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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 $R$?

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I'm not sure if this provides a complete (heh) answer, but here goes:

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.

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  • $\begingroup$ 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? $\endgroup$ – sawe Jun 13 at 16:10
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    $\begingroup$ 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 $\mathcal{B}_1\otimes\mathcal{B}_1.$ $\endgroup$ – cmk Jun 13 at 16:13

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