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.

  • $\begingroup$ Any comments from a downvoter? $\endgroup$ – Ilya Nov 19 '15 at 8:15

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.)

enter image description here

  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.