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.