Measurable Function

Functions discussed in this chapter are defined on measurable sets, whose value could be generalized real number, i.e. $\mathbb{R}$ including $\pm \mathrm{\infty }$.

Definition of Measurable Function

Assume function $f$ has defined on measurable set $D$. If $\mathrm{\forall }\alpha \in \mathbb{R}$, set

$$\{f>\alpha \}=D(f>\alpha ):=\{x\in D:f(x)>\alpha \}$$

is also measurable, then we call $f$ is a measurable function on $D$.

Example. Continuous function on an interval $D\subset \mathbb{R}$ is a measurable function.

Proof

• For an open interval $D$, this is easy.

• For an arbitrary interval $D$, the original image of $f>\alpha$ would differ from open sets with at most $2$ points.

Characteristic Function

Assume $A\subset X$, then define a characteristic function $A$ as

$${\chi }_A(x)=\{\begin{array}{l}1,x\in A\\ 0,x\in X\mathrm{\setminus }A\end{array}$$

Measurability of Characteristic Function

For measurable set $D$, denote ${\chi }_D$ as its characteristic function, then

$$\{{\chi }_D>\alpha \}=\{\begin{array}{l}\varnothing ,\alpha \ge 1\\ D,0\le \alpha <1\\ \mathbb{R},\alpha <0\end{array}$$

So we could prove ${\chi }_D$ is measurable function.

Properties

Properties of Measurable Function

Assume $f(x)$ is measurable on $E$, then the following sets are all measureable.

(i) $\{f\le \alpha \}$ (ii) $\{f\ge \alpha \}$

(iii) $\{f<\alpha \}$ (iv) $\{f=\alpha \}$

(v) $\{f<+\mathrm{\infty }\}$ (vi) $\{f=+\mathrm{\infty }\}$

(vii) $\{f>-\mathrm{\infty }\}$ (viii) $\{f=-\mathrm{\infty }\}$

Proof

• (i) $=E-\{f>\alpha \}$

• (ii) $=\bigcap _{k=1}^{\mathrm{\infty }}\{f>\alpha -\frac{1}{k}\}$ (length decrease, so we choose inferior)

• (iii) using (ii) with same logic of proving (i)

• (iv) using (i) amd (ii)

• (v) $=\bigcup _{k=1}^{\mathrm{\infty }}\{f<k\}$ (length increase, so we choose superior)

• (vi) use (v) with same logic of proving (i)

• (vii) same logic of proving (i)

• (viii) use (vii) with same logic of proving (i).

The above (i), (ii) and (iii) are equivalent definition of measurable function, but (iv) is not.

Example. $f$ is measurable on measurable set $D$, iff $\mathrm{\forall }r\in \mathbb{Q}$, $\{f>r\}$ is measurable.

Proof

• "$\Rightarrow$". Let $\mathrm{\forall }\alpha$, $\alpha =r$.

• "$\Leftarrow$". Notice that $\mathrm{\forall }\alpha ,\mathrm{\exists }\{r_n\}$, such that $r_n\in \mathbb{R}$, and $r_1>r_2>\cdots$, and $r_n\to \alpha$, so

$$\{f>\alpha \}=\bigcup _{n=1}^{\mathrm{\infty }}\{f>r_n\}$$

which means $f$ is measurable.

When $\{f=r\}$ is measurable, we could not deduce the same result. Bacause let $A$ to be a non-measurable set, define

$$f=\{\begin{array}{l}\sqrt{3},x\in A\\ \sqrt{2},x\notin A\end{array}$$

then $\mathrm{\forall }r\in \mathbb{Q}$, $\{f=r\}=\varnothing$ is measurable, but let $\alpha =1.5$,

$$\{f>1.5\}=A$$

is not measurable, so $f$ is not measurable.

Region Extension

(i) Assume generalized real function $f(x)$ is defined on $D_1\cup D_2$. If $f(x)$ is measurable on $D_1$ and $D_2$, then $f(x)$ is also measurable on $D_1\cup D_2$.

(ii) Assume $f(x)$ is measurable on $D$. If $A\subset D$ is also measurable, then $f(x)$ is also measurable on $A$.

Proof

(i) $\{x\in D_1\cup D_2:f>\alpha \}=\{x\in D_1:f>\alpha \}\cup \{x\in D_2:f>\alpha \}$ and by measure operation.

(ii) $\{x\in A:f>\alpha \}=\{x\in D:f>\alpha \}\cap A$ and by measure operation.

Lemma of functions

Assume $f(x)$, $g(x)$ are measurable on $E$, then $\{f>g\}$ is also measurable.

Proof

$\mathrm{\forall }{\epsilon }_r=\frac{1}{r},r\in {\mathbb{N}}^{+}$, we have

$$\{f>g\}=\bigcup _{r=1}^{\mathrm{\infty }}\{f>{\epsilon }_r\}\cap \{{\epsilon }_r>g\}$$

so $\{f>g\}$ is measurable.

