# Borel sets and product spaces.

Let $$(F_1,M_1)$$ and $$(F_2,M_2)$$ be two measurable space, the product $$M_1\otimes M_2$$ is by definition the smallest $$\sigma$$-algebra which contains all rectangle $$A\times B$$, with $$A \in M_1$$ and $$B \in M_2$$. If we add measures $$\mu_1$$ and $$\mu_2$$, there is a way to construct a measure space $$(F_1\times F_2,M_1\otimes M_2,\mu)$$ such that $$\mu(A\times B) = \mu_1(A)\mu_2(B)$$, and this construction is unique in the $$\sigma$$-finite case. The main application of all this abstract stuff was the introduction of Lebesgue measure on $$R^2$$, which however did require an additional completion step (even if $$M_1$$ and $$M_2$$ were Lebesgue, $$M_1 \otimes M_2$$ turned out not to be Lebesgue).

Now for my question: what exactly is $$M_1\otimes M_2$$ before completion? If $$M_1$$ and $$M_2$$ are the collections of the Borel sets of the line, is $$M_1\otimes M_2$$ the class of Borel set of the plane? My intuition suggest that this is the case, but I can't think of a simple way to prove it. And if $$M_1$$ and $$M_2$$ are the classes of all Lebesgue measurable subsets of $$R$$?

For the first question in your final paragraph, it's exactly what it's defined to be; I'm not sure what else to really say about that. For your second question in the final paragraph, it depends. If $$X$$ and $$Y$$ are $$\sigma$$-compact metric spaces, then $$\mathcal{B}_{X\times Y}=\mathcal{B}_X\otimes\mathcal{B}_Y.$$ so if $$\mathcal{B}_1$$ is the Borel $$\sigma$$-algebra on $$\mathbb{R},$$ then $$\mathcal{B}_n=\bigotimes_{j=1}^n\mathcal{B}_1,$$ where $$\mathcal{B}_n$$ denotes the Borel $$\sigma$$-algebra on $$\mathbb{R}^n.$$ However, this is not true for Lebesgue measure; if $$\mathcal{L}_1$$ is the Lebesgue $$\sigma$$-algebra on $$\mathbb{R}$$, then $$\mathcal{L}_n\neq \bigotimes_{j=1}^n\mathcal{L}_1,$$ so we need to complete this to get $$\mathcal{L}_n$$. Also, this completion is the same as completing $$\mathcal{B}_n.$$ Note that we can define the Lebesgue measure on $$\mathbb{R}^n$$ in the Caratheodory manner, so this is a way to view the completion.
• Thanks for your answer, is there a simple proof I can look up for the relation $B_2=B_1⨂B_1$? Also, does $B_2=L_1⨂L_1$ hold? – sawe Jun 13 at 16:10
• To answer your first question, I was referencing Michael Taylor's book, Measure Theory and Integration (the 6th chapter). To answer your second question, no, $\mathcal{L}_1\otimes\mathcal{L}_1$ is larger than $\mathcal{B}_1\otimes\mathcal{B}_1.$ – cmk Jun 13 at 16:13