4
$\begingroup$

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!

$\endgroup$
  • 1
    $\begingroup$ Use $\sigma$-finiteness of $\mathbb{R}^d$. $\endgroup$ – copper.hat Apr 14 '13 at 2:26
  • $\begingroup$ Need more details... $\endgroup$ – van abel Apr 14 '13 at 3:18
  • $\begingroup$ You can prove this in stages, first intervals, then open sets, then $G_\delta$ sets, then... $\endgroup$ – leo Apr 14 '13 at 5:16
4
$\begingroup$

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.

| cite | improve this answer | |
$\endgroup$
  • 1
    $\begingroup$ 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 $\endgroup$ – LeastSquaresWonderer Mar 26 '18 at 3:46
  • $\begingroup$ 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 at 15:47

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.