# Borel measurability of a subset of a product space

Let $X$ and $Y$ be compact metric spaces and let $\mathcal B_X$ and $\mathcal B_Y$ be their respective Borel $\sigma$-algebras.

Let $\mu$ be a Borel probability measure on $X$ and let $\mathcal B^*_X$ be the $\mu$-completion of $\mathcal B_X$.

If $E$ belongs to the product $\sigma$-algebra $\mathcal B^*_X\otimes\mathcal B_Y$, does there exist a $\mu$-null set $A$ in $\mathcal B_X$ such that $E\cup(A\times Y)\in\mathcal B_X\otimes\mathcal B_Y$?

Let $\mathscr{C}$ be the collection of $E\in\mathcal{B}_X^*\otimes \mathcal{B}_Y$ for which:

1. There exists $A\subseteq X$ null with $E\cup (A\times Y)\in\mathcal{B}_X\otimes\mathcal{B}_Y$.
2. There exists $A\subseteq X$ null with $E\setminus (A\times Y)\in\mathcal{B}_X\otimes\mathcal{B}_Y$.

Then it suffices to show that $\mathscr{C}$ is a $\sigma$-algebra containing the rectangles $A\times B$, $A\in\mathcal{B}_X^*$, $B\in\mathcal{B}_Y$.

For rectangles, use the fact that we can approximate, both on the inside and the outside, elements of $\mathcal{B}_X^*$ by elements of $\mathcal{B}_X$.

For countable unions, property 1. is easy to deal with. For 2., suppose $E_n\in\mathscr{C}$, and for each $n$ choose $A_n$ satisfying $2$. Then $\bigcup_m A_m$ is null in $\mathcal{B}_X$ and $$\bigg(\bigcup_n E_n\bigg)\setminus\bigg(\bigg(\bigcup_m A_m\bigg)\times Y\bigg)=\bigcup_n\bigg(E_n\setminus \bigg(\bigcup_m \big(A_m\times Y\big)\bigg)\bigg)$$

For a given $n$, $E_n\setminus\big(\bigcup_m (A_m\times Y)\big)=\big(E_n\setminus(A_n\times Y)\big)\setminus\big(\bigcup_m(A_m\times Y)\big)$, which belongs to $\mathcal{B}_X\otimes\mathcal{B}_Y$, so their union also belongs to it.

For complements, given $E\in\mathscr{C}$, use property 1. for $E$ to obtain property 2. for $E^c$ and vice-versa, so $E^c\in\mathscr{C}$.

• Thanks for this. Am I correct that it remains to show that $\mathcal C$ is closed under countable intersections? – mo15 May 22 '16 at 9:19
• @mo15 This is not necessary, in general, since any class of sets which is closed under countable unions and complements is closed under countable intersections (because $\bigcap_n A_n=\left(\bigcup_n A_n^c\right)^c$), but you should mention this, if your definition of $\sigma$-algebra has this condition as well. – Luiz Cordeiro May 24 '16 at 6:59
• You should probably add that the null sets $A$ in properties 1. and 2. should be Borel. Up to that: nice answer! – PhoemueX May 24 '16 at 7:10
• @Luiz. I see. In this case the answer seems perfectly fine to me. – mo15 May 25 '16 at 14:39
• @PhoemueX. I do not see why the null sets should be added to properties 1 and 2. – mo15 May 25 '16 at 14:40