# Why is it valid to take uncountable infimum of one dimension of a multivariate function of random variables?

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 $$$$g: \omega \mapsto \inf \bigg\{ f(x,\eta(\omega)) \bigg| x \in \mathbb R\bigg\}$$$$ 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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• 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? Jun 22 at 17:37
• 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. Jun 22 at 18:09
• 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 $\inf_x f(x,y)=\mathbb 1_A$. Jun 22 at 21:34
• Do you have a response to the answer below? Jun 24 at 21:19
• @IosifPinelis Sorry I forgot to select accepted answer. The construction is very neat and solves my problem. Jun 25 at 9:24

$$\newcommand\si\sigma\newcommand\om\omega\newcommand\Om\Omega\newcommand\R{\mathbb R}\newcommand\F{\mathcal F}\newcommand\B{\mathcal B}$$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.