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.

  • $\begingroup$ 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? $\endgroup$ Jun 22 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. $\endgroup$ Jun 22 at 18:09
  • 3
    $\begingroup$ 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 $\inf_x f(x,y)=\mathbb 1_A$. $\endgroup$ Jun 22 at 21:34
  • $\begingroup$ Do you have a response to the answer below? $\endgroup$ Jun 24 at 21:19
  • $\begingroup$ @IosifPinelis Sorry I forgot to select accepted answer. The construction is very neat and solves my problem. $\endgroup$ Jun 25 at 9:24

1 Answer 1


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


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.