Basic Operations

Basic Operations of Measurable Function

Assume $f(x)$, $g(x)$ are measurable function on $D$, then the following functions are measurable.

(i) $cf(x)(c\in \mathbb{R})$. (ii) $f(x)+g(x)$. (iii) $f(x)\cdot g(x)$.

Proof

(i) move $c$ to the side of $\alpha$.

(ii) same logic with Lemma.

(iii) use $4f(x)\cdot g(x)=[f(x)+g(x){]}^2-[f(x)-g(x){]}^2$. You only need to consider whether $f^2(x)$ are measurable, which is easy to see. Notice that $\mathrm{\forall }\alpha \ge 0$,

$$\{f^2>\alpha \}=\{f>\sqrt{\alpha }\}\cup \{f<-\sqrt{\alpha }\}$$

is measurable, and $\mathrm{\forall }\alpha <0$,

$$\{f^2>\alpha \}=D$$

is measurable.

Example. Assume $f$ is defined on measurable set $D$, prove: if $f^2$ is measurable on $D$ and $\{f>0\}$ is measurable, then $f$ is measurable on $D$.

Proof

• $\mathrm{\forall }\alpha \ge 0$,

$$\{f>\alpha \}=\{f^2>{\alpha }^2\}\cap \{f>0\}$$

is measurable.

• $\mathrm{\forall }\alpha <0$,

$$\{f>\alpha \}=\{f>0\}\cup \{\alpha <f\le 0\}=\{f>0\}\cup [\{f^2<{\alpha }^2\}\cap \{f\le 0\}]$$

is measurable.

Example. Assume $f$ is a measurable function on $D$, prove: for all open set $G\subset \mathbb{R}$ and closed set $F\subset \mathbb{R}$,

$$f^{-1}(G),f^{-1}(F)$$

are measurable.

Proof

Known from configuration of set, $\mathrm{\forall }$ open set $G\subset \mathbb{R}$, there exists $\{a_n\}$ and $\{b_n\}$, such that $a_n<b_n$, and $G=\bigcup _{n=1}^{\mathrm{\infty }}(a_n,b_n)$, so

$$f^{-1}(G)=\bigcup _{n=1}^{\mathrm{\infty }}\{a_n<f<b_n\}$$

is measurable. Notice that $f^{-1}(F)=f^{-1}(\mathbb{R}-F^c)=D-f^{-1}(F^c)$ is also measurable.

Limit superior & limit inferior

Limit superior & limit inferior

Sequence of Function.

Similarly, given a sequence of function $\{f_n(x){\}}_{n\ge 1}$ we have limit superior

$$\underset{n\to \mathrm{\infty }}{\overset{―}{lim}}{f}_n=\underset{n\ge 1}{inf}\{\underset{k\ge n}{sup}\{f_k(x)\}\}$$

and limit inferior

$$\underset{n\to \mathrm{\infty }}{\underset{―}{lim}}{f}_n=\underset{n\ge 1}{sup}\{\underset{k\ge n}{inf}\{f_k(x)\}\}$$

Theorem for measurability of Limit

Assume $\{f_n(x){\}}_{n\ge 1}$ is a sequence of measurable functions on measurable set $D$, then the following functions are all measurable.

$$\begin{array}{}\text{(1)}& \underset{n\ge 1}{sup}{f}_n(x),\underset{n\ge 1}{inf}{f}_n(x)\text{(2)}& \underset{n\to \mathrm{\infty }}{\overset{―}{lim}}{f}_n(x),\underset{n\to \mathrm{\infty }}{\underset{―}{lim}}{f}_n(x)\end{array}$$

Proof

• For expression $\text{1}$, we can write it as

$$\{\underset{n\ge 1}{sup}{f}_n(x)>\alpha \}=\bigcup _n^{\mathrm{\infty }}\{f_n(x)>\alpha \}$$

where the latter one is measurable.

• For expression $\text{2}$, we could use conclusion from $\text{1}$.

Example. Assume $f_n$ is a measurable function on $D$. Prove: Set

$$\{x\in D:f_n\text{ converges}.\}$$

is measurable.

Proof

$f_n$ converges, iff its limit superior and inferior are of the same. So we define

$$F(x)=\underset{n\to \mathrm{\infty }}{\overset{―}{lim}}{f}_n,f(x)=\underset{n\to \mathrm{\infty }}{\underset{―}{lim}}{f}_n,$$

both of which are measurable by Theorem for limit. Then consider

$$\{F=f\}=D-\{F>f\}$$

since the latter is measurable according to Lemma of functions, $\{F=f\}$ is measurable.

Almost Everywhere

Definition of Almost Everywhere of a Proposition

Assume $D$ is measurable, and proposition $P(x)$ is related with points in $D$. If there exists a zero-measure set $E\subset D$, such that $P(x)$ holds on $D\mathrm{\setminus }E$, then we call $P(x)$ holds almost everywhere on $D$.

