Ex4-13

Chapter 4, Question 13 
Question: One difficulty encountered in trying to extend the proof of Egorov’s theorem to the continuous parameter case f_y(x)\to f(x) as y\to y_0 is showing that the analogues of the sets E_m in Lemma 4.18 are measurable. This difficulty can often be overcome in individual cases. Suppose, for example, that f(x,y) is defined and continuous in the square 0\leq x\leq 1, 0<y\leq 1 and that f(x)=\lim_{y\to 0}f(x,y) exists and is finite for x in a measurable subset E of [0,1]. Show that if \varepsilon and \delta satisfy 0<\varepsilon,\delta<1, the set E_{\varepsilon,\delta}=\{x\in E: |f(x,y)-f(x)|\leq \varepsilon \text{ for all }y<\delta\} is measurable. 
Proof: Fix \varepsilon,\delta\in(0,1), let \{y_k\}_{k=1}^\infty be a dense subset of (0,\delta) (for example, take \{y_k\}_{k=1}^\infty = \mathbb Q\cap (0,\delta)). Define E_k=\{x\in E: |f(x,y_k)-f(x)|\leq \varepsilon\}. We claim that each E_k is measurable and the following identity
E_{\varepsilon,\delta}=\bigcap_{k=1}^\infty \{x\in E: |f(x,y_k)-f(x)|\leq \varepsilon\}holds. Then the result immediately shows E_{\varepsilon,\delta} is measurable. 
First we have to show f(x) is a measurable function on E. Consider the sequence \{1/n\}_{n=1}^\infty , we see f(x,1/n) is a continuous function of x relative to E, hence measurable. Moreover, f(x)=\lim_{n\to\infty} f(x,1/n) for all x\in E, by Theorem 4.12 f is measurable on E as a limit of sequence of measurable functions. We also know that for every k, f(x,y_k) is a measurable function of x, hence E_k=\{|f(\cdot, y_k)-f(\cdot)|\leq\varepsilon\} is measurable. 
If x\in E_{\varepsilon,\delta}, then |f(x,y)-f(x)|\leq\varepsilon for all y\in(0,\delta), which in turn is particularly true for y=y_k. This shows E_{\varepsilon,\delta}\subseteq E_k\; \forall k\in\mathbb N, and thus E_{\varepsilon,\delta}\subseteq \bigcap_k E_k. On the other hand if x\in \bigcap_k E_k and y\in(0,\delta), we may choose a sequence \{z_n\}_{n=1}^\infty from \{y_k\}_{k=1}^\infty (treating \{y_k\}_{k=1}^\infty as an unordered set) such that z_n\to y. Then since for a fixed x, f(x,y) is continuous on y\in(0,\delta), we find that|f(x,y)-f(x)|=\lim_{n\to\infty} |f(x,z_n)-f(x)|\leq \varepsilon.This shows x\in E_{\varepsilon, \delta}. Finally, we conclude E_{\varepsilon,\delta} is measurable because it is a countable intersection of measurable sets.