Ex4-14

Solution to Chapter 4, Exercise 14

''Question: Let $$f(x,y)$$be as in Exercise 13. Show that given'' $$\epsilon>0$$, there exists a closed $$F\subset E$$with $$|E-F|<\epsilon$$such that $$f(x,y)$$converges uniformly for $$x\in F$$to $$f(x)$$as $$y\to 0$$. (Follow the proof of Egorov's theorem, using the sets $$E_{\epsilon,1/m}$$in Exercise 13 in place of the sets $$E_m$$in the proof of Lemma 4.18.)

Solution: Suppose $$f(x,y)$$is defined and continuous in the square $$0\leq x \leq 1, 00,\eta>0$$, there exists a closed subset $$F$$of $$E$$and integer $$K>1$$such that $$|E-F|<\eta$$ and

$$|f(x,y)-f(x)|\leq \epsilon$$

for $$x\in F$$ for all $$y<1/K$$.

Proof. Given $$\epsilon>0,\eta>0$$. Since $$f(x)=\lim_{y\to 0}f(x,y)$$exists for $$x$$in $$E$$, for each $$x\in E$$, there exists $$\delta>0$$such that if $$y<\delta$$, then

$$|f(x,y)-f(x)|< \epsilon$$

By Question 13, for any integer $$m>1$$, the set

$$E_{\epsilon,1/m}=\left\{ x\in E~\bigg|~|f(x,y)-f(x)|\leq \epsilon~\text{for all}~ y< \frac{1}{m}\right\}$$

is measurable. We claim that

$$\bigcup_{m}E_{\epsilon,1/m}=E$$

By definition, each $$E_{\epsilon,1/m}$$is a subset of $$E$$. Now for any $$x\in E$$, there exists $$\delta>0$$such that

$$|f(x,y)-f(x)|< \epsilon$$

for $$y<\delta$$.

Choose $$m'>1$$so that $$1/m'<\delta$$, then $$x$$is in the set

$$E_{\epsilon,1/m'}=\left\{ x\in E~\bigg|~|f(x,y)-f(x)|\leq \epsilon~\text{for all}~ y< \frac{1}{m'}\right\}$$

so $$x\in \bigcup_{m}E_{\epsilon,1/m}$$and hence $$E \subseteq \bigcup_{m}E_{\epsilon,1/m}$$.

Thus, $$\bigcup_{m}E_{\epsilon,1/m}= E$$. We also have$$E_{\epsilon,1/m}=\left\{ x\in E~\bigg|~|f(x,y)-f(x)|\leq \epsilon~\text{for all}~ y< \frac{1}{m}\right\}$$$$\subseteq E_{\epsilon,1/(m+1)}=\left\{ x\in E~\bigg|~|f(x,y)-f(x)|\leq \epsilon~\text{for all}~ y< \frac{1}{m+1}\right\}$$

since if $$x$$already satisfy $$|f(x,y)-f(x)|\leq \epsilon~\text{for all}~ y< 1/m$$, then $$x$$will also satisfy $$|f(x,y)-f(x)|\leq \epsilon~\text{for all}~ y< 1/(m+1)$$.

This means that $$E_{\epsilon,1/m}\nearrow E$$. Then since $$E_{\epsilon,1/m}$$are measurable, by Theorem 3.26,

$$\lim_{m\to\infty}|E_{\epsilon,1/m}|=|E|$$

Since $$|E_{\epsilon,1/m}| < |E| < 1 < \infty$$, for any $$\eta>0$$, there exists $$m_0$$such that

$$|E-E_{\epsilon,1/m_0}| = |E|-|E_{\epsilon,1/m_0}| < \frac{\eta}{2}$$

We know there exists a closed subset $$F$$of $$E_{\epsilon,1/m_0}$$such that

$$|E_{\epsilon,1/m_0}-F|<\frac{\eta}{2}$$

by Lemma 3.22 since $$E_{\epsilon,1/m_0}$$is measurable. Then

$$E-F=(E-E_{\epsilon,1/m_0})\cup(E_{\epsilon,1/m_0}-F)$$$$|E-F|<\frac{\eta}{2}+\frac{\eta}{2}=\eta$$

Also, in $$F$$,

$$|f(x,y)-f(x)|\leq \epsilon~\text{for all}~ y< \frac{1}{m_0}$$

Therefore, we have shown that for any $$\epsilon>0,\eta>0$$, there exists a closed subset $$F$$of $$E$$and integer $$K>1$$such that $$|E-F|<\eta$$ and

$$|f(x,y)-f(x)|\leq \epsilon$$

for $$x\in F$$for all $$y<1/K$$.

Now we proceed to prove the statement of the question.

Given $$\epsilon>0$$, choose closed sets $$F_n\subset E$$, and integers $$K_n>1$$for each $$n\in \mathbb{N}$$so that

$$|E-F_n|<\frac{\epsilon}{2^n}$$

and

$$|f(x,y)-f(x)|\leq \frac{1}{n}$$

for all $$y<1/K_n$$and $$x\in F_n$$.

Define $$F=\bigcap_n F_n$$. Then $$F$$is a closed set and $$F\subset F_n$$for all $$n$$.

Now we want to show that $$f(x,y)$$converges uniformly to $$f(x)$$as $$y\to 0$$in $$F$$. For any $$\epsilon'>0$$, choose $$n'$$so that $$1/n'<\epsilon'$$. Then take $$\delta=1/K_{n'}$$. For all $$x\in F$$, $$x$$is in $$F_{n'}$$. Then if $$y<\delta$$, we have

$$|f(x,y)-f(x)|\leq \frac{1}{n'}<\epsilon'$$

for all $$x\in F$$.

This shows that $$f(x,y)$$converges uniformly to $$f(x)$$as $$y\to 0$$in $$F$$. Also,

$$E-F=E-\bigcap_{n=1}^\infty F_n=\bigcup_{n=1}^\infty(E-F_n)$$

and so

$$\begin{align} &<\frac{\epsilon}{2}+\frac{\epsilon}{4}+\frac{\epsilon}{8}+...\\ &=\epsilon \end{align}$$
 * E-F|&\leq|E-F_1|+|E-F_2|+|E-F_3|+...\\