Suppose that $X$ is a Polish space and $\mathcal{E}$ is the $\sigma $-algebra of subsets of $X$ with the property of Baire. Consider the product $\sigma $-algebra $\mathcal{E}\otimes \mathcal{E}$ on $X\times X$, which is the coarsest $\sigma $-algebra on $X\times X$ making the canonical projections $\mathcal{E}$-measurable.

QUESTION: Is it true that $\mathcal{E}\otimes \mathcal{E}$ contains all meager subsets of $X\times X$? (This would imply that $\mathcal{E}\otimes \mathcal{E}$ coincides with the $\sigma $-algebra of subsets of $X\times X$ with the property of Baire.)


Assuming that $X$ is an uncountable Polish space, the desired conclusion that $\mathcal{E} \otimes \mathcal{E}$ contains all sets with the property of Baire is not true. In fact, analytic sets have the property of Baire and it is a variant of a result due to Mansfield and Rao that no universal analytic set belongs to $\mathcal{E} \otimes \mathcal{E}$. See Miller, Measurable rectangles, Theorem 1 for a proof of this. In my answer to a related question on math.SE there are more explanations and further references.

  • $\begingroup$ NB: this also tells us how to find a meager set that doesn't belong to $\mathcal{E} \otimes \mathcal{E}$: take a universal analytic set $A$ in $X \times X$ and take an open set $U$ such that $M = A \mathbin{\Delta} U$ is meager. Then $M \notin \mathcal{E} \otimes \mathcal{E}$. $\endgroup$
    – Martin
    Mar 20 '13 at 23:10

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