# $A\times B$ are in the same measure if A and B are in different measures

So I have $A \in M_{m_1}$ and $B \in M_{m_2}$ and I am trying to prove that $A\times B \in M_{m_1+m_2}$. I have that $A\times B \in \mathbb R^{m_1+m_2}$. I also have that $A \times B = (A \times \mathbb R^{m_1})\cap(\mathbb R^{m_2} \times B)$ and that $A=H\cup T$ where $|T|=0$ and H is an $F_\sigma$ set.

I have started by using the definition of measurable which, in short, is if $A \subset B$ and $A,B\in \mathbb R^n$ then there is $\epsilon > 0$ such that $|B-A|<\epsilon$ then $A\epsilon M_n$, but I am struggling to get past that. If I use the definition then I can say there is some $X\in \mathbb R^{m_1+m_2}$, but then I have to somehow show that $|X-A\times B|<\epsilon$.

Does anyone have some suggestions as how to proceed?

Edit: $M_m$ is the Lebesgue measure and $A \in M_{m}$ means $A$ is Lebesgue measurable.

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 What are $M_{m_1}$ and $M_{m_2}$? Also, you should proofread this carefully since I think you are mixing up $\in$ and $\subset$. – Nate Eldredge Sep 18 '12 at 23:39