let $\xi,\eta: \Omega \to \mathbb R$ be i.i.d. random variables on a measurable space $(\Omega , \mathcal F,\mathbb P)$, and let $f: \mathbb R^2 \to \mathbb R$ be a bivariate measurable function (say under Borel $\sigma$-algebra). Clearly, $\omega \mapsto f(\xi(\omega),\eta(\omega))$ is also a random variable.


In many books the authors treat equations like $\inf\_{x \in \mathbb R} f(x , \eta)$ as random variables without further explaination about measurability.
However, this is equivalent to \begin{equation}g: \omega \mapsto \inf \bigg\{ f(x,\eta(\omega)) \bigg| x \in \mathbb R\bigg\}\end{equation}
which is an uncountable infimum of random variables.


Generally speaking, taking uncountable infimum of random variables is not valid. A counter example is shown in [Uncountable infimum of measurable functions](https://mathoverflow.net/questions/316651/uncountable-infimum-of-measurable-functions). The counter example works here if $x$ is subscript instead of a dimension, i.e. $g(w) = \inf\_{x\in \mathbb R} f\_x(\eta(\omega))$. However, since $f$ is a bivariate measurable function, it seems its measurability prevents the construction in that counter example being valid.


I believe $g$ is always measurable, but I can not come up with a proof. I wonder if there's a clear proof (or even better, a generalization) of this claim.



Why is it valid to take uncountable infimum of one dimension of a multivariate function of random variables?
$\newcommand\si\sigma\newcommand\om\omega\newcommand\Om\Omega\newcommand\R{\mathbb R}\newcommand\F{\mathcal F}\newcommand\B{\mathcal B}$No, $g$ is not in general Borel measurable, even if $f$ is Borel measurable.


E.g., let $\Om:=\R$ with $\F:=\B(\R)$, the Borel $\si$-algebra over $\R$. Let $\eta(\om):=\om$ for all $\om\in\R$. Let
$$f:=1-1\_B,$$
where $B$ is a Borel measurable subset of $\R^2$ such that the projection set
$$A:=\{\om\in\R\colon\,\exists\,x\in\R\ \,(x,\om)\in B\}
$$
[is not Borel measurable](https://en.wikipedia.org/wiki/Projection_(measure_theory)#Measurable_projection_theorem). Then $f$ is Borel measurable, whereas
$$A=g^{-1}(\{ 0 \})$$
and hence $g$ is not Borel measurable.


