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This is an interesting question that has stumped the entirety of my measure theory class, including the professor:

Prove or disprove:

Let $(X,\mathcal A)$, $(Y,\mathcal B)$ be measure spaces.

Let $f$ be an $\mathcal A$ $\times$ $\mathcal B$ measurable nonnegative function, and let $g(x)$ $=$ $sup_{y \in Y}f(x,y)$, with $g(x)<\infty$ for all $x$. Is $g(x)$ necessarily an $\mathcal A$-measurable function?

We all feel the answer is no, given that slices are measurable, and sups of measurable functions are only guaranteed to be measurable over a countable index. We think the correct answer is to start with a nonmeasurable set $S$ in $X$, and to try to build a set $T$ in $Y$ that makes $S \times T$ measurable in $\mathcal A \times \mathcal B$, but we suspect this is quite difficult with no further guidance.

Any ideas?

Johnny Apple
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2 Answers2

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Here is an example in which the resulting function $g$ is nonmeasurable, but only slightly so:

We let $X=Y=[0,1]$, endowed with the Borel $\sigma$-algebra. We let $A\subseteq [0,1]^2$ be a Borel set such that the projection $\pi_X(A)$ to the first coordinate is not Borel (and hence only analytic) and $f=1_A$ be the indicator function of $A$. Then $g$ is the indicator function of $\pi_X(A)$ and hence not Borel measurable. However, $g$ will be measurable with respect to every complete probability measure on $X$.

This problem is actually a typical example for why one uses analytic sets in optimization problems. For a nice introduction, see Some Measurability Results for Extrema of Random Functions over Random Sets by Stinchcombe and White.

Michael Greinecker
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  • Can you find A explicitly? – Johnny Apple Jan 29 '14 at 19:32
  • @JohnnyApple Basically yes, but one usually constructs such a set in a different space, such as the Baire space $\mathbb{N}^\mathbb{N}$ and then transfers them to other Polish spaces using isomorphism theorems. Constructions can be found in any book on descriptive set theory, which however is not an easy subject in itself. – Michael Greinecker Jan 29 '14 at 21:49
  • An explicit example that is at least easy to state is taking $X=Y$ to be $C_{[0,1]}$. Then there is a Borel set with projection being the set of all continuous functions which fail to be differentiable at some point. The original reference can be found here (in German). – Michael Greinecker Jan 29 '14 at 21:55
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The following is Corollary 2.13 from Crauel, Random Probability Measures on Polish Spaces:

Suppose that $f\colon X\times\Omega\to \mathbb{R}$ is product measurable, where $(\Omega,\mathcal{F})$ is a measurable space and $X$ is a Polish space equipped with its Borel $\sigma$-algebra. Let $C\colon\Omega\to 2^X$, $\omega\mapsto C(\omega)$, be any set valued mapping such that $\textrm{graph}(C)$ is product measurable, then $\omega\mapsto\sup_{x\in C(\omega)}f(x,\omega)$ is measurable with respect to the universal completion of $\mathcal{F}$.

In particular, one may then take $C(\omega)\equiv X$.

Mark
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