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In Section 13 of Probability and Measure by Billingsley, it has been shown that for a measurable space $(F, \mathcal{F})$, $g:F\rightarrow \mathbb{R}^m$ and $g_i: F\rightarrow \mathbb{R}$ with $g(x) = [g_1(x),\cdots, g_m(x)], \forall x \in F$, $g$ is $\mathcal{F}/\mathcal{B}(\mathbb{R}^m)$ measurable if and only if $g_i$ is $\mathcal{F}/\mathcal{B}(\mathbb{R})$ measurable.

More generally, suppose $\{ (G_i, \mathcal{G}_i), i=1,\cdots,m \}$ are measurable space, $(\prod_{i=1}^m G_i, \mathcal{G})$ is also a measurable spaces, but $\mathcal{G}$ may not be $\prod_{i=1}^m \mathcal{G}_i$. for $g:F \rightarrow \prod_{i=1}^m G_i$, and $g_i: F \rightarrow G_i$ with $g(x) = [g_1(x),\cdots, g_m(x)]$. I wonder what are some conditions under which $g$ is measurable if and only if $g_i: F \rightarrow G_i, i=1, \cdots, m$ are measurable? Can the fact $g(x)=[g_1(x),⋯,g_m(x)]$ be used in your conditions?

How will your conditions be used to explain the example when $(G_i, \mathcal{G_i}) = (\mathbb{R}, \mathcal{B}(R))$ and $(\prod_{i=1}^m G_i, \mathcal{G}) = (\mathbb{R}^m, \mathcal{B}(\mathbb{R}^m))$ mentioned earlier? Thanks and regards!

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    You cannot define $f_i$ in terms of $f$ this way! And I see no reason to define $g_i$ this way either. It makes more sense to define $g_i$ and then define $g = g_1 \times \dotsb \times g_m$.2011-10-11
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    Sorry, it does make sense to define $g_i$ in terms of $g$. It is a matter of taste. But the $f_i$ are in fact ill-defined.2011-10-11
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    @AndréCaldas: Edited that part out. Thanks!2011-11-02
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    See the first pages in Folland's "Real Analysis" for an interesting (and somewhat tricky) extension of this statement to infinite product of measurable spaces.2011-11-02
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    @MarkSchwarzmann: Is it somewhere in Chapter 2 Integration? I can't find it.2011-11-02
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    It is proposition 2.4 in chapter 2. You'll find, however, that the tricky part is the definition of the product sigma-algebra for infinitely many measurable spaces. This is done in chapter 1, somewhere in the beginning.2011-11-03
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    @MarkSchwarzmann: Thanks! It seems to be what I have been looking for.2011-11-03

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