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I want to propose to you this exercise. Let $f:[0,1]\times \mathbb{R}\to\mathbb{R}$ a function with these properties:

1)For every $x\in\mathbb{R}$ the map $t\mapsto f(t,x)$ is measurable.

2)For almost every $t\in [0,1]$ the maps $x\mapsto f(t,x)$ is continuous.

Let $\varphi:[0,1]\to\mathbb{R}$ a continuous function. Then show that the map $g(t):=f(t,\varphi(t))$ is measurable. I've tried to solve it, i have no good ideas. Thank you for the help.

DDT
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  • Perhaps this can help: https://math.stackexchange.com/questions/331169/a-question-concerning-measurability-of-a-function –  Jan 29 '16 at 20:36
  • Hi. Thanks for your answer. But my problem is slightly different: The map $x\mapsto f(t,x)$ is continuous for ALMOST every t, not for all, so we cannot take the limit under f. And we have a more strong assumption on $\varphi$. It's continuous, not just measurable. – DDT Jan 29 '16 at 21:29

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First replace $f$ by $\tilde{f}$, where $\tilde{f}(t,x) = 0$ if $f(t, \cdot)$ is not continuous. By your assumption, we have $\tilde{f}(t,x) = f(t,x)$ for almost all $t$ for all $x$. Hence, $g(t) = \tilde{f}(t, \varphi(t))$ for almost all $t$.

Now, the map $t \mapsto \tilde{f}(t, \varphi(t)$ is measurable by the answer here (A question concerning measurability of a function). Thanks to @maxerize for providing the link to this answer in the comments.

PhoemueX
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