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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?)

flag	asked Jun 16 2011 at 13:20 zfzf21 43●2

Now... the question makes sense in a measure space with no topology. – Gerald Edgar Jun 16 2011 at 14:18

Juris Steprans · Accepted Answer · 2011-06-16 13:52:08Z UTC

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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 ess sup of the vertical section of f. So modifying F on a null set yields that g is measurable.

answered Jun 16 2011 at 13:52

Juris Steprans
2,338●4●13

	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? – zfzf21 Jun 16 2011 at 20:39
	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. – Juris Steprans Jun 16 2011 at 22:34
	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 – zfzf21 Jun 17 2011 at 4:13
	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. – Juris Steprans Jun 17 2011 at 12:05

Johannes · Answer 2 · 2013-02-25 09:37:57Z UTC

I am not sure, which one you mean by $\mathrm{ess}\inf$?

1) 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:

2) 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.

Measurability of essential supremum of function of two variables

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