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Suppose that $E_1,E_2$ are two (Lebesgue) measurable subsets of $R^d.$ Define $E:=E_1\times E_2=\left\{(x,y)|x\in E_1,y\in E_2\right\}.$ Can we say that $E$ is a Lebesgue measurable subset of $R^{2d}$?

Recall (cf. Stein's Real analysis), that a subset $E$ of $R^d$ is called measurable if, for any $\varepsilon>0$, there exists an open set $O\supset E$, such that $$m_*(O\setminus E)<\varepsilon,$$ where $m_*(\cdot)$ is the outer measure, which is defined by $m_*(E)=\inf\sum\limits_{i=1}^\infty |Q_i|$, the infimum is taken over all almost countable closed cubic coverings $\bigcup_{i=1}^\infty Q_i\supset E.$

I have tried to do it by definition, since $$(O_1\times O_2)\setminus(E_1\times E_2)=(O_1\setminus E_1)\times O_2\bigsqcup E_1\times(O_2\setminus E_2)$$ we can see that the conclusion is positive for $O_1,O_2$ with finite measure, but since $E_1, E_2$ may have an infinite measure, this is what I am asking for solving!

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    $\begingroup$ Use $\sigma$-finiteness of $\mathbb{R}^d$. $\endgroup$
    – copper.hat
    Apr 14, 2013 at 2:26
  • $\begingroup$ Need more details... $\endgroup$
    – van abel
    Apr 14, 2013 at 3:18
  • $\begingroup$ You can prove this in stages, first intervals, then open sets, then $G_\delta$ sets, then... $\endgroup$
    – leo
    Apr 14, 2013 at 5:16

2 Answers 2

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I'll expand copper.hat's point into an answer. To prove that the product of measurable sets in $\mathbb{R}^d$ is measurable, it suffices to show that the product of measurable set of finite measure in $\mathbb{R}^d$ is measurable (this generalizes to arbitrary $\sigma$-finite measure spaces).

Proof: Let $E_{1},E_{2}$ be given as above. Define $E_{1,N} = E_{1} \cap B(0,N)$, the intersection of $E_{1}$ with the ball of radius $N$. This is still measurable, as it is the intersection of two measurable sets, and it is has finite measure by monotonicity of measure, as $B(0,N)$ has finite measure. Similarly define $E_{2,N}$. As the poster demonstrated, we know that $E_{1,N} \times E_{2,N}$ is measurable for any choice of $N$. But now we note that

$$E_{1} \times E_{2} = \bigcup_{N \in \mathbb{N}}(E_{1,N} \times E_{2,N})$$

So $E_1 \times E_2$ is the countable union of measurable sets, and hence measurable.

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    $\begingroup$ Why can you say : By hypothesis we have proved that $E_{1,N}\times E_{2,N}$ is measurable for any $N$? Is it because the cartesian product of the two intersections is contained in the cartesian product of the two balls, which is itself measurable? Or why? To what hypothesis are you referring? Also, why does it generalize to non-finite? Thank you $\endgroup$ Mar 26, 2018 at 3:46
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    $\begingroup$ I have the same question @LeastSquaresWonderer, were you able to understand why it is ok to ay "by hypothesis we have proved that ..." $\endgroup$
    – funmath
    Feb 5, 2020 at 15:47
  • $\begingroup$ Yes, why by hypothesis is $E_1 \cap B_N(0) \times E_2 \cap B_N(0)$ measurable ? $\endgroup$
    – homosapien
    Jul 13, 2022 at 20:08
  • $\begingroup$ @elchanon Solomon how do you have by hypothesis that set is measurable ? $\endgroup$
    – homosapien
    Aug 17, 2022 at 20:36
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    $\begingroup$ I changed it from "by hypothesis" to "as the poster has shown," which is less confusing. $\endgroup$ Aug 18, 2022 at 8:28
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It would appear I'm over a decade late(I have taken some inspiration from Solomon)

Take x $\in \mathbb{R}^n$ and y $\in \mathbb{R}^m$ and $E_1 \subset \mathbb{R}^n$ , $E_2 \subset \mathbb{R}^m$ to be measurable.

We know that if a characteristic function is Lebesgue measurable then the set that defines it must also be Lebesgue measurable. We will use this fact and the following equivalence.

$\chi_{E_1 \times E_2}(x,y)$ =$\chi_{E_1 \times \mathbb{R}^m}(x,y).\chi_{\mathbb{R^n} \times E_2}(x,y)$

First, lets see that $\chi_{E_1 \times \mathbb{R}^m}(x,y)$ is measurable. We begin by re-writing the funciton as a pointwise limit

$\chi_{E_1 \times \mathbb{R}^m}(x,y)$ = $\lim_{k\to\infty} \chi_{E_1 \times B(0,k)}(x,y)$

If $E_1$ is an open set, then we are finished, since $E_1 \times B(0,k)$ is a Borel set and is therefore measurable. If $E_1$ is $G_\delta$ then $E_1$ = $\cap_{l \in \mathbb{N}}$ $U_l$ where $U_l$ is a open set for all $l \in \mathbb{N}$. In this case $\chi_{E_1 \times \mathbb{R}^m}^{-1}(\{ 1 \})$ = $\cap_{l \in \mathbb{N}}$ $\chi_{U_l \times \mathbb{R}^m}^{-1}(\{1\})$ which is a countable intersection of measurable sets and therefore measurable in and of itself.

The case in which m($E_1$) = 0 is reduced to the case of a $G_\delta$. Finally, we arrive at the general case: Take $E_1$ to be a measureable set, we can re-write it as follows: $E_1 = G-F$ where $G$ is a $G_\delta$ such that m($E_1$)=m($G$), $E_1$ $\subset$ G and F is a measure zero set. Now, using the previous cases we can say that :

$\chi_{E_1 \times \mathbb{R}^m} ^{-1}(\{ 1 \}) = \chi_{G-F \times \mathbb{R}^m} ^{-1}(\{1\})$ = $\chi_{G \times \mathbb{R}^m - F \times \mathbb{R}^m} ^{-1}(\{1\})$ = $\chi_{G \times \mathbb{R}^m}^{-1}(\{ 1 \}) \cap \chi_{F \times \mathbb{R}^m}^{-1}(\{ 1 \})^{C}$

is a measurable set by properties of $\sigma$-algebras.

Therefore $\chi_{E_1 \times B(0,k)}(x,y)$ is a Lebesgue measurable function. The proof for $\chi_{B(0,k)\times E_2}(x,y)$ is analog. there fore $\chi_{E_1 \times \mathbb{R}^m}(x,y).\chi_{\mathbb{R^n} \times E_2}(x,y)$ is a measurable function since it is the product of measurable functions over the same domain. This proves that $E_1 \times E_2$ is measurable

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