Suppose that $E_1,E_2$ are two measurable (Lebesgue) 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 $\epsilon>0$, there exist an open set $O\supset E$, such that $$m_*(O\setminus E)<\epsilon,$$ 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 infinite measure, this is what I am asking for solving!

  • 2
    $\begingroup$ 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

1 Answer 1


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 $E_{2,N}$. By hypothesis, we have proved 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.

  • 3
    $\begingroup$ 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
  • 2
    $\begingroup$ 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

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