# Measurability and induced probability distribution of an uncountable family of random variables

I've been trying to figure this out for a long time and I cannot seem to wrap my head around it:

a) We have an index set $$\mathcal{I}=[0,1]$$, and for each $$i\in\mathcal{I}$$, $$x_i$$ is drawn i.i.d. from $$\{0,1\}^3$$ according to some distribution $$F_x$$. We have a measurable map $$a_i(x_i):\{0,1\}^3\to A$$ where $$A$$ is a finite set. We define the set $$Q=\{i\in\mathcal{I}:a_i(x_i)=a\}$$ for some $$a\in A$$ at a given realization of the random process $$(x_i)_{i\in\mathcal{I}}$$. Essentially then, $$Q$$ is a random set that depends on the draws $$(x_i)_{i\in\mathcal{I}}$$. The question i'm trying to address is whether or not I can be sure that the process induces a probability distribution over the family of the sets of the form of $$Q$$.

b) Pretty related, if $$s_\alpha:S\to\mathbb{R}$$ is a random variable, and $$\mathcal{A}$$ is an uncountable index set, how can we make sure that a given realization of the process $$(s_\alpha)_{\alpha\in\mathcal{A}}$$ generate a $$Y=\{s:s=s_\alpha, i\in\mathcal{A}\}$$ that is a measurable set and also induces a probability distribution on it?

If the family indices for these two questions were countable, I know how to do it, using the product topology by having the measurability of each of the coordinates with the sigma algebra generated by cylinders, which I would have due to the sets being formed by realizations of random variables and mapping them through measurable functions, however, when I introduce the uncountability factor in the family of realizations, I want to be sure that the measurability is still preserved. I have seen this in the context of Brownian Motion and more generally, continuous stochastic processes, but I'm not sure if it translates perfectly in what I have, given that I don't necessarily have a filtration.

Many thanks

• What do you mean by measurability of a random set? – d.k.o. Feb 15 at 3:30
• In this case, every realization $(x_i)_{i\in\mathcal{I}}$ generates a set $Q$ defined as above, so ex ante, before such realization, the set $Q$ depends on the underlying sampling space. What I'm interested in knowing is when that realization generates a set that is measurable, for instance, with respect to the Lebesgue Measure. I'm afraid that there could exist some realizations of the process $(x_i)_{i\in\mathcal{I}}$ that could generate a set $Q$ that is actually not measurable. – Zeky Murra Feb 15 at 11:55
• Intuitively for me, if each $x_i$ is a random variable, each of them maps a Borel set back into the sigma algebra generated by $x_i$, so if that is true for each $i$, I think that every set on $\sigma(\cup_{i\in\mathcal{I}} \mathcal{B}(\mathbb{R}))$ should map back into $\sigma(x_i:i\in\mathcal{I})$. If $\mathcal{I}$ is a countable set, then i know that $\sigma(\cup_{i\in\mathcal{I}} \mathcal{B}(\mathbb{R}))=\prod_{i\in\mathcal{I}}\mathcal{B}(\mathbb{R})$, and then I don't have a problem. I'm just struggling to extend this notion to an uncountable index set. – Zeky Murra Feb 15 at 12:02
• Also what do you mean by the distribution over the family of sets of the form $Q$? Are you interested in the distribution of $Q$ for particular value of $a$? – d.k.o. Feb 15 at 19:11
• Yes, Indeed, I'm interested, on computing the probability of a specific set $Q$ happening. – Zeky Murra Feb 17 at 15:31

Let $$(\Omega,\mathcal{F},\mathsf{P})$$ be the underlying probability space. Then $$Q(\omega)$$ need not be (Borel/Lebesgue) measurable for all $$\omega\in \Omega$$. A trivial example would be $$A=\{0,1\}$$ and $$a_i=1\{i\in N\}$$, where $$N$$ is a non-measurable subset of $$[0,1]$$. In this case each $$a_i$$ is measurable (a constant function) and $$Q=N$$ for $$a=1$$.
Let $$\xi_i=a_i\circ x_i$$. Then $$(\xi_i:i\in \mathcal{I})$$ is a stochastic process consisting of mutually independent random variables. You are interested in the sections of $$\Xi=\{(i,\omega):\xi_i(\omega)=a\}$$ for some $$a\in A$$. A sufficient condition ensuring that $$Q(\omega)=\Xi^{\omega}$$ is measurable is the (joint) measurability of $$\xi$$ w.r.t. $$\mathcal{B}\otimes \mathcal{F}$$, where $$\mathcal{B}$$ is a $$\sigma$$-algebra on $$\mathcal{I}$$. Unfortunately, a process like $$\xi$$ is not necessarily measurable (see , for example, Section 19.5 in Stoyanov, Counterexamples in Probability).
By the way, when $$\mathcal{I}$$ is countable, each $$\Xi^{\omega}$$ is automatically measurable ($$\because \mathcal{B}=2^{\mathcal{I}}$$).