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Let $f:X\times Y \mapsto R$ be a measurable function on product space $X\times Y$, where $X$ and $Y$ both are some metric spaces. Define $g(x) := \sup_{y\in Y} f(x,y)$.

[Q.] Is $g$ a measurable function on $X$? If not, what is the sufficient condition to be measurable?

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3 Answers 3

up vote 6 down vote accepted

Lack of measurability of $g$ is a commonly known issue in stochastic optimal control, so this question have been studied quite extensively. The theory is the most rich in case $X$ and $Y$ are not general metric spaces, but are homeomorhpic to Borel subsets of complete separable metric spaces. One usually says that $X,Y$ are (standard) Borel spaces.

Note that $g$ is Borel-measurable on $X$ iff $\{x:g(x)>c\}$ is Borel-measurable for any real $c$, but $$ \{x:g(x)>c\} = \pi_X\{(x,y):f(x,y)>c\} \tag{1} $$ where $\pi_X$ is a projection map onto $X$. Although Lebesgue believed differently, projections of Borel sets may fail to Borel sets (discovered by Souslin and Luzin). As a result, although the example of Nate uses $Y$ as not a Borel space, we could construct a similar example for a Borel $Y$. Take $A$ be any Borel subset of $X\times Y$ such that $A:=\pi_X(B)$ is not Borel, then measurable $f = 1_B$ gives us non-measurable $g = 1_A$.

At the same time, projection of Borel subset of a Borel space is always an analytic set. These sets are in fact often defined as images of arbitrary Borel sets under Borel maps. Analytic subsets of Borel spaces are universally measurable: that is for any Borel probability measure $p$ on $X$, any analytic set $A$ is $p$-measurable, so we can define $p(A)$ unambiguously.

Following $(1)$, let's say that $f$ is upper semianalytic whenever $\{(x,y):f(x,y)>c\}$ is analytic for all $c\in \Bbb R$. Then by $(1)$ we obtain that $g$ is also upper semianalytic, so this class of functions is closed under taking such suprema. Clearly, every Borel function is upper semianalytic, but not vice-versa - e.g. $g$ from the second paragraph.

If you are still interested just in the Borel measurability, you need some continuity assumptions. For example, if $f$ is lower semicontinuous, then so is $g$, and if $f$ is upper semicontinuous and then so is $g$ if $Y$ is compact. Semicontinuous functions are Borel measurable, of course. I think you may want to take a look at Section 7.5 of the Stochastic Optimal Control book by Bertsekas and Shreve, freely available at MIT.

FWIW, this situation is very much related to the following one. A map $\phi:X\to Y$ between two Borel spaces is Borel measurable iff its graph is Borel measurable. However, for some Borel $B\subseteq X\times Y$ there may not be Borel $\phi$ whose graph is contained in $B$ over $\pi_X(B)$. There still does exist a universally measurable $\phi$ with such property, even if $B$ just analytic, not necessarily Borel. For a Borel $\phi$ to exist, certin continuity assumptions are needed such as compactness of $B$.

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Thanks for the above comprehensive answer on this matter. Section 7.5 of the book [Bertsekas, Shreve] is helpful. However, the book is on discrete case of optimal control. I wonder if there is any analogous one, proving dynamic programming in continuous case of stochastic optimal control. –  user79963 Jul 26 at 15:40

Nope. Take, say, $X = [0,1]$ and $Y$ your favorite non-measurable subset of $[0,1]$. Let $f(x,y) = 1$ if $x=y$ and $0$ otherwise. Then $g = 1_Y$ which is not a measurable function on $X$.

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Hi Nate, I just wonder in this case what is the measurable set in $X\times Y$? –  user79963 Jul 25 at 7:10
@user79963: What measurable set are you talking about? Are you wondering why $f$ is measurable? Note that $f = 1_A$ where $A = \{(y,y) : x \in Y\}$. You can easily check that $A$ is actually closed in $X \times Y$, and in particular measurable. –  Nate Eldredge Jul 25 at 7:29

Inspired from Nate's answer, the answer is NO in general, and here is one example. Let $X = Y = [0,1]$, and $f(x,y) = I_{\{x = y\}} I_{\{y\in M\}}$, where $M$ is some non-Lebesgue-measurable set and $I$ is the indicator. $f$ is Lebesgue measurable since $\{f=1\} \subset \{(x,y): x = y\}$ is zero set. However, $g(x) = I_M(x)$, which makes it non-Lebesgue-measurable.

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I guess this depends on what is meant by "measurable". Your function $f$ is Lebesgue measurable but not Borel. –  Nate Eldredge Jul 25 at 7:30
@NateEldredge Thanks, it is corrected. –  user79963 Jul 26 at 9:18
@NateEldredge If $M$ is taken non-Borel as before, is it easy to show $f$ of above construction is non-Borel? –  user79963 Jul 26 at 9:32
Define $h : [0,1] \to [0,1]^2$ by $h(x)=(x,x)$, which is continuous, hence Borel. Then $f \circ h= I_M$ which is not Borel. –  Nate Eldredge Jul 26 at 17:28
@NateEldredge yes, thanks a lot –  user79963 Jul 27 at 12:07

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