Let $A \subset \mathbb{R^m}$, $B \subset \mathbb{R^n}$ and $\lambda_n$ the Lebesgue measure on $\mathbb{R^n}$.

How to prove that if $A$ and $B$ are Lebesgue measurable sets then also $A \times B$ is Lebesgue measurable.

For showing that the cartesian product is Lebesgue measurable, I tried to use:

$A \times B$ is Lebesgue measurable if $\varepsilon>0$, and an open set $U$ and a closed set $V$ exist such that $V \subset A \times B \subset U$ and $\mathcal{L^{n+m}}(U$\ $V)<\varepsilon$.

I don't know how to continue from here and how to conclude that the cartesian product is also Lebesgue measurable.

  • $\begingroup$ You are probably discussing measurable with respect to $\mathcal{L}{n+m}$ measure on $\mathbb{R}^{n+m}$, or am I wrong? $\endgroup$ – Keen-ameteur Nov 26 '18 at 17:27
  • $\begingroup$ Let $A \subset \mathbb{R^m}$, $B \subset \mathbb{R^n}$ and $\lambda_n$ , thats all what is given, so I suppose it could be right. $\endgroup$ – Olsgur Nov 26 '18 at 18:54
  • $\begingroup$ What does $\mathcal{L}^{n+m}(U\setminus V)< \epsilon$ then? It must be the Lebesgue measure on $\mathbb{R}^{n+m}$. $\endgroup$ – Keen-ameteur Nov 26 '18 at 19:15
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    $\begingroup$ By the way are you familiar with the terms $\sigma$ algebra and product $\sigma$ algebra? $\endgroup$ – Keen-ameteur Nov 26 '18 at 19:22
  • $\begingroup$ I know the definitions, but I haven't really worked with them yet, especially not with the product sigma algebra. $\endgroup$ – Olsgur Nov 26 '18 at 19:54

As a prelude, let me first say that $A\times B$ is clearly in the product $\sigma$ algebra, since the product sigma algebra $\mathcal{F}\otimes \mathcal{G}$ of two measurable spaces is the sigma algebra generated by measurable 'rectangles'. i.e:

$\mathcal{F}\otimes \mathcal{G}= \sigma\Big( \{ E\times F: E\in \mathcal{F}, F\in \mathcal{G} \} \Big)$

Going by this definition, your set $A\times B$ is Lebesgue measurable in $\mathbb{R}^{n+m}$. However going by your criterion, recall first that for $A\in \mathcal{L}^n$ and $B\in \mathcal{L}^m$, we have:

$\lambda_n \otimes \lambda_m (A\times B)= \lambda_n(A)\times \lambda_m(B)$

Since $A,B$ are Lebesgue measurable, there exist $U_1,U_2$ open and $V_1,V_2$ closed such that:

$U_1\supseteq A \supseteq V_1$ , $U_2\supseteq B \supseteq V_2$ while $\lambda_n(U_1\setminus V_1)\leq \frac{1}{2} \tilde{\epsilon}$ and $\lambda_m(U_2\setminus V_2)\leq \frac{1}{2} \tilde{\epsilon}$

Notice that $U_1\times U_2\supseteq A\times B \supseteq V_1\times V_2$ while $U_1\times U_2$ is open and $V_1 \times V_2$ is closed. Since you can write:

$U_1\times U_2= \Big( U_1\times V_2 \Big) \sqcup \Big( U_1 \times (U_2 \setminus V_2) \Big)= \Big( V_1\times V_2 \Big) \sqcup \Big( (U_1\setminus V_1)\times V_2 \Big) \sqcup \Big( U_1 \times (U_2\setminus V_2) \Big) $


$\lambda_{n+m}\Big( (U_1\times U_2) \setminus (V_1 \times V_2) \Big)= \lambda_{n+m}\Big( (U_1\setminus V_1)\times V_2 \Big) + \lambda_{n+m}\Big( U_1 \times (U_2\setminus V_2) \Big)= $

$= \lambda_n(U_1 \setminus V_1)\cdot \lambda_m(V_2)+ \lambda_n(U)\cdot \lambda_m(U_2 \setminus V_2)$

If you assume $\lambda_n(U_1), \lambda_m(U_2)\leq M$, then for $\tilde{\epsilon}= \dfrac{\epsilon}{2M}$, you've shown that the difference is of measure less than $\epsilon$. Otherwise, there is a standard argument (which I will elaborate if you request) of going from the finite measured case to the $\sigma$-finite case.

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  • $\begingroup$ Thanks for the explanation. If there is a standard way to show it, how would it look like in this case? $\endgroup$ – Olsgur Nov 27 '18 at 13:56
  • $\begingroup$ I'm unclear on what exactly do you mean by showing in a standard way? The 'standard argument'? $\endgroup$ – Keen-ameteur Nov 27 '18 at 18:34
  • $\begingroup$ Yes, I mean the standard argument with the $\sigma$-finite case. $\endgroup$ – Olsgur Nov 27 '18 at 23:23
  • $\begingroup$ Can you elaborate here. Why can we assume that the measures of $U_1$ and $U_2$ are finite? $\endgroup$ – user439126 Oct 7 at 3:33
  • $\begingroup$ @user439126 This follows from the sigma-finite argument which is not written. Since $\mathbb{R}^{n+m}$ is sigma finite, we can partition the space into a countable union of finite measures spaces, and use sigma sub-addittivity. $\endgroup$ – Keen-ameteur Oct 7 at 6:02

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