Question: Prove that if $F$ is a measurable function on $\mathbb{R}^2$ into $\mathbb{R}$, such that for each $x_{0} \in \mathbb{R}$, the equation $F(x_0, y) = 0$ has a countable number of solutions y, then $Z = \{(x,y) : F (x, y) = 0\}$ has measure $0$.
Solution attempt: we can write $Z$ as $\cup_{x \in \mathbb{R}}\{(x,y): F (x, y) = 0\}$ and each such subset is now countable by hypothesis. I am trying to write this as the the countable union of countable sets, then the result follows since we can bound the measure with a zero series. But how to write $Z$ as the countable union of countable sets? Suggestions?
$Z$ is measurable as $Z=F^{-1}\{0\}$ and $F$ is measurable. This implies that $\chi_Z$ is measurable. Now by Fubini-Tonelli Theorem we can write: $m(Z)=\int_{\mathbb{R}^2} \chi_Z dm=\int_{\mathbb{R}}(\int_{\mathbb{R}} \chi_{Z'} dy)dx= \int_{\mathbb{R}}m(Z')dx=\int_{\mathbb{R}}0=0$, where $Z'$ is $Z$ but for a fixed $x_0$.