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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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    $\begingroup$ It holds if $X$ and $Y$ are (Lebesgue) measurable. $\endgroup$ – Daniel Fischer Nov 13 '13 at 23:17
  • $\begingroup$ 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? $\endgroup$ – Luis Goncalvez Nov 13 '13 at 23:26
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    $\begingroup$ 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. $\endgroup$ – Daniel Fischer Nov 13 '13 at 23:37
  • $\begingroup$ @DanielFischer sorry for commenting this old post but do you mind expanding on how you should that $\mathcal{A}_R$ is closed for taking complements? $\endgroup$ – Luigi M Sep 7 '17 at 4:34
  • $\begingroup$ @LuigiM Let $Q_k = [-k,k]^r$. Then $$\lambda_n(R\times Q_k) = \lambda_n R\times (Q_k\cap Y)) + \lambda_n(R\times (Q_k\setminus Y)) = \lambda_l(R)\cdot \lambda_r(Q_k) = \lambda_r(R)\cdot\bigl(\lambda_r(Q_k\cap Y) + \lambda_r(Q_k\setminus Y)\bigr).$$ Let $k\to \infty$ and use the continuity from below of measures to obtain $\lambda_n(R\times (\mathbb{R}^r\setminus Y)) = \lambda_l(R)\cdot \lambda_r(\mathbb{R}^r\setminus Y)$. $\endgroup$ – Daniel Fischer Sep 7 '17 at 22:35

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