# Measurability of a mapping to a product space and of its component mappings

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 $g = g_1 \times \dotsb \times g_m$. – André Caldas Oct 11 '11 at 20:17
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. – André Caldas Oct 11 '11 at 20:35
@AndréCaldas: Edited that part out. Thanks! – Tim Nov 2 '11 at 20:16
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. – Mark Nov 2 '11 at 22:28
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. – Mark Nov 3 '11 at 0:23

## 1 Answer

For question 1:

The mapping $g$ is measurable iff $g^{-1}(\mathcal{G}) \subset \mathcal{F}_1$.

The same way, $g_i$ is measurable iff $g_i^{-1}(\mathcal{G}_i) \subset \mathcal{F}_1$. So, to have measurability of $g$ implying that of $g_i$, you need to have $$g_i^{-1}(\mathcal{G}_i) \subset g^{-1}(\mathcal{G})$$ for all $i = 1, \dotsc, m$. That is, $$\sigma\left(g_i^{-1}(\mathcal{G}_i),\, i = 1, \dotsc, m\right) \subset g^{-1}(\mathcal{G}).$$

The same way, in order to have measurability of $g_i$ implying that of $g$, you need to have $$g^{-1}(\mathcal{G}) \subset \sigma\left(g_i^{-1}(\mathcal{G}_i),\, i = 1, \dotsc, m\right).$$

In question 2, $f_i$ is not well defined.

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Thanks, Andre! In summary, "$g_i$ are measurable iff $g$ is measurable", iff $g^{-1}(\mathcal{G}) = \sigma\left(g_i^{-1}(\mathcal{G}_i),\, i = 1, \dotsc, m\right).$ I wonder if this condition can be represented in terms of both $\mathcal{G}$ and $\prod_{i=1}^m \mathcal{G}_i$, in a similar way to the Euclidean space example. – Tim Oct 12 '11 at 0:11
Forget my previous comment if it doesn't make sense. But: (1) Your conditions haven't used the fact that $g(x_1) = [g_1(x_1),\cdots, g_m(x_1)]$. So how can it be used to form some conditions? (2) How can your conditions be used to explain "it is true when $(G_i, \mathcal{G_i}) = (\mathbb{R}, \mathcal{B}(R))$ and $(\prod_{i=1}^m G_i, \mathcal{G}) = (\mathbb{R}^m, \mathcal{B}(\mathbb{R}^m))$"? – Tim Nov 2 '11 at 19:43