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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?

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    $\begingroup$ What do you mean by "simple" sets? The cartesian product is a subset of ${\bf R}^{q+p}$ just as any other set. If anything, it is simpler than most other sets. $\endgroup$ – tomasz May 29 '17 at 19:22
  • $\begingroup$ Yes, you are right. For some reason I didn't think of that. So this then holds? I was under the impression given the structure of the question that the proof wouldn't be so trivial. $\endgroup$ – Jocelyn Franzen May 29 '17 at 22:06
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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.

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