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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

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  • $\begingroup$ What do you mean by measurability of a random set? $\endgroup$ – d.k.o. Feb 15 at 3:30
  • $\begingroup$ 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. $\endgroup$ – Zeky Murra Feb 15 at 11:55
  • $\begingroup$ 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. $\endgroup$ – Zeky Murra Feb 15 at 12:02
  • $\begingroup$ 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 $a$? $\endgroup$ – d.k.o. Feb 15 at 19:11
  • $\begingroup$ Yes, Indeed, I'm interested, on computing the probability of a specific set $Q$ happening. $\endgroup$ – Zeky Murra Feb 17 at 15:31
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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 ($\because \mathcal{B}=2^{\mathcal{I}}$).

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