Exercise 43 chapter 2 in Real Analysis of Folland I got stuck on this problem and couldn't find any clue to solve it. Can anyone give me some hint or give me some solution for it. I really appreciate!


Suppose that $\mu(X) < \infty$ and $f: X \times [0,1] \rightarrow \mathbb{C}$ is a function such that $f(.,y)$ is measurable for each $y \in [0,1]$ and $f(x, .)$ is continuous for each $x \in X$.
    a) If $0 < \epsilon, \delta < 1$, then $E_{\epsilon, \delta} = \{x: |f(x, y) - f(x, 0)| \le \epsilon, \forall y \lt \delta \}$ is measurable
    b) For any $\epsilon \gt 0$, there is a set $E \subset X$ such that $\mu(E) < \epsilon$ and $f(., y) \rightarrow f(., 0)$ uniformly on $E^c$ as $y \rightarrow 0$


 A: Uhm, I only have a suggestion for part a. Notice that $E_{\epsilon,\delta}$ almost look like the inverse image of an open set. I know, $\leq$ does not make it look like the inverse image of an open set. But if you divide the set in $E^1$ and $E^2$, where for $E^1$ the strict inequality holds, and for $E^2$ equality holds, then $E^1\cap E^2=\emptyset$, and $E^1$ IS the inverse image of an open set. As for $E^2$, well, can there actually be any element in there?
Edit: let me explain better.
1) If $E^2$ is not empty, it means $\exists x$ such that $|f(x,y)-f(x,0)|=\epsilon, \forall y<\delta$. But if $f(x,\cdot)$ is continuous, this cannot happen.
2) This is more complicate than I thought. And David's hint is really important. Define $F_y(x)=f(x,y)-f(x,0)$. Then $A_y = F_y^{-1}(-\epsilon,\epsilon)$ is measurable. As David says, the definition of $E_{\epsilon,\delta}$ can be replaced by
$$
E_{\epsilon,\delta}=\{x:|F_y(x)|<\epsilon,\forall y<\delta, y\in\mathbb{Q}\}
$$
Therefore, it follows that $E_{\epsilon,\delta}=\cap A_y$, where the intersection is over all $y<\delta,y\in\mathbb{Q}$. Since each $A_y$ is measurable and the intersection is countable, we obtain that $E_{\epsilon,\delta}$ is measurable.
