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It has been shown in here that a set of positive measure may not contain a product set of positive measure. However, the examples are typically quite irregular. I was wondering whether the statement holds if the original set is a preimage of some regular function.

More precisely, let $f:[0,1]\times \mathbb{R} \to \mathbb{R}$ be a Caratheodory function, in the sense that $f$ is Borel measurable with respect to the first component, and continuous with respect to the second component. If $E=\{(x,y)\mid f(x,y)\not=0\}$ has a positive product measure, then can we find $A\subset [0,1]$ and $B\subset \mathbb{R}$ with positive measures such that $A\times B\subset E$?

If the above statement is still false, could we add more regularity of $f$ in $x$ such that the statement holds? It is clear that $f$ being continuous is a sufficient condition.

John
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1 Answers1

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Your claim is correct.

Theorem 28.7 from Kechris descriptive set theory states: If $X$ is standard Borel, $Y$ is Polish, $A \subset X \times Y$ Borel s.t. every $x$-section of $A$ is open. If $(V_n)_n$ is any basis for the open sets of $Y$, it holds $A = \bigcup_n B_n \times V_n$, with $B_n$ Borel in X.

Clearly, your $E$ is Borel (as Caratheodory functions are jointly Borel) and its $x$-sections are open. Hence we can write $E = \bigcup_n B_n \times V_n$, with $B_n$ as in the theorem.

If you assertion was wrong, in particular $\mu(B_n \times V_n)=0$ for all $n \in \mathbb{N}$. This implies $\mu(E)=0$, which is a contradiction to your assumptions.

user940347
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