We denote this statement as

$$P(x)a.e.x\in D.$$

Approximation

Approximation with Simple Function

Simple Function

Assume $f$ is a function on measurable set $D$. If $f(D)$ is composed of finite points $\{a_k{\}}_{k=1}^n$, and sets $E_k=\{f=a_k\}(k=1,2,\cdots ,n)$ are all measurable, then we call $f$ is a simple function on $D$.

Apparently, here $f(x)$ could be expressed by finite linear combination of characteristic functions, i.e.

$$f(x)=\sum _{k=1}^n{a}_k{\chi }_{E_k}(x)$$

Approximate measurable function with sequence of simple functions

If $f(x)$ is measurable on $D$, then there exists a sequence of simple functions $\{f_k{\}}_{k=1}^{\mathrm{\infty }}$, such that for every point $x\in D$, $\{f_k{\}}_{k\ge 1}$ converges to $f(x)$. Specially, if

(i) $f$ is non-negative, then $\{f_k{\}}_{k\ge 1}$ is monotonically increasing.

(ii) $f$ is bounded, then $\{f_k{\}}_{k\ge 1}$ converges uniformly to $f(x)$.

HintsProof

Use a special construction.

Formulate a sequence of function which converges to $f(x)$. This is a little tricky.

Define

$$f_n(x)=\{\begin{array}{l}n,f(x)>n\\ \frac{k-1}{2^n},\frac{k-1}{2^n}\le f(x)<\frac{k}{2^n},k=-n{2}^n+1,-n{2}^n\cdots ,n{2}^n\\ -n,f(x)<n\end{array}$$

Look the following image for intuitional impression.

Now we show that $f_n(x)$ corvergese to $f(x)$. For fixed point $x\in D$, if

(i) $f(x)=\mathrm{\infty }$. Then $f_n(x)=n\to \mathrm{\infty }=f(x)$. The same for $f(x)=-\mathrm{\infty }$.

(ii) $-\mathrm{\infty }<f(x)<+\mathrm{\infty }$. Then for sufficient large number $n$, there exists unique $k_n$ such that $-n{2}^n+1\le k_n\le n{2}^n$ and

$$\frac{k_n-1}{2^n}\le f(x)\le \frac{k_n}{2^n}$$

which means $f_n(x)=\frac{k_n-1}{2^n}$, satisfying

$$0\le f(x)-f_n(x)\le \frac{1}{2^n}.$$

Let $n\to \mathrm{\infty }$, we have $f_n(x)$ converges to $f(x)$.

• Specially, if $f(x)$ is non-negative, for $f(x)=+\mathrm{\infty }$, then $f_n(x)=n$ is monotonically increasing.

For $f(x)<\mathrm{\infty }$, if $f(x)<n$, we have $f(x)=\frac{k-1}{2^n}$ for some $k$ such that $\frac{k-1}{2^n}<f(x)<\frac{k}{2^n}$, then for $f_{n+1}(x)$, we partition this interval into two parts, which are

$$[\frac{2k-2}{2^{n+1}},\frac{2k-1}{2^{n+1}}),[\frac{2k-1}{2^{n+1}},\frac{2k}{2^{n+1}})$$

the former part $f_{n+1}(x)=f_n(x)=\frac{k-1}{2^n}$, while the latter part gives $f_{n+1}(x)=\frac{2k-1}{2^{n+1}}>f_n(x)$, so it is monotonically increasing.

for $\mathrm{\infty }>f(x)\ge n$, this is similar to $\mathrm{\infty }$.

• Specailly, if $f(x)$ is bounded, denote its upper bound as $M$, so when $n>M$, we have $|f(x)-f_n(x)|<\frac{1}{2^n}$ on all $x\in D$, which means $f_n(x)\rightrightarrows f(x)$.

Example. Assume $f$ is differentiable on $\mathbb{R}$, prove $f^{\prime }(x)$ is measurable.

Proof

$f$ is differentiable, so $f$ and $f(x+\frac{1}{n})$ is continuous, thus measurable. So

$$f^{\prime }(x)=\underset{n\to \mathrm{\infty }}{lim}\frac{f(x+\frac{1}{n})-f(x)}{\frac{1}{n}}\stackrel{\mathrm{\Delta }}{=}{f}_n(x)$$

$f_n$ is measurable, so $f^{\prime }$ is also measurable.

Uniform Convergnce a.e.

