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Let $(X,\mathcal{M})$ be a measurable space and $(\mathbb{R},\mathcal{B}(\mathbb{R}))$ be a measurable space of Borel sets on the real line. Let $f_i:X\rightarrow \mathbb{R}$ be given for $i=1,2,\ldots n$. Define $g:X\rightarrow\mathbb{R}^n$ as $g(x)=(f_1(x),\ldots,f_n(x))$. I want to show that $g$ is measurable $\iff$ each $f_i$ is.

I guess for at least one part of this problem, my question would be: If $E\subseteq\mathbb{R}^n$ is open in the product topology, is it true that $E$ is the countable union of open rectangles in $\mathbb{R}^n$? I know this is true in $\mathbb{R}$ but how do I convince myself it holds also in $\mathbb{R}^n$? With this, I can write $g^{-1}(E)=g^{-1}(\cup_{k=1}^\infty R_k)=\cup_{k=1}^\infty g^{-1}(R_n)=\cup_{k=1}^\infty\cap_{j=1}^n\{x\;|\; f_j(x)\in(a_j^k,b_j^k)\}$, hence measurable. And I can also show that if a sigma-algebra is generated by a collection, showing the above only for members in the collection suffices.

mathemagician
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  • Hint: for one implication. If $f$ is measurable and $g$ is continuous, then $g \circ f$ is measurable, that is, continuity doesn't spoils anything. The projections $\pi_j$ are continuous, so if $f$ is measurable, then $f_j = \pi_j \circ f $ is too. – Ivo Terek Oct 20 '14 at 00:30
  • Look at the proofs here: http://math.stackexchange.com/questions/318299/any-open-subset-of-bbb-r-is-a-countable-union-of-disjoint-open-intervals-co

    This can be modified to work for $\mathbb{R}^n$.

     – Alex R. Oct 20 '14 at 00:33

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