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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 $y \rightarrow 0$

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    $\begingroup$ Hint: Since $f$ is continuous in $y$, the definition of $E_{\epsilon,\delta}$ is equivalent to the same definition except restricted to rational $y<\delta$. $\endgroup$ – David C. Ullrich Aug 4 '15 at 16:22
  • $\begingroup$ That is a very important point. It allows to switch from uncountable to countable! $\endgroup$ – bartgol Aug 5 '15 at 16:04
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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.

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  • $\begingroup$ Hm, very interesting. That's what I wondered when I read your hint. I guess what's remained now is to prove that the two definitions of $E_{\epsilon, \delta}$ is the same. $\endgroup$ – le duc quang Aug 5 '15 at 16:17
  • $\begingroup$ Yeah, but that's really easy. One direction is immediate, the other follows from the fact that $|..|<\delta$ is open. $\endgroup$ – bartgol Aug 5 '15 at 16:26
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    $\begingroup$ I know, or we can just state that for all $t < \delta$, there exists $q$ rational which satisfies $t < q < \delta$, then it's easy to prove that. Do you have any idea for part b??? $\endgroup$ – le duc quang Aug 5 '15 at 17:03
  • $\begingroup$ Uhm, not really. But I suspect you will need Egoroff's theorem, which has to do with relation between pointwise and uniform convergence on finite measure sets. The statement has a big overlap with your problem $\endgroup$ – bartgol Aug 5 '15 at 17:36
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    $\begingroup$ I think I found the solution. We just need to change the definition in Egoroff theorem of $E_n(k) = \bigcup_{0 < r < 1/n, r \in \mathcal{Q}} \{x: |f(x, y) - f(x, 0)| > k^{-1}\}$. I will post the whole solution when having free time then ^^ $\endgroup$ – le duc quang Aug 6 '15 at 17:56

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