Prove $B_{m+n}=B_m * B_n$ if $B_k$ is the sigma-algebra of all Borel sets in $\mathbb R^k$ Let $B_k$ be the sigma-algebra of all Borel sets in $\mathbb R^k$. How can we prove that $B_{m+n}=B_m * B_n$? I am a beginner in analysis, hope to seek help here.
 A: Welcome!
I think it should follow from the fact that the "boxy" open sets of the form $$(a_1, b_1) \times \ldots \times (a_n, b_n)$$for $a_i, b_i \in \mathbb{R}$ generate the sigma algebra on $B_n$. This follows for example from showing that they generate the (standard) topology. 
We combine that with a lemma: let $X,Y$ be sets, $A_i, B_j$ collections of subsets of $X,Y$. Then $$\sigma( \sigma(A_i) \times \sigma(B_j)) = \sigma( A_i \times B_j)$$where $\sigma(-)$ denotes sigma algebra generated by, and $U_\alpha \times V_\beta$ denotes all pairwise products, viewed naturally as subsets of $X \times Y$. 
$\supseteq$ isn't terrible. For $\subseteq$, if suffices to show that anything of the form $\sigma(A_i) \times \sigma(B_j)$ lies in $\sigma( A_i \times B_j)$. To see that in turn, by intersecting it suffices to show anything of the form WLOG $\sigma(A_i) \times Y$ lies in $\sigma(A_i \times B_j)$. To see this, $\sigma(A_i \times B_j)$ contains $\sigma(A_i \times \{Y\})$ which should be $\sigma(A_i) \times \{Y\}$ by minimality. 
