Borel and σ-algebra related question

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SUMMARY

The discussion focuses on proving that the Borel σ-algebra of the product space, denoted as B_(m+n), is equal to the product of the Borel σ-algebras of the individual spaces, B_m and B_n. The proof utilizes the properties of open sets in R^k, specifically that U_m × U_n is contained within U_(m+n), leading to the conclusion that B_m × B_n is a subset of B_(m+n). Conversely, it is established that every open subset of R^(m+n) can be expressed as a countable union of sets from U_m × U_n, confirming that B_(m+n) is also a subset of B_m × B_n.

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Juliayaho
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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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Juliayaho said:
Let $$B_k$$ be the σ-algebra of all Borel sets in $$R^k$$. Prove that $$B_{m+n} = B_m \times B_n$$.
Let $$U_k$$ be the set of all open subsets of $$R^k$$. Then $$U_m\times U_n \subset U_{m+n} \subset B_{m+n}$$ (where $$U_m\times U_n$$ 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 $$B_m\times B_n \subset B_{m+n}$$.

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 $B_{m+n}\subset B_m\times B_n$.
 

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