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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.

Luis Goncalvez
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    It holds if $X$ and $Y$ are (Lebesgue) measurable. – Daniel Fischer Nov 13 '13 at 23:17
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    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
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    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
  • @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? – Luigi M Sep 07 '17 at 04:34
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    @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 $\lambda_n(R\times (\mathbb{R}^r\setminus Y)) = \lambda_l(R)\cdot \lambda_r(\mathbb{R}^r\setminus Y)$. – Daniel Fischer Sep 07 '17 at 22:35
  • @DanielFischer Thanks for the nice answer. Is there a paper or book for reference to such result? – vulture Jun 22 '19 at 21:52
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    Hi @DanielFischer, I apologize for the bring up the old post again, but I am not entirely sure how you obtained your result in the end for $\lambda_n(R \times (\mathbb{R}^r \setminus Y)) = \lambda_l(R) \cdot \lambda_r(\mathbb{R}^r\setminus Y)$. In particular, without assuming finiteness, how do you get the cancellations from the equation above? – Mathematics_Beginner Oct 18 '22 at 17:42

1 Answers1

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The answer to your particular question is yes, it is true. Two facts are involved:

  1. The product of two Borel $\sigma$-algebras is a Borel $\sigma$-algebra of the product space, i.e., $$\mathcal{B}[\mathbb{R}^n] \times \mathcal{B}[\mathbb{R}^m] = \mathcal{B}[\mathbb{R}^{n\times m}].$$

  2. Using the fact that (1) result holds for elementary sets and (2) Lebesgue measure is constructed using elementary sets, one can prove that the hypothesis holds for all Borel measurable (and even Lebesgue measurable) sets.

I will leave this and this link to Prof. Tao's exercise from his measure theory book in case you are wondering about the tedious proof.

There are several nice results regarding the generalisation of the topic you are speaking of to abstract measure spaces. One can arrange this results in a neat way. Interestingly, to prove them, you must use your clever observation about rectangles.

  1. Having a pre-measure $\lambda_0$ (i.e., a measure defined on simpler set, e.g. rectangles), you can always extend this measure to a full-blown outer measure by covering any set by elementary sets (old good Pythagoras). $$\lambda^*(X) = \inf \left\{\sum_{n=1}^{\infty}\lambda_0(X_n) \left| \begin{array}{l} X \subseteq \bigcup_{n=1}^{\infty}X_n\\ \text{$X_n$ are elementary} \end{array} \right. \right\}$$

  2. Having an outer measure, you can always restrict your attention to the "nice" measurable sets (Caratheodory-Hahn-Kolmogorov Extension) and the resulting restriction will be a measure.

  3. You can repeat the above for the product pre-measures (i.e. your elementary sets are now products of the sets from the original spaces). The resulting product measure (which is a Hahn-Kolmogorov extension of the elementary product measure) will be a unique measure satisfying $$\lambda_{X \times Y}(E \times F) = \lambda_{X}(E) \times \lambda_{Y}(F)$$ and by the Extension Theorem it always exists.

Since you were asking for more resources on this topic: You can read more about the technicalities of the construction and uniqueness of product measures on this blog.

togrultopal
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