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Suppose $$F(t) = \lVert {f(t)} \rVert_{L^2(\Omega_t))} \tag{1}$$ is a continuous as a function of $t$ for each $t \in [0,T].$

A continuous function between measure spaces is measurable, so (1) is measurable.

I would like an explanation of this. What is the measure spaces involved? I guess $[0,T]$ and $\mathbb{R}.$ Is it that simple??

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up vote 2 down vote accepted

Every continuous function $F:[a,b]\to\mathbb{R}$ is measurable. Indeed for every $c\in\mathbb{R}$, the set $(c,+\infty)$ is open. Since $F$ is continuous $F^{-1}((c,+\infty))$ is open, so it is measurable. Since $c$ is arbitrary $F$ is measurable.

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