Lemma: use language of Set to describe convergence

Assume $\{f_k(x){\}}_{k\ge 1}$, $f(x)$ are measurable function on $a.e.x\in E$, and $m(E)<\mathrm{\infty }$. If $f_k(x)\to f(x),a.e.x\in E$, then $\mathrm{\forall }\epsilon >0$, let $E_k(\epsilon )=\{x\in E:|f_k(x)-f(x)|\ge \epsilon \}$, then we have

$$\underset{j\to \mathrm{\infty }}{lim}m(\bigcup _{k=j}^{\mathrm{\infty }}{E}_k(\epsilon ))=0.$$

Proof

We have the following equivalent statement for point-wise convergence.

$$\begin{array}{rl}& f_k(x)\to f(x),a.e.x\in E\\ \iff & \mathrm{\forall }\epsilon >0,\mathrm{\exists }N>0,\mathrm{\forall }k\ge N,|f_k(x)-f(x)|<\epsilon \\ \iff & \mathrm{\forall }\epsilon >0,\mathrm{\exists }N>0,x\in \bigcap _{k=N}^{\mathrm{\infty }}\{x\in E:|f_k(x)-f(x)|<\epsilon \}\\ \text{(3)}& \iff & \mathrm{\forall }\epsilon >0,x\in \bigcup _{N=1}^{\mathrm{\infty }}\bigcap _{k=N}^{\mathrm{\infty }}\{x\in E:|f_k(x)-f(x)|<\epsilon \}\end{array}$$

Note that we could have an equivalent statement at the last

$$x\in \bigcap _{\epsilon >0}\bigcup _{N=1}^{\mathrm{\infty }}\bigcap _{k=N}^{\mathrm{\infty }}\{x\in E:|f_k(x)-f(x)|<\epsilon \}$$

but $\bigcap _{\epsilon >0}$ is not denumerable, so the in the following proof, we have to choose a denumerable set like $\mathbb{Q}$ or $\frac{1}{r}$.

Now we have a equivalent statement of $f_k(x)\to f(x),a.e.x\in E$, that is, expression $\text{3}$. Then its opponent proposition $f_k(x)\nrightarrow f(x)$ could be expressed by

$$\mathrm{\exists }\epsilon >0,x\in \bigcap _{N=1}^{\mathrm{\infty }}\bigcup _{k=N}^{\mathrm{\infty }}\{x\in E:|f_k(x)-f(x)|\ge \epsilon \}$$

That is,

$$\begin{array}{}\text{(4)}& \{x\in E:f_k(x)\nrightarrow f(x)\}=\bigcup _{\epsilon >0}\bigcap _{N=1}^{\mathrm{\infty }}\bigcup _{k=N}^{\mathrm{\infty }}{E}_k(\epsilon )\end{array}$$

which has zero-measure according to $a.e.x\in E$, so by limit operations in measure, we have

$$\underset{N\to \mathrm{\infty }}{lim}m(\bigcup _{k=N}^{\mathrm{\infty }}{E}_k(\epsilon ))=m(\underset{N\to \mathrm{\infty }}{lim}\bigcup _{k=N}^{\mathrm{\infty }}{E}_k(\epsilon ))=m(\bigcap _{N=1}^{\mathrm{\infty }}\bigcup _{k=N}^{\mathrm{\infty }}{E}_k(\epsilon ))=0.$$

The first equation holds because $E_k(\epsilon )$ is measurable ($f_k,f$ are all measurable functions) and $\bigcup _{k=N}^{\mathrm{\infty }}{E}_k(\epsilon )$ is monotonically decreasing when $N\to \mathrm{\infty }$. The last equation holds because the left side of equation $\text{4}$ has zero-measure (which means the item to be unioned should have zero-measure).

Egoroff Theorem

Assume $\{f_k(x){\}}_{k\ge 1}$ is a measurable function which is bounded on $a.e.x\in E$, and $m(E)<\mathrm{\infty }$. If $f_k(x)\to f(x),a.e.x\in E$, then $\mathrm{\forall }\delta >0$, $\mathrm{\exists }\text{ measurable set }{E}_{\delta }\subset E$, such that $m(E_{\delta })\le \delta$ and

$$f_k(x)\rightrightarrows f(x),\mathrm{\forall }x\in E-E_{\delta }.$$

Proof

Choose $\epsilon =\frac{1}{r}$ and define

$$E_k\left(\frac{1}{r}\right)=\{x\in E:|f_k(x)-f(x)|\ge \frac{1}{r}\},$$

then by condition of the theorem and Lemma, $\mathrm{\forall }\delta >0$, $\mathrm{\forall }r\in \mathbb{R}$, $\mathrm{\exists }{N}_r$, $\mathrm{\forall }N\ge N_r$,

$$m\left[\bigcup _{k=N}^{\mathrm{\infty }}{E}_k\left(\frac{1}{r}\right)\right]<\frac{\delta }{2^r}.$$

Then we have to sum up there measure. That is, let $E_{\delta }=\bigcup _{r=1}^{\mathrm{\infty }}\bigcup _{k=N}^{\mathrm{\infty }}{E}_k\left(\frac{1}{r}\right)$, which satisfies

$$m(E_{\delta })\le \sum _{r=1}^{\mathrm{\infty }}m\left[\bigcup _{k=N}^{\mathrm{\infty }}{E}_k\left(\frac{1}{r}\right)\right]<\delta .$$

Check that

$$E\mathrm{\setminus }{E}_{\delta }=\bigcap _{r=1}^{\mathrm{\infty }}\bigcap _{k=N}^{\mathrm{\infty }}{E}_k^c\left(\frac{1}{r}\right)=\bigcap _{r=1}^{\mathrm{\infty }}\bigcap _{k=N}^{\mathrm{\infty }}\{x\in E:|f_k(x)-f(x)|<\frac{1}{r}\}$$

Notice that $f_k\rightrightarrows f,a.e.x\in A$, iff $\underset{x\in A}{sup}|f_k(x)-f(x)|\to 0(k\to \mathrm{\infty })$. If we let $A=E\mathrm{\setminus }{E}_{\delta }$, this holds apparently.

