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Let $X$ and $Y$ be measurable spaces, and $A \subseteq X\times Y$ is a measurable subset of the product space. For any $y\in Y$ let $A_y = \{x\in X: (x,y)\in A\}$ be the $y$-section of $A$. Under which condition for any probability measure $p$ on $X$ there exists $y\in Y$ such that $p(A_y) > 0$?

Motivation: I was thinking of a zero-sum game with a payoff of $(p\otimes q)(A)$, and would like to consider cases when neither of players can fix this measure to be $0$ over $A$ just by choosing his own marginal measure.

What I did: For simplicity I assume that any singleton is measurable. Necessary condition is that $A$ has the full projection, otherwise on can put $p$ to be a Dirac measure outside of projection of $A$. Furthermore, let's sat that $A$ has a countably complete projection property (CCPP) if there exists $y_1, y_2, \dots\in Y$ such that $\bigcup_n A_{y_n} = X$. Clearly, CCPP is a sufficient condition, e.g. if $X$ is Lindelof and $A$ has full projection and open sections. This condition is not necessary, though. For example, consider $X = Y = [0,1]$, and let $A_y = F + y$ where $C$ is a nowhere dense set with positive Lebesgue measure. Here $+$ is a cyclic shift addition over $[0,1]$. Then any countable union of $A_y$ is nowhere dense again, hence never $X$ itself. At the same time, one can use Fubini's theorem to show the desired property.

What I expect from the bounty: answer to the original question - Under which condition for any probability measure $p$ on $X$ there exists $y\in Y$ such that $p(A_y) > 0$? I am looking for a complete/partial characterization besides the fact already stated here, or a reference on the subject.

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  • $\begingroup$ Any comments from a downvoter? $\endgroup$ – Ilya Nov 19 '15 at 8:15
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I have some not very checked ideas which may help someone to obtain the bounty.

I have two approaches to the sufficient condition.

The first is to show that the family $\mathcal A=\{A_y: A_y$ is a measurable subset of $X\}$ is sufficiently big. For instance, Theorem 17.10 from “Classical Descriptive Set Theory” by Alexander S. Kechris, implies that for each probability Borel measure $p$ on a metrizable space $X$ and each measurable set $M\subset X$ such that that $p(M)>0$ there exist a closed subset $N’$ of the set $M$ such that $p(N’)>0$. It seems that we can remove from $N$ a dense open (in $X$) subset $D$ with a small measure, so the obtained set $N=N’\setminus D$ will be a closed nowhere dense subset of the space $X$ such that $p(N)>0$. So, the sufficient condition would be “each closed nowhere dense subset $N$ of the space $X$ can be covered by a countable subfamily of the family $\mathcal A$”. We can strengthen this results if the set $N$ with $p(N)>0$ can be chosen even smaller (in some sense).

The second is based on Fubini Theorem and seems to be a generalization of Example constructed by Ilya and PhoemueX.

For each point $x\in X$ let $A^x = \{y\in Y: (x,y)\in A\}$ be the $x$-section of $A$. Let $q$ be a probability measure on $Y$. Since $A$ is a measurable subset of the product space $X\times Y$, for almost all points $x\in X$ and $y\in Y$ the sets $A^x$ and $A_y$ are measurable and

$$ \int_Y p(A_y)dq(y)=\int_{X\times Y} 1_A dp(x)dq(y)= \int_X q(A^x)dp(x).$$

If $p(A_y)=0$ for each $y$ then the left hand side equals $0$, if $q(A^x)>0$ for each $x$ then the right hand side is greater than $0$ and we obtain a contradiction. Thus a sufficient condition for the positive answer is an existence of a probability measure $q$ on $Y$ such that $q(A^x)>0$ for each $x\in X $. It seems that in Ilya and PhoemueX’s Example is for the case when $q$ is Lebesgue measure. Another sufficient condition can be obtained by atomic measure, that is when there exits a countable subset $S$ of $Y$ such that $A^x\cap S\ne\varnothing$ for each $x\in X$ (or, at least, for each $x\in X$ such that the Lebesgue measure of the set $A^x$ is zero.)

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  • $\begingroup$ Thanks for generalizing mentioned ideas. If no better answer appears, you'll get the bounty. Also, thanks for sharing the question on the seminar - any interesting discussions there? $\endgroup$ – Ilya Nov 19 '15 at 8:43
  • $\begingroup$ @Ilya Thanks. There were three questions at the seminar, and yours was the last. So, as usually, I did not fit in time. Therefore I am going to talk about your question tomorrow. So if you have some related problems, I can consider them at the seminar. Unfortunately, I have no big expectations for this talk, because none of the seminar participants has deep knowledge of descriptive set theory (we mainly deal with topological algebra). Also I asked my scientific consultant, who is more acquainted with the subject, about your question, but this was not very helpful. $\endgroup$ – Alex Ravsky Nov 24 '15 at 11:56
  • $\begingroup$ So I think that you may ask your question at MathOverflow. Also I think that some restrictions on the set $A$ may simplify the problem. $\endgroup$ – Alex Ravsky Nov 24 '15 at 11:56
  • $\begingroup$ That's fine, it was just our of idle interest, and there was some progress on the problem. Perhaps, indeed I'll ask this on MO once I have time. $\endgroup$ – Ilya Nov 24 '15 at 12:15

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