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Can anyone please suggest how to justify widely used formula for interchange of integral and infimum:

$ \inf_{u(t)\in U}\int_{t_0}^{t_1}g(t,u(t))dt=\int_{t_0}^{t_1}\inf_{u\in U}g(t,u)dt, $

where $ U\subset\mathbb{R}^n$ is a compact set and the function $g$ is Caratheodory? In a Theorem by R. Tyrell Rockafellar the infimum on the right-hand side is taken over all $ u\in\mathbb{R}^n$:

$ \inf_{x\in X}\int_{S} f(s,x(s))\mu(ds)=\int_{S} \inf_{x\in \mathbb{R}^n}f(s,x)\mu(ds), $

where $ X $ is a decomposable linear space of measurable functions, and $ f$ is a normal integrand on $S\times \mathbb{R}^n$.

A possible workaround could be redefining the $ g$ function to be infinite for $ u\notin U$ and all $ t$. But I'm not sure if the Rockafellar theorem assumptions on normal integrand will be satisfied in this case?

Any help appreciated!

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    $\begingroup$ Clearly there is an inequality. So the problem becomes: for each $t$, choose $u(t)$ so that $g(t,u(t))$ is close to the infimum. And make the choice so that $u(t)$ is measurable. This may be known as a "measurable selection" theorem. $\endgroup$
    – Gerald Edgar
    Commented Jan 8, 2016 at 13:41
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    $\begingroup$ Hi can someone post a reference to this specific theorem of Rockafellar, I cant seem to find one. Thanks. $\endgroup$
    – ABIM
    Commented Jan 20, 2017 at 19:09
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    $\begingroup$ @AIM_BLB I think is this one sites.math.washington.edu/~rtr/papers/… $\endgroup$
    – sound wave
    Commented Aug 10, 2021 at 11:00

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