Note that the condition $m(E)<\mathrm{\infty }$ in the above theorem could not be removed.

Example. $f_n(x)=x^n$ converges to $f(x)=\{\begin{array}{l}0,0<x<1\\ 1,x=1\end{array}$ point-wisely, but not uniformly. If we remove a sufficiently small set $(1-\delta ,1]$, then $f_n(x)\rightrightarrows f(x)$ uniformly.

Extension of Measurable Function

The following theorem is the main result of this part.

Lusin Theorem

Assume $f(x)$ is a measurable function $a.e.$ on $E$, then $\mathrm{\forall }\delta >0$, $\mathrm{\exists }$ closed set $F\subset E$, s.t. $m(E\mathrm{\setminus }F)<\delta$, and $f(x)$ is a continuous function on $F$.

HintsProof

Using Theorem in Mathematical Analysis for uniform convergent function, that is, the convergent function of uniformly convergent function sequence is continuous(Here notice we have to cut off a closed set from $E$, which might be infinite). So we have to make use of Egoroff Theorem.

(i) $f$ is a simple function on $E$.

From conclusion in Property of Simple Function, there exists measurable sets $E_i(i=1,2,\cdots ,N)$, such that $E=\bigcup _{i=1}^N{E}_n$, $E_i\cap E_j=\varnothing$, and

$$f(x)=\sum _{i=1}^N{\alpha }_i{\chi }_{E_i}(x)$$

From Theorem of Approximation for measurable sets, $\mathrm{\forall }\delta >0$, because $E_i$ is measurable, $\mathrm{\exists }$ closed set $F_i\subset E_i$, such that $m(E_i\mathrm{\setminus }{F}_i)<\delta /N$, and ${\chi }_{E_i}(x)\in C(F_i)$, or to be more specific, is constant on $F_i$. So from $F:=\bigcup _{i=1}^N{F}_i\subset E=\bigcup _{i=1}^N{E}_i$, we have

$$\begin{array}{rl}m(E\mathrm{\setminus }F)& =m\left[\left(\bigcup _{i=1}^N{E}_i\right)\mathrm{\setminus }\left(\bigcup _{i=1}^N{F}_i\right)\right]\\ & \le m\left[\bigcup _{i=1}^N\left(E_i\mathrm{\setminus }{F}_i\right)\right]\\ & \le \sum _{i=1}^{N}m(E_i\mathrm{\setminus }{F}_i)<\delta \end{array}$$

(ii) $m(E)<\mathrm{\infty }$.

From conclusion in Approximation with simple function, there exists a sequence of simple function $\{{\psi }_k{\}}_{k\ge 1}$ such that ${\psi }_k(x)\to f(x)$.

Then since $m(E)<\mathrm{\infty }$, we could use Egoroff Theorem for the above function sequence. That is, $\mathrm{\forall }\delta >0$, $\mathrm{\exists }\text{ measurable function }{E}_{\delta }\subset$, such that $m(E\mathrm{\setminus }{E}_{\delta })<\delta /2$ and

$${\psi }_k(x)\rightrightarrows f(x),\mathrm{\forall }x\in E\mathrm{\setminus }{E}_{\delta }.$$

The following method is a little tricky. $\mathrm{\forall }k$, from (i) we could know for simple function ${\psi }_k(x)$, there exists closed set $F_k\subset E\mathrm{\setminus }{E}_{\delta }$, such that $m((E\mathrm{\setminus }{E}_{\delta })\mathrm{\setminus }{F}_k)<\delta /2^{k+1}$ and ${\psi }_k(x)\in C(F_k)$. Define $F=\bigcap _{k=1}^{\mathrm{\infty }}{F}_k$ which is still a closed set.

