In the book of David Williams on "Probablity with Martingales" it is proven that a real-valued function which is $\sigma(X)$- measurable, is in fact a measurable function of $X$, where $X$ is any random variable on some probability space. The proof of this is fairly simple, and presentend in an Appendix.

I have a problem with a remark on this fact. Let us consider an index set $I$. Then, for some random variables $$Y_i : (\Omega, \mathcal{F}) \rightarrow (S, \mathcal{S}) , i \in I$$ we can construct the random vector $Y: (\Omega, \mathcal{F}) \rightarrow (S^I, \mathcal{S}^I),$ with $\pi_i \circ Y = Y_i,$ where $\mathcal{S}^I$ is the product sigma-algebra. Now in general (also for uncountable $I$), it seems to me that $\sigma(Y) =\sigma(Y_i | i \in I),$ by the nature of the product sigma-algebra. So I would find that if $X$ is $\sigma(Y_i | i \in I)$ - measurable, than it can be written as function of the $Y_i$'s (i.e. of the random vector $Y$).

So if $(S, \mathcal{S}) = (\mathbb{R}, \mathcal{B}(\mathbb{R})),$ when $I$ is finite, this function would be a borel function on $\mathbb{R}^I,$ because the borel sigma-algebra on $\mathbb{R}^I$ is the product sigma-algebra. If $I$ is uncountable it is stated that this is not true anymore.

My question is very simple. In what way do we define the borel sigma-field on $\mathbb{R}^I,$ when $I$ is not finite? I thought the only reasonable way to do that is considering the product topology on the product set. But why don't I find the product sigma-algebra by doing this?


The Borel $\sigma$-algebra on $\Bbb R^I$ is $\sigma(\mathcal T^I)$, where $\mathcal T^I$ is the product topology on $\Bbb R^I$. When $I$ is uncountable, this is strictly larger than the product $\sigma$-algebra $\mathcal B(\Bbb R)^I$.

  • $\begingroup$ Thank you. That answers my question. $\endgroup$ – Kore-N Jan 8 '16 at 7:25

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