N\right)\)是零测集.
如果我们能在\(\displaystyle E-\left(\bigcup_{k=1}^\infty N_k \cup N\right)\)上找到\(E_\delta\)使命题成立，那么自然在\(\displaystyle E_\delta \cup \left(\bigcup_{k=1}^\infty N_k \cup N\right)\)上我们可以找到\(E_\delta\)使命题成立. 因此，在证明过程中，我们不妨假设\(f_k\)和\(f\)在\(E\)上为有限函数.
If \(m(E)=\infty\), then the Egorov's theorem does not hold.
6.2. Convergence in Measure
Definition 6.2 Suppose \(f_n\) and \(f\) are measurable functions defined on \(E\), where \(f_n\) and \(f\) are finite almost everywhere. If \[\forall \sigma>0, \lim_{n \to \infty} m(E[|f-f_n| \geq \sigma])=0,\] then \(\{f_n\}\) converges in measure to \(f\), denoted as \(f_n \Rightarrow f\).
Theorem 6.2 If \(f_n \Rightarrow f\), \(f_n \Rightarrow g\), then \(f(x)=g(x)\ \text{a.e.}\ x \in E\).
Proof. We have \(|f(x)-g(x)| \leq |f(x)-f_k(x)|+|g(x)-f_k(x)|\). If \(\displaystyle |f-f_k|<\frac{1}{2n}\) and \(\displaystyle |g-f_k|<\frac{1}{2n}\), then \(\displaystyle |f-g|<\frac{1}{n}\), i.e., \[E\left[|f-g|<\frac{1}{n}\right] \supset E\left[|f-f_k|<\frac{1}{2n}\right] \cap E\left[|g-f_k|<\frac{1}{2n}\right],\] or equivalently, \[E\left[|f-g| \geq \frac{1}{n}\right] \subset E\left[|f-f_k| \geq \frac{1}{2n}\right] \cup E\left[|g-f_k| \geq \frac{1}{2n}\right].\] Therefore, \[m\left(E\left[|f-g| \geq \frac{1}{n}\right]\right) \leq m\left(E\left[|f-f_k| \geq \frac{1}{2n}\right]\right)+m\left(E\left[|g-f_k| \geq \frac{1}{2n}\right]\right).\] Let \(k \to \infty\), then \(\displaystyle m\left(E\left[|f-g| \geq \frac{1}{n}\right]\right)=0\).
Since \(\displaystyle E[f \neq g]=\bigcup_{n=1}^\infty E\left[|f-g| \geq \frac{1}{n}\right]\), then \(m(E[f \neq g])=0\), i.e., \(f(x)=g(x)\ \text{a.e.}\ x \in E\).\(\square\)
Convergence in measure cannot derive almost everywhere convergence; almost everywhere convergence cannot derive convergence in measure.
Theorem 6.3 (Lebesgue's Theorem) Suppose \(f_n\) and \(f\) are measurable functions defined on \(E\), where \(f_n\) and \(f\) are finite almost everywhere, and \(m(E)<\infty\). If \(f_n(x) \to f(x)\ \text{a.e.}\ x \in E\), when \(n \to \infty\), then \(f_n \Rightarrow f\).
Proof. By lemma 6.1.1, we know \[\lim_{n \to \infty} m\left(\bigcup_{k \geq n} E[|f_k-f| \geq \varepsilon]\right)=0.\] Since \(\displaystyle E[|f_n-f| \geq \varepsilon] \subset \bigcup_{k \geq n} E[|f_k-f| \geq \varepsilon]\), then \[m(E[|f_n-f| \geq \varepsilon]) \leq m\left(\bigcup_{k \geq n} E[|f_k-f| \geq \varepsilon]\right).\] Take \(n \to \infty\), then \[\lim_{n \to \infty} m(E[|f_n-f] \geq \varepsilon])=0,\] i.e., \(f_n \Rightarrow f\).\(\square\)
Theorem 6.4 (Riesz's Theorem) Suppose \(\{f_n\}\) converges in measure to \(f\) on \(E\), then there exists a subsequence \(\{f_{n_s}\}\) such that \(f_{n_s}(x) \to f(x)\ \text{a.e.}\ x \in E\).
Proof. Since \(f_n \Rightarrow f\), then \[\exists N \geq 1\ \text{s.t.}\ \forall n_s \geq n \geq N, m\left(E\left[|f_{n_s}-f| \geq \frac{1}{s}\right]\right)<\frac{1}{2^s}.\]
Take an arbitrary \(\displaystyle x \in \bigcup_{N \geq 1}\bigcap_{s \geq N} E\left[|f_{n_s}-f|<\frac{1}{s}\right]\), then \[\exists N \geq 1\ \text{s.t.}\ \forall s \geq N, |f_{n_s}(x)-f(x)|<\frac{1}{s}.\] Since for any \(\varepsilon>0\), there exists an \(N \geq 1\) such that \(\displaystyle \varepsilon>\frac{1}{N}\). Hence, \[\forall \varepsilon>0, \exists N \geq 1\ \text{s.t.}\ \forall s \geq N, |f_{n_s}(x)-f(x)|<\frac{1}{s} \leq \frac{1}{N}<\varepsilon,\] i.e., \(f_{n_s}(x) \to f(x)\) as \(s \to \infty\) when \(\displaystyle x \in \bigcup_{N \geq 1}\bigcap_{s \geq N} E\left[|f_{n_s}-f|<\frac{1}{s}\right]\).