Now we have to prove that the part we remove is small enough. That is, since

$$E\mathrm{\setminus }F=E\mathrm{\setminus }\left(\bigcap _{k=1}^{\mathrm{\infty }}{E}_k\right)\subset [E_{\delta }]\cup [(E\mathrm{\setminus }{E}_{\delta })\mathrm{\setminus }\left(\bigcap _{k=1}^{\mathrm{\infty }}{E}_k\right)]$$

so

$$\begin{array}{rl}m\left[E\mathrm{\setminus }F\right]& \le m\left[E_{\delta }\right]+m[(E\mathrm{\setminus }{E}_{\delta })\mathrm{\setminus }\left(\bigcap _{k=1}^{\mathrm{\infty }}{E}_k\right)]\\ & \le \frac{\delta }{2}+m\left[\bigcup _{k=1}^{\mathrm{\infty }}((E\mathrm{\setminus }{E}_{\delta })\mathrm{\setminus }{F}_k)\right]\\ & \le \frac{\delta }{2}+\sum _{k=1}^{\mathrm{\infty }}m\left[((E\mathrm{\setminus }{E}_{\delta })\mathrm{\setminus }{F}_k)\right]\\ & \le \frac{\delta }{2}+\frac{\delta }{2}=\delta \end{array}$$

Thus, ${\psi }_k(x)$ uniformly converges to $f(x)$ on $F$, by Theorem in Mathematical Analysis for uniform convergent function, meaning $f(x)\in C(F)$.

(iii) Last situation. $m(E)=\mathrm{\infty }$.

We have to transform this problem into (ii), that is, use finite-measure set to intersect.

Define $E_k:=E\cap \{x\in E:k-1\le |x|<k\}$. Then $E=\bigcup _{k=1}^{\mathrm{\infty }}{E}_k$, and $E_i\cap E_j=\varnothing$.

So $\mathrm{\forall }k$, there exists closed sets $F_k\subset E_k$, such that $m(E_k\mathrm{\setminus }{F}_k)<\delta /2^{k+1}$, and $f\in C(F_k)$. Consider define $E_{\delta }=\bigcup _{k=1}^{\mathrm{\infty }}{F}_k$ (see difference with (ii)) which is just a measurable set, or to be more specific, a $F_{\sigma }$ set. But again by using Theorem of Approximation for measurable sets, there exists a closed set $F\subset E_{\delta }$, such that $m(E_{\delta }-F)<\delta /2$.

Now we check the measure of the removal part. Since

$$E\mathrm{\setminus }F\subset (E\mathrm{\setminus }{E}_{\delta })\cup (E_{\delta }\mathrm{\setminus }F),$$

we have

$$\begin{array}{rl}m(E\mathrm{\setminus }F)& \le m\left[E\mathrm{\setminus }{E}_{\delta }\right]+m\left[E_{\delta }\mathrm{\setminus }F\right]\\ & =m\left[E\mathrm{\setminus }\left(\bigcup _{k=1}^{\mathrm{\infty }}{F}_k\right)\right]+m\left[E_{\delta }\mathrm{\setminus }F\right]\\ & \le m[\bigcup _{k=1}^{\mathrm{\infty }}(E_k\mathrm{\setminus }{F}_k)]+\frac{\delta }{2}\\ & \le \sum _{k=1}^{\mathrm{\infty }}m(E_k\mathrm{\setminus }{F}_k)+\frac{\delta }{2}\\ & =\frac{\delta }{2}+\frac{\delta }{2}=\delta \end{array}$$

Thus we are done.

Continuous Extension Theorem

Assume $f(x)$ is a measurable function bounded $a.e.$ on measurable set $E\subset \mathbb{R}$, then $\mathrm{\forall }\delta >0$, there $\mathrm{\exists }$ closed set $F_{\delta }\subset E$ and continuous function $g(x)$ on whole space $\mathbb{R}$, such that $m(E-F_{\delta })<\delta$ and $g(x)=F(x),\mathrm{\forall }x\in F_{\delta }$, and

$$\begin{array}{}\text{(5)}& \underset{x\in \mathbb{R}}{sup}g(x)=\underset{x\in F_{\delta }}{sup}f(x),\underset{x\in \mathbb{R}}{inf}g(x)=\underset{x\in F_{\delta }}{inf}f(x).\end{array}$$

which naturally satisfies the condition in the theorem.

Proof

By Lusin Theorem, $\mathrm{\forall }\delta >0$, there exists closed set $F\subset E$, such that $m(E-F)<\delta$, and $f\in C(F)$. Now the only problem is, how to extend this continuoud function on $F$ to all space $F$ and satisfy the so-called boundary condition $\text{5}$.

By Conposition of closed set, $F$ could be expressed by $\mathbb{R}-G$, where $G$ is a union of denumerable open intervals which are mutually disjoint. That is,

$$F^c=\mathbb{R}-F=\bigcup _{i=1}^{\mathrm{\infty }}(a_i,b_i).$$

where at most $2$ open intervals are infinite intervals. From here we formulate a function on $\mathbb{R}$

$$g(x)=\{\begin{array}{ll}f(x),& x\in F,\\ f(a_i)+\frac{f(b_i)-f(a_i)}{b_i-a_i}(x-a_i),& x\in (a_i,b_i),\\ f(a_i),& x\in (a_i,+\mathrm{\infty }),\\ f(b_i),& x\in (-\mathrm{\infty },b_i).\end{array}$$

Convergence in Measure

Example. $\mathrm{\forall }n\in {\mathbb{N}}^{+}$, $\mathrm{\exists }!k,i\in {\mathbb{N}}^{+}$, such that $n=2^k+i$, where $0\le u\le 2^k$. Formulate a function sequence

$$f_n(x)={\chi }_{[\frac{i}{2^k},\frac{i+1}{2^k}]}(x),n=1,2,\cdots ,x\in [0,1]$$

So for every point $x\in [0,1]$, $f_n(x)$ does not converge.

