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Suppose $f:[0,1]^2\to\mathbb{R}$, $(t,x)\mapsto f(t,x)$, is such that for each $t\in[0,1]$ $f(t,\cdot)$ is Lebesgue measurable on $[0,1]$, and for each $x\in[0,1]$ $f(\cdot,x)$ is continuous everywhere on $[0,1]\ni t$.

1. Does this imply that $f(t,x)$ is measurable on $[0,1]^2$?

2. Does this imply that the function $g(x)=\min\limits_{t\in[0,1]}f(t,x)$ is measurable on $[0,1]$?

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up vote 6 down vote accepted
  1. Yes, by continuity in the $t$ variable $f(t,x)=\lim_n f(\lfloor n t\rfloor/n,x)$, which expresses $f$ as the pointwise limit of a sequence of measurable functions.

  2. Yes, by continuity in the $t$ variable we have $\min_{t\in[0,1]} f(t,x)=\min_{t\in[0,1]\cap {\mathbb Q}} f(t,x)$, where $\mathbb Q$ means the rational numbers.

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Yeah, right. I was struggling for two days already and couldn't think of such a simple solution :( Thanks!! –  dergachev Sep 20 '10 at 18:57
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