Let $(X,d)$ be a separable metric space with Borel measure $\mu$. Let $f:X \times X \to \mathbb{R}$ be Borel measurable with respect to the product measure on $X \times X$, and let $g(x)=\operatorname{ess sup}_{y \in X} f(x,y)$. Is $g(x)$ necessarily measurable? (Is there some argument that can be pieced together using separability of $X$ and Lusin's Theorem, if we assume that $\mu$ is a Radon measure?)
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$\begingroup$ Now... the question makes sense in a measure space with no topology. $\endgroup$– Gerald EdgarCommented Jun 16, 2011 at 14:18
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3 Answers
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You are right. For each $n$ choose a set of measure less than $1/n$ on the complement of which $f$ is continuous. Now take the actual sup on each vertical section of this restricted function. This yields a measurable function $f_n$ for each $n$ defined on $X$. The sup of the increasing sequence of $f_n$ will also be a measurable function $F$. Except for a null set, $F$ will give the $\operatorname{esssup}$ of the vertical section of $f$. So modifying $F$ on a null set yields that $g$ is measurable.
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$\begingroup$ I have not been able to get this argument to work. The supremum over each vertical section can be expressed as a supremum over countably many points $y$ by continuity and separability, but the countable set on which you take the supremum should be dense in the section and varies across different vertical sections, hence it does not imply that $f_n$ is measurable. Is there a way to fix the argument? $\endgroup$– zfzf21Commented Jun 16, 2011 at 20:39
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$\begingroup$ You are right that separabilty is no help here. Instead, use the fact that analytic sets are measurable. Let $F_n$ be the continuous functions defined on a subset of $X\times X$ except for a set of measure $1/n$. Given $t\in \mathbb R$, $f_n^{-1}(t,\infty)$ is equal to the set of all $x\in X$ such that there exists $y\in X$ such that $F_n(x,y)> t$. The single existential quantifier ensures that this set is analytic and, hence, measurable. $\endgroup$ Commented Jun 16, 2011 at 22:34
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$\begingroup$ This is very helpful, thank you. There is still one part of the argument I do not understand---why can we take the actual sup on each vertical section, as opposed to the ess sup? We know $f$ is continuous on each vertical section, but the actual sup is not necessarily equal to the ess sup if the section contains a point such that a neighborhood of that point in the section has zero measure. Can we construct our closed set so that almost every section does not contain such points? (If we simply remove all such points from the domain of $F_n$, is the resulting set still Borel measurable?) Thanks $\endgroup$– zfzf21Commented Jun 17, 2011 at 4:13
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$\begingroup$ Yes, this has to be taken into account. But, as you suggest, a Cantor Bendixon argument allows you to remove the isolated points and change the domain by only a countable set. This, of course, relies on the fact that X is second countable. $\endgroup$ Commented Jun 17, 2011 at 12:05
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For the record, here is a simple answer based on Tonelli/Fubini's theorem.
For arbitrary $c \in \mathbb R$, the set $M_c := \{(x,y) \in X \times X \mid f(x,y) > c\}$ is measurable, thus the indicator function $\chi_{M_c}$ is measurable (everything w.r.t. the product measure). Tonelli/Fubini's theorem tells us that $h \colon X \to [0,\infty]$, $$h_c(x) := \int_X \chi_{M_c}(x,y) \, \mathrm d y \qquad \forall x \in X,$$ is measurable. Finally, $$ \{ x \in X \mid g(x) > c \} = \bigl\{ x \in X \bigm| \mu(\{y \in X \mid f(x,y) > c \}) > 0 \bigr\} = \{ x \in X \mid h_c(x) > 0 \} $$ is measurable for all $c \in \mathbb R$. Thus, $g$ is measurable.
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$\begingroup$ Because $g$ takes values in the extended real number line, I think we also have to prove that the set $\{ x \in X \mid g(x) = +\infty \}$ is measurable. $\endgroup$– AkiraCommented Jul 6, 2022 at 17:43
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1$\begingroup$ @Akira: This set is just the intersection of $\{x \in X \mid g(x) > n\}$ over $n \in \mathbb{N}$. $\endgroup$– gerwCommented Jul 7, 2022 at 5:52
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I am not sure, which one you mean by $\mathrm{ess}\inf$?
- The essential infimum of the (parametric) function $h_i: x\mapsto f(i,x)$, i.e. the element in $X$, that is the almost-sure greatest lower bound of $h_i$?
I think this is the case you tried to prove. I do not know however, how it compares to the second case:
- Or the essential infimum of the set of functions $\{g_i\}_{i\in X}$ with $g_i: x\mapsto f(x,i)$? I.e. the measureable function from $X$ to $\mathbb{R}$, that is the almost-sure greatest lower bound of the set of functions $\{g_i\}_{i\in X}$?
I think in this case, by the definition of the essential infimum of a collection of measurable functions, it is always measureable.
