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)