Let $A$ be a Borel-measurable subset of $\mathbb{R}^n$, and denote by $A_t:=\{u\in \mathbb{R}^{n-1} \mid (t,u)\in A\}$ ($t\in\mathbb{R}$) the $t$-slices of $A$.
If $N$ is a Lebesgue nullset of $\mathbb{R}$, can we conclude that the subset
$$ A_N \, := \, \bigcup_{t\in N}(t,A_t) \quad \text{of } \ \ A $$
is a Lebesgue nullset of $\mathbb{R}^n$? [If necessary, it may be assumed that $A$ is compact.]
(This is not a homework question.)
Yes. This easily follows from Fubini's theorem (or, even easier, from Tonelli's theorem).