Is the product of two measurable subsets of $R^d$ measurable in $R^{2d}$?

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!

Solution 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.