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 Ewith |E-F|<\epsilonsuch that f(x,y)converges uniformly for x\in Fto 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_min the proof of Lemma 4.18.)
Solution: Suppose f(x,y)is defined and continuous in the square 0\leq x \leq 1, 0<y\leq 1and that f(x)=\lim_{y\to 0} f(x,y)exists and is finite for xin a measurable subset Eof [0,1].
We first prove a lemma.
Lemma: For any \epsilon>0,\eta>0, there exists a closed subset Fof Eand integer K>1such 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 xin E, for each x\in E, there exists \delta>0such 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>0such that
|f(x,y)-f(x)|< \epsilon
for y<\delta.
Choose m'>1so that 1/m'<\delta, then xis 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 haveE_{\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 xalready satisfy |f(x,y)-f(x)|\leq \epsilon~\text{for all}~ y< 1/m, then xwill 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_0such that
|E-E_{\epsilon,1/m_0}| = |E|-|E_{\epsilon,1/m_0}| < \frac{\eta}{2}
We know there exists a closed subset Fof 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 Fof Eand integer K>1such that |E-F|<\eta and
|f(x,y)-f(x)|\leq \epsilon
for x\in Ffor 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>1for 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_nand x\in F_n.
Define F=\bigcap_n F_n. Then Fis a closed set and F\subset F_nfor all n.
Now we want to show that f(x,y)converges uniformly to f(x)as y\to 0in F. For any \epsilon'>0, choose n'so that 1/n'<\epsilon'. Then take \delta=1/K_{n'}. For all x\in F, xis 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 0in F. Also,
E-F=E-\bigcap_{n=1}^\infty F_n=\bigcup_{n=1}^\infty(E-F_n)
and so
\begin{align}
|E-F|&\leq|E-F_1|+|E-F_2|+|E-F_3|+...\\
&<\frac{\epsilon}{2}+\frac{\epsilon}{4}+\frac{\epsilon}{8}+...\\
&=\epsilon
\end{align}