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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. 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.

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  • $\begingroup$ Do you know anything about $f$ (e.g. continuity)? That would allow you to replace an uncountable infimum by a countable infimum. Also by "binary", do you mean that the range of $f$ is $\{0,1\}$ or do you just mean that $f$ is a measurable function of 2 variables? $\endgroup$
    – Anthony Quas
    Commented Jun 22, 2022 at 17:37
  • $\begingroup$ Let's suppose we don't know other properties of $f$ except measurability. And sorry for confusing statement. By "binary" I mean a function of two variables. I'll edit the quetions statement. $\endgroup$
    – Yongyi Yang
    Commented Jun 22, 2022 at 18:09
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    $\begingroup$ Iosif Pinelis answers your question below, but let me give a simple answer to a different (but related problem): when the sigma-algebra is the completion of the Borel sigma-algebra (the Lebesgue sigma-algebra). In this case, let $A$ be a non-Lebesgue measurable subset and set $f(x,y)=0$ if $x=0$ and $y\in A^c$ and $f(x,y)=1$ otherwise. Then $f(x,y)$ is Lebesgue measurable (since $A\times\{0\}$ has measure 0), but $\inf_x f(x,y)=\mathbb 1_A$. $\endgroup$
    – Anthony Quas
    Commented Jun 22, 2022 at 21:34
  • $\begingroup$ Do you have a response to the answer below? $\endgroup$
    – Iosif Pinelis
    Commented Jun 24, 2022 at 21:19
  • $\begingroup$ @IosifPinelis Sorry I forgot to select accepted answer. The construction is very neat and solves my problem. $\endgroup$
    – Yongyi Yang
    Commented Jun 25, 2022 at 9:24

1 Answer 1

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$\newcommand\si\sigma\newcommand\om\omega\newcommand\Om\Omega\newcommand\R{\mathbb R}\newcommand\F{\mathcal F}\newcommand\B{\mathcal B}$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. Then $f$ is Borel measurable, whereas $$A=g^{-1}(\{ 0 \})$$ and hence $g$ is not Borel measurable.

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