Since for all \(n \geq 1\), \(\displaystyle \bigcap_{N \geq 1}\bigcup_{s \geq N} E\left[|f_{n_s}-f| \geq \frac{1}{s}\right] \subset \bigcup_{s \geq n} E\left[|f_{n_s}-f| \geq \frac{1}{s}\right]\), then \[\begin{aligned} m\left(\bigcap_{N \geq 1}\bigcup_{s \geq N} E\left[|f_{n_s}-f| \geq \frac{1}{s}\right]\right)&\leq m\left(\bigcup_{s \geq n} E\left[|f_{n_s}-f| \geq \frac{1}{s}\right]\right) \\&\leq \sum_{s \geq n} m\left(E\left[|f_{n_s}-f| \geq \frac{1}{s}\right]\right) \\&< \sum_{s \geq n} \frac{1}{2^s}=\frac{1}{2^{n-1}}. \end{aligned}\] Hence, \[m\left(\bigcap_{N \geq 1}\bigcup_{s \geq N} E\left[|f_{n_s}-f| \geq \frac{1}{s}\right]\right)=0.\] Therefore, \(f_{n_s}(x) \to f(x)\ \text{a.e.}\ x \in E\).\(\square\)
7. Measurable Function and Continuous Function
从拓扑的角度看，一个函数是可测的如果这一函数能使Borel集的原像是可测集，一个函数是连续的如果这一函数能使开集的原像是开集. 因此不难看到，对于一个定义在可测集上的连续函数一定是可测的. 然而，一个可测函数不一定是连续函数. 因此，我们想探究这两类函数之间的区别.
Theorem 7.1 (Lusin's Theorem) Suppose \(f\) is a measurable function defined on \(E\), where \(f\) is finite almost everywhere, then for all \(\delta>0\), there exists a closed set \(F_\delta \subset E\) such that \(m(E-F_\delta)<\delta\), and \(f\) is continuous on \(F_\delta\).
Proof. First, we consider simple function. Suppose \(\displaystyle E=\bigcup_{i=1}^s E_i\), where \(E_i\) is measurable, and \(f(x)=c_i\), for all \(x \in E_i\). For all \(\delta>0\), there exists a closed set \(F_i \subset E_i\) such that \(\displaystyle m(E-F_i)<\frac{\delta}{s}\). Let \(\displaystyle F_\delta=\bigcup_{i=1}^s F_i\). We know \(F_\delta\) is closed and \(\displaystyle m(E-F_\delta)<\delta\). Consider \(f\) defined on \(F_\delta\). For any closed set \(F\) on image, \[f^{-1}(F)=f^{-1}(F) \cap F_\delta=f^{-1}(F) \cap \bigcup_{i=1}^s F_i=\bigcup_{i=1}^s (f^{-1}(F) \cap F_i),\] which is closed. Therefore, \(f\) is continuous on \(F_\delta\).
Then we consider bounded and measurable function. There exists a sequence of simple functions \(\{\varphi_k\}\) such that \(\{\varphi_k\}\) is uniformly convergent. For all \(\delta>0\), there exists a closed set \(F_k \subset E\) such that \(\displaystyle m(E-F_k)<\frac{\delta}{2^k}\) and \(\varphi_k\) is continuous on \(F_k\). Let \(\displaystyle F_\delta=\bigcap_{k=1}^\infty F_k\), then \(\varphi_k\) is continuous on \(F_\delta\). We know \[\begin{aligned} m(E-F_\delta)&=m\left(E-\bigcap_{k=1}^\infty F_k\right) \\&=m\left(\bigcup_{k=1}^\infty (E-F_k)\right) \\&\leq \sum_{k=1}^\infty m(E-F_k)<\delta. \end{aligned}\] Since \(\varphi_k\) is continuous on \(F_\delta\), and \(\{\varphi_k\}\) is uniformly convergent to \(f\), then \(f\) is continuous on \(F_\delta\).
Finally, we consider measurable function. Since \(m(\{x: |f(x)|=\infty\})=0\), then we can assume without loss of generality that \(f\) is finite on \(E\). Let \[g(x)=\frac{f(x)}{1+|f(x)|}, x \in E.\] We know \(g\) is bounded and measurable on \(E\). Therefore, there exists a closed set \(F_\delta \subset E\) such that \(m(E-F_\delta)<\delta\) and \(g\) is continuous on \(F_\delta\). Hence, \[f(x)=\frac{g(x)}{1-|g(x)|}, |g(x)|<1\] is continuous on \(E\).\(\square\)