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[SOLVED] Borel and σ-algebra related question

Juliayaho

New member
Apr 11, 2013
13
Let B_k be the σ-algebra of all Borel sets in R^k. Prove that B_(m+n) = B_m x B_n.

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Opalg

MHB Oldtimer
Staff member
Feb 7, 2012
2,702
Let \(\displaystyle B_k\) be the σ-algebra of all Borel sets in \(\displaystyle R^k\). Prove that \(\displaystyle B_{m+n} = B_m \times B_n\).
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 \(\displaystyle B_m\times B_n \subset B_{m+n}\).

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 $B_{m+n}\subset B_m\times B_n$.