For the above example, we could consider its convergence in measure, or probability.

Still the above example.

If we define a set

$$E_n(\epsilon )=\{x\in [0,1]:|f_n(x)-0|\ge \epsilon \},$$

then its measure

$$m(E_n(\epsilon ))=\frac{1}{2^k}\to 0$$

which means for sufficient large $n$, the frequency of $0$ is close to $1$. So we call $\{f_n(x)\}$ converges to $f(x)=0$ in measure.

Definition of convergence in measure

Assume function $f$ and function sequence $\{f_n{\}}_{n\ge 1}$ are measurable function bounded $a.e$ on $D$. If $\mathrm{\forall }\delta >0$, we have

$$\underset{n\to \mathrm{\infty }}{lim}m(\{|f_n-f|\ge \delta \})=0$$

then we call $\{f_n\}$ converges to $f$ in measure on region $D$.

Using $\epsilon -N$ language, we have

$$\mathrm{\forall }\epsilon >0,\sigma >0,\mathrm{\exists }N\in {\mathbb{N}}^{+},\mathrm{\forall }n\ge N,m(\{|f_n-f|>\sigma \})<\epsilon .$$

Generally speaking, "convergence point-wisely" has no apparent relationship with "convergence in measure", readers could see the following exmaple.

Example. Define $f_n(x)={\chi }_{(n,\mathrm{\infty })}(x)$, $f(x)=0$, then $\mathrm{\forall }x\in \mathbb{R}$, we have $f_n(x)\to f(x)$. But if we let $\delta =1/2$, then $\{|f_n-f|\ge 1/2\}=(n,\mathrm{\infty })$, whose measure $m(\{|f_n-f|\ge 1/2\})=\mathrm{\infty }$, which does not converges to $f(x)$ in measure.

Uniqueness of Convergent Function

Uniqueness of Convergent Function

If $f_n(x)$ converges to $f(x)$ and $g(n)$ on $E$ in measure, then

$$f(x)=g(x),a.e.x\in E.$$

Proof

Because $|f(x)-g(x)|\le |f(x)-f_n(x)|+|f_n(x)-g(x)|$ (at least one set of the right side should be larger then $\epsilon /2$ if $|f(x)-g(x)|>\epsilon$), so

$$\{|f(x)-g(x)|>\epsilon \}\subset \{|f(x)-f_n(x)|>\frac{\epsilon }{2}\}\cup \{|g(x)-f_n(x)|>\frac{\epsilon }{2}\}$$

So by subadditivity, we have

$$m(\{|f(x)-g(x)|>\epsilon \})\le m\left(\{|f(x)-f_n(x)|>\frac{\epsilon }{2}\}\right)+m\left(\{|g(x)-f_n(x)|>\frac{\epsilon }{2}\}\right)\to 0,n\to \mathrm{\infty }$$

Relationship between two Convergences

Lebesgue's Theorem: Relationship between two Convergences

Assume $m(E)<\mathrm{\infty }$, $\{f_n(x)\}$ is a measurable function sequence $a.e.$ on $E$. If $\{f_n\}$ converges to function $f(x)$ $a.e.$ on $E$, where $f(x)$ is a bounded function $a.e.x\in E$, then $f_n(x)$ converges to $f(x)$ in measure.

HintsProof

Use $m(E)<\mathrm{\infty }$ to show that $f_n(x)\rightrightarrows f(x)$, which could gives convergence in measure.

By Egoroff Theorem, because $m(E)<\mathrm{\infty }$, we have point-wise convergent function sequence $\{f_n\}$ also converges uniformly to $f(x)$. That is, we have $\mathrm{\forall }\delta >0$, $\mathrm{\exists }$ measurable set $E_{\delta }$, such that $m(E_{\delta })<\delta$ and

$$f_n(x)\rightrightarrows f(x),\mathrm{\forall }x\in E\mathrm{\setminus }{E}_{\delta }.$$

Then by definition of uniform convergence, we have $\mathrm{\forall }\epsilon >0$, $\mathrm{\exists }N>0$, $\mathrm{\forall }n\ge N$,

$$|f_n(x)-f(x)|<\epsilon ,x\in E\mathrm{\setminus }{E}_{\delta }$$

which means the set

$$\{x\in E:|f_n(x)-f(x)|\ge \epsilon \}\subset E_{\delta }$$

so

$$m(\{x\in E:|f_n(x)-f(x)|\ge \epsilon \})\le m(E_{\delta })<\delta .$$

and we are done.

Note that $m(E)<\mathrm{\infty }$ is not necessary, see that we use the uniform convergence to prove, which is the most important.

