# Cartesian product of Lebesgue measurable sets is Lebesgue measurable

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.

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

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

Then:

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

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

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.

• Thanks for the explanation. If there is a standard way to show it, how would it look like in this case? – Olsgur Nov 27 '18 at 13:56
• I'm unclear on what exactly do you mean by showing in a standard way? The 'standard argument'? – Keen-ameteur Nov 27 '18 at 18:34
• Yes, I mean the standard argument with the $\sigma$-finite case. – Olsgur Nov 27 '18 at 23:23
• Can you elaborate here. Why can we assume that the measures of $U_1$ and $U_2$ are finite? – user439126 Oct 7 at 3:33
• @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. – Keen-ameteur Oct 7 at 6:02