If $\mu(B)=0$ then $\mu_y(B_x)=0$

Question

Let $B_x$ be the $x$-section of a $\mu_x\otimes \mu_y$-measurable set $B$, where $\mu_x\otimes \mu_y$, which I will call $\mu$, is the Lebesgue extension of the product measure $\mu_x\times \mu_y$ (both measures being $\sigma$-additive complete measures defined on $\sigma$-algebras of subsets of $X$ and $Y$, respectively), defined by $$B_x=\{y\in Y:(x,y)\in B\}$$ I read that if $\mu(B)=0$ then for almost all $x\in X$ $\mu_y(B_x)=0$ but, although it sounds very intuitive, I cannot prove it to myself. Could anybody explain this interesting fact? I $\infty$-ly thank you!

Answer 1

Recall that $\mu(B)=\int_X\mu_Y(B_x)\mathrm d\mu_X(x)$ hence, if $\mu(B)=0$ then $\mu_Y(B_x)=0$ for $\mu_X$-almost every $x$.

To show this, consider $A_n=\{x\in X\mid n\mu_Y(B_x)\geqslant1\}$ and note that, for every $n$, $0=n\mu(B)\geqslant\int_{A_n}n\mu_Y(B_x)\mathrm d\mu_X(x)\geqslant\mu_X(A_n)$ hence $\mu_X(A_n)=0$. This implies that $\mu_X(A_0)=0$ where $A_0=\bigcup\limits_{n\geqslant1}A_n=\{x\in X\mid\mu_Y(B_x)\ne0\}$, QED.