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If $X \subseteq \mathbb{R}^{l}$ and $Y \subseteq \mathbb{R}^{r}$ with $l + r = n$, is it true that $\lambda_{n}(X \times Y) = \lambda_{l}(X) \cdot \lambda_{r}(Y)$ (where $\lambda_{m}$ is the Lebesgue measure in $\mathbb{R}^{m}$)? I know the result holds if $X$ and $Y$ are rectangles, but would like to know if it is true for arbitrary sets.

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It holds if $X$ and $Y$ are (Lebesgue) measurable. – Daniel Fischer Nov 13 '13 at 23:17
Any idea on how to prove it? It follows directly for rectangles from the definition, but I don't know any property to handle Cartesian products in general? – Luis Goncalvez Nov 13 '13 at 23:26
For a rectangle $R$, consider $\mathcal{A}_R = \{ Y : \lambda_n(R\times Y) = \lambda_l(R)\cdot \lambda_r(Y)\}$. That contains all rectangles, and is a $\sigma$-algebra, so it's the set of all Lebesgue-mesurable subsets of $\mathbb{R}^r$. Then for a measurable $Y\subset \mathbb{R}^r$, consider $\mathcal{L}_Y = \{ X : \lambda_n(X\times Y) = \lambda_l(X)\cdot \lambda_r(Y)\}$. By the previous, it contains all rectangles, also, it is a $\sigma$-algebra. – Daniel Fischer Nov 13 '13 at 23:37

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