Q: If N has Lebesgue measure zero, does the completeness of Lebesgue measure imply NxB is Lebesgue measurable if B is not measurable? My professor gave us the following problem:
Let $N$ be a Lebesgue null set in $\mathbb{R}^q$. We want to show for any arbitrary $B\subseteq \mathbb{R}^p$, the cartesian product $N \times B$  is a Lebesgue null set in $\mathbb{R}^{q+p}$.
In a previous homework I showed that $N \times \mathbb{R}^p$ is a Lebesgue null set if $N$ is a Lebesgue null set. I know that the Lebesgue measure is complete, namely if $A$ is a set of Lebesgue measure zero and if $C \subset A$, then $C$ is Lebesgue measurable with measure zero as well. However, we have only ever said this about "simple" sets (not cartesian products).  
My question is namely:
Even if $B$ is not Lebesgue measurable, does it still hold because of $N \times B \subseteq N \times \mathbb{R}^p$ that $N \times B$ is Lebesgue measurable with measure zero since it is the subset of a null set? Or does this claim not apply to products of sets?

A: Yes, it holds.
Generally, for any measure, a subset of a null set is measurable in the sense of Carathéodory by virtue of having outer measure $0$ (trivially, by monotonicity of outer measure). Where the measure is defined and what sort of set it might be is immaterial.
