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- Thread starter Juliayaho
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- Feb 7, 2012

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Let \(\displaystyle U_k\) be the set of all open subsets of \(\displaystyle R^k\). Then \(\displaystyle U_m\times U_n \subset U_{m+n} \subset B_{m+n}\) (where \(\displaystyle U_m\times U_n\) means all sets of the form $S\times T$ with $S\in U_m$ and $T\in U_n$). From that, you should be able to deduce that \(\displaystyle B_m\times B_n \subset B_{m+n}\).Let \(\displaystyle B_k\) be the σ-algebra of all Borel sets in \(\displaystyle R^k\). Prove that \(\displaystyle B_{m+n} = B_m \times B_n\).

For the reverse inclusion, use the fact that every open subset of $R^{m+n}$ is a countable union of sets in $U_m\times U_n$. Thus $U_{m+n}\subset B_m\times B_n$ and hence $B_{m+n}\subset B_m\times B_n$.