6 Convergence of Measurable Functions

Egorov's Theorem

Suppose $(X,\mathcal{A},\mu )$ is a measure space with $\mu (X)<\infty$. Suppose $f_1,f_2,\dots$ is a sequence of measurable functions from $X$ to $\mathbb{R}$ that converges pointwise on $X$ to a measurable function $f:X\to \mathbb{R}$. Then for every $ϵ>0$, there exists a set $E\in \mathcal{A}$ such that $\mu (X-E)<ϵ$ and $f_n\to f$ uniformly on $E$.

\begin{proof} For any $n\in {\mathbb{N}}_{+}$, we have that

$${\cup }_{m=1}^{\infty }{\cap }_{k=m}^{\infty }\{x:|f_k(x)-f(x)|<\frac{1}{n}\}=X.$$

Define $A_{m,n}={\cap }_{k=m}^{\infty }\{x:|f_k(x)-f(x)|<\frac{1}{n}\}$, then $A_{m,n}\uparrow X$ as $m\to \infty$. For any $ϵ>0$, there exists $m_n$ such that $\mu (A_{m_n,n})>\mu (X)-ϵ/2^n$. Let $X_0={\cap }_{n=1}^{\infty }{A}_{m_n,n}$. Note that

$$\mu (X-X_0)=\mu (X-{\cap }_{n=1}^{\infty }{A}_{m_n,n})=\mu ({\cup }_{n=1}^{\infty }(X-A_{m_n,n}))⩽\sum _{n=1}^{\infty }\frac{ϵ}{2^n}=ϵ.$$

For any ${ϵ}^{\prime }>0$, take $N$ such that $1/N<{ϵ}^{\prime }$. If $x\in X_0\subseteq A_{m_N,N}$, for any $k>m_N$, we have $|f_k(x)-f(x)|<\frac{1}{N}<{ϵ}^{\prime }$ for all $x\in X_0$. Therefore, $f_n\to f$ uniformly on $X_0$. \end{proof}

Remark. $\mu (X)<\infty$ is necessary. Let $X=\mathbb{N}$, $\mathcal{A}=\mathcal{P}(X)$, and let $\mu$ be counting measure. Define $f_n={\chi }_{1,\cdots ,n}$, then $f_n\to f$ where $f\equiv 1$. Let $ϵ=\frac{1}{2}$. For any $X_0\in \mathcal{P}(X)$ with $\mu (X-X_0)<1/2$, we aim to show $f_n$ does not converge uniformly on $X_0$. Since $\mu (X-X_0)⩾1$ if $X\ne X_0$, then $X=X_0$. Claim that $f_n\to f$ does not converge uniformly on $X$, that is, there exists ${ϵ}_0>0$, for any $N>0$, there exists $x\in X$ and $n⩾N$ such that $|f_n(x)-f(x)|⩾{ϵ}_0$. Take ${ϵ}_0=1/2$, then for any $N>0$ there is $|f_N(N+1)-f(N+1)|=1⩾ϵ$. Hence, $f_n\to f$ is not uniformly convergent.

simple function

Let $X$ be a set. A function $s:X\to \mathbb{R}$ is called a simple function if it takes on only finitely many values, i.e., the image $s(X)$ is a finite subset of $\mathbb{R}$.

approximation by simple functions

Suppose $(X,\mathcal{A})$ is a measurable space and $f:X\to [-\infty ,\infty ]$ is measurable. Then there exists a sequence $f_1,f_2,\dots$ of functions from $X$ to $\mathbb{R}$ such that

• each $f_k$ is a simple measurable function;
• $\left|f_k(x)\right|\le \left|f_{k+1}(x)\right|\le |f(x)|$ for all $k\in \mathbb{N}$ and all $x\in X$;
• ${\lim }_{k\to \infty }{f}_k(x)=f(x)$ for every $x\in X$;
• $f_1,f_2,\dots$ converges uniformly to $f$ on any set on which $f$ is bounded.

\begin{proof} Define $f_k$ as the diagram below.

Then it is easy to prove the theorem. \end{proof}

Definition

Let $(X,\mathcal{A},\mu )$ be a measure space and let ${\left\{f_n\right\}}_{n=1}^{\infty }$ be a sequence of real-valued measurable functions defined on $X$. Let $f$ be a real-valued measurable function defined on $X$. We say that $\left\{f_n\right\}$ converges in measure to $f$, with notation $f_n\stackrel{\mu }{\to }f$, if for every $ϵ>0$, we have

$$\underset{n\to \infty }{\lim }\mu \left(\left\{x\in X:\left|f_n(x)-f(x)\right|\ge ϵ\right\}\right)=0$$

We say that $\left\{f_n\right\}$ is Cauchy in measure if for every $ϵ>0$ and for every $\delta >0$, there exists $N\in \mathbb{N}$ such that for all $n,m\ge N$, we have

$$\mu \left(\left\{x\in X:\left|f_n(x)-f_m(x)\right|\ge ϵ\right\}\right)<\delta .$$

Theorem

Let $(X,\mathcal{A},\mu )$ be a measure space and $\mu (X)<\infty$. Let ${\left\{f_n\right\}}_{n=1}^{\infty }$ be a sequence of real-valued measurable functions, defined on $X$, converging a.e. to a real-valued measurable function $f$. Then $f_n\stackrel{\mu }{\to }f$.

\begin{proof} Let $D=\{x\in X:|f_n(x)-f(x)|\nrightarrow 0\}$. Then $\mu (D)=0$. For any $ϵ>0$, let $A_n=\{x:|f_n(x)-f(x)|⩾ϵ\}$. Note that $\lim {\sup }_{n\to \infty }{A}_n\subseteq D$ and so $\mu (\lim \sup A_n)=0$. Define $B_n={\cup }_{k=n}^{\infty }{A}_k$, then $B_n\downarrow 0$, which yields that $f_n\stackrel{\mu }{\to }f$. \end{proof}

Remark. Let $X=\mathbb{R}$, and let $f_n={\chi }_{(n,n+1)}$. Note that $f_n\to 0$ almost everywhere, and $\mu (\{x\in X:|f_n(x)-0|⩾1/2\})=1\nrightarrow 0$. Therefore, $f_n/\stackrel{\mu }{\to }0$. Furthermore, this is also an example.

Proposition

Let $(X,\mathcal{A},\mu )$ be a measure space, and let $f$ and $f_1,f_2,\dots$ be real-valued measurable functions on $X$. If $f_n\stackrel{\mu }{\to }f$, then there is a subsequence of $\left\{f_n\right\}$ that converges a.e. to $f$.

\begin{proof} Since ${\lim }_{n\to \infty }\mu (\{x:|f_n(x)-f(x)|⩾1\})=0$, there exists $n_1$ such that $\mu (\{x:|f_{n_1}(x)-f(x)|⩾1\})<\frac{1}{2}$. Repeat this procedure,

$$\mu (\{x:|f_{n_k}(x)-f(x)|⩾\frac{1}{k}\})<\frac{1}{2^k}.$$

Now we get a subsequence $\{f_{n_k}\}$. Define $A_k=\{x:|f_{n_k}(x)-f(x)|⩾1/k\}$, then $D={\cap }_{j=1}^{\infty }{\cup }_{k=j}^{\infty }{A}_k$. It is easy to prove that $\mu (D)=0$. If $x\notin D$, then there exists $j$ such that $x\notin {\cup }_{k=j}^{\infty }{A}_j$ and so $|f_{n_k}(x)-f(x)|<1/k$ for all $k⩾j$. Hence $f_{n_k}\to f$ on $D^c$. \end{proof}