Proving M1 x M1 ⊆ M2 using Lebesgue Measure

[Tedjn]

This is in the context of a homework problem but not directly related.

If M_n is the collection of measurable sets of Rⁿ under Lebesgue measure, what would be the first step in showing that M₁ x M₁ ⊆ M₂. I'm quite convinced it's true, but my knowledge of and ability to work with the Lebesgue measure is very poor, so I have no clue where to begin.

 

[micromass]

OK, the first thing we need to know. How did you define $$M_1\times M_1$$ ??

 

[Tedjn]

Definition of M₁ x M₁ would be the collection of all subsets of R² that is a Cartesian product of two sets in M₁. Sorry I left that out.

 

[micromass]

OK, it's pretty hard to know what assumptions you already know. I'll assume you know the following (correct me if I'm wrong):

1) Let $$\mathcal{B}_1$$ be the Borel sets of $$\mathbb{R}$$. Set $$\mathcal{B}_2$$ to be the $$\sigma$$-algebra generated by all sets of the form $$B\times B^\prime$$, with $$B,B^\prime\in \mathcal{B}_1$$

2) There is a unique $$\sigma$$-finite measure $$\lambda_2$$ on $$\mathcal{B}_2$$ such that $$\lambda_2(B\times B^\prime)=\lambda_1(B)\lambda_1(B^\prime)$$.


Now take E and F two Lebesque measurable sets in $$\mathbb{R}$$. Then there exist Borel sets $$E_i, F_i$$ such that $$E_1\subseteq E\subseteq E_2$$ and $$F_1\subseteq F\subseteq F_2$$ with $$\lambda_1(E_1)=\lambda_1(E_2)$$ and $$\lambda_1(F_1)=\lambda_1(F_2)$$.

Thus we have $$E_1\times F_1\subseteq E\times F\subseteq E_2\times F_2$$ and $$\lambda_2(E_1\times F_1)=\lambda_1(E_1)\lambda_1(F_1)=\lambda_1(E_2)\lambda_1(F_2)=\lambda_2(E_2\times F_2)$$. This implies that $$E\times F\in M_2$$.

I hope this satisfies. Otherwise, just tell me what you don't know or what parts you don't get... (the more information you give me, the more I can help)