Riesz Theorem

Assume function $f$ and function sequence $\{f_n\}$ are measurable function $a.e.$ on $E$, where $f$ and $\{f_n\}$ are bounded $a.e.$ on $E$. Then

(i) If $\{f_n\}$ converge to $f$ in measure, then there exists subsequence $\{f_{n_j}\}$ convergent to $f$ $a.e.x\in E$.

(ii) If $m(E)<\mathrm{\infty }$, and $\{f_n\}$ converge to $f$ $a.e.x\in E$, then $\{f_n\}$ converge to $f$ in measure.

Proof for (i)Proof for (ii)

We hope to use $\delta$ to get a subsequence of $f_n$.

$\mathrm{\forall }k\in {\mathbb{N}}^{+}$, let ${\delta }_k=\frac{1}{2^k}$, so $f_n\stackrel{m}{\to }f$ implies

$$\underset{n\to \mathrm{\infty }}{lim}m(\{|f_n-f|\ge {\delta }_k\})=0.$$

So there exsits $n_k$, such that

$$m(\{|f_{n_k}-f|\ge {\delta }_k\})<\frac{1}{2^k}.$$

Notice $\{m(\{|f_n-f|\ge {\delta }_k\}){\}}_{n\ge 1}$ is a number sequence, so $n_1<n_2<\cdots$.

Define

$$E:=\bigcap _{p=1}^{\mathrm{\infty }}\bigcup _{k=p}^{\mathrm{\infty }}\{|f_{n_k}-f|\ge \frac{1}{2^k}\}$$

and for every $p$, measure

$$m(E)\le m\left(\{|f_{n_k}-f|\ge \frac{1}{2^k}\}\right)<\sum _{k=p}^{\mathrm{\infty }}\frac{1}{2^k}=\frac{1}{2^{p-1}}.$$

let $p\to \mathrm{\infty }$, we have $m(E)=0$. Thus for every $x\in D-E$, i.e.

$$x\in \bigcup _{p=1}^{\mathrm{\infty }}\bigcap _{k=p}^{\mathrm{\infty }}\{|f_{n_k}-f|<\frac{1}{2^k}\}$$

which means, $\mathrm{\exists }{p}_0\ge 1$, such that

$$|f_{n_k}-f|<\frac{1}{2^k},\mathrm{\forall }k\ge p_0$$

meaning $\underset{k\to \mathrm{\infty }}{lim}{f}_{n_k}=f$.

$\mathrm{\forall }\delta >0$, $\mathrm{\forall }\epsilon >0$, since $m(D)<\mathrm{\infty }$, by Egoroff Theorem, there exists measurable set $E\subset D$, such that $m(D-E)<\epsilon$, and $f_n\rightrightarrows f,a.e.x\in E$. That is, there exsits $N$, such that

$$|f_n(x)-f(x)|<\delta ,x\in E,n\ge N.$$

which means

$$m(\{|f_n(x)-f(x)|\ge \delta \})=M(D-E)<\epsilon ,n>N$$

meaning

$$\underset{n\to \mathrm{\infty }}{lim}m(\{|f_n(x)-f(x)|\ge \delta \})=0.$$

Lemma: Sufficient and Necessary Condition for convergence in measure

Assume function $f$ and function sequence $\{f_n\}$ are measurable on $E$ where $m(E)<\mathrm{\infty }$. Then $f_n(x)$ converges to $f(x)$ in measure, iff $\mathrm{\forall }$ subsequence $\{f_{n_j}\}$, $\mathrm{\exists }$ a subsequence $\{f_{n_{j_i}}\}$ such that

$$f_{n_{j_i}}\to f(x),a.e.x\in E.$$

Proof

• "$\Rightarrow$".

We prove $\mathrm{\forall }$ subsequence $\{f_{n_j}\}$, $f_{n_j}$ converges to $f$ in measure, which holds apparently because $f_n(x)$ converges to $f(x)$ in measure.

Then apply Riesz Theorem to $f_{n_j}$ and we are done.

• "$\Leftarrow$". We make use of contradiction. That is, if $\{f_n\}$ does not converges to $f$ in measure, i.e. $\mathrm{\exists }{\epsilon }_0>0$,

$$\underset{n\to \mathrm{\infty }}{lim}m(\{|f_n-f|\ge {\epsilon }_0\})\ne 0.$$

So $\mathrm{\forall }j,\mathrm{\exists }{n}_j$, s.t.

$$\begin{array}{}\text{(6)}& m(\{|f_{n_j}-f|\ge {\epsilon }_0\})>0.\end{array}$$

But if there exists a subsequence $\{f_{n_{j_i}}\}$ such that

$$f_{n_{j_i}}\to f(x),a.e.x\in E.$$

then by Lebesgue's Theorem ($m(E)<\mathrm{\infty }$), we have $f_{n_{j_i}}\stackrel{m}{\to }f$, which contradicts inequation $\text{6}$.