# Uncountable family of random variables

Let $\{ \xi _a \}_{a \in [0;1]}$ be a family of independent uniformly distributed on $[0;1]$ random variables on some probability space $(\Omega, \mathscr{F},P)$, indexed by a continuous parameter. Let $u$ be an independent of $\{ \xi _a \}_{a \in [0;1]}$ uniformly distributed on $[0;1]$ random variable. For $\omega \in \Omega$, define the map

$$\alpha : \Omega \to \mathbb{R}, \ \ \\ \ \alpha (\omega) = \xi_{u(\omega)} (\omega).$$

Is $\alpha$ a random variable?

I think the answer is negative, since the family $\{ \xi _a \}_{a \in [0;1]}$ is uncountable. How could I prove this? I asked this question on MSE, link.

Motivation. Let $\lambda$ be the Lebesgue measure on $[0;1]$, and define another measure $\#$ by

$$\# A = \text{ the number of points in } A, \ \ \ \ A \in \mathscr{B}([0;1]).$$ The measure $\#$ is not $\sigma$-finite.

I was wondering whether one could define a random variable of the form

$$N(\eta, [0;1]),$$

where $N$ is a Poisson point process on $[0;1]^2$ with intensity $\# \times \lambda$, and $\eta$ is an independent of $N$ Poisson random measure on $[0;1]$ with the intensity $\lambda$. I think if the answer on the question about $\alpha$ is negative, then $N(\eta, [0;1])$ is not a random variable either.

• This doesn't answer your question directly, but even leaving the problem of the existence of non-measurable functions to logicians, we can, at least, conclude that even if the RV in question exists, nothing can be said about it from just the information given. Indeed, just take any RV $f(\omega)$ and replace $\xi_a$ by $f$ on the set $u(\omega)=a$ of zero measure. Nothing will change but we now get $\alpha=f$, so the definition is pretty useless as given no matter what. Jul 1 '14 at 12:17

$\alpha$ need not be a random variable.
The most natural choice for $(\Omega, \mathcal{F}, P)$ is product space: let $\Omega = [0,1]^{[0,1] \cup \{2\}}$ and for $A \subset [0,1] \cup 2$, let $\pi_A : \Omega \to [0,1]^A$ be the projection map. (For $a \in [0,1] \cup 2$, we let $\pi_a$ denote $\pi_{\{a\}} : \Omega \to [0,1]^{\{a\}} = [0,1]$.) Let $\mathcal{F}$ be the product $\sigma$-field on $\Omega$, which is by definition the smallest $\sigma$-field that makes all $\pi_a : \Omega \to [0,1]$ measurable. It is then not hard to show that every $B \in F$ is of the form $B = \pi_A^{-1}(C)$ for some countable $A \subset [0,1] \cup \{2\}$ and some $C \subset [0,1]^A$ which is measurable with respect to the product $\sigma$-field on $[0,1]^A$. That is, a measurable subset of $\Omega$ can only look at countably many coordinates.
Now for each countable $A \subset [0,1] \cup \{2\}$, let $\mu_A$ be the measure on $[0,1]^A$ which is the infinite product of Lebesgue measure, and for $B = \pi_A^{-1}(C) \in \mathcal{F}$, set $\mu(B) = \mu_A(C)$. It's easy to verify that $\mu$ is well defined and countably additive (note that $\bigcup_n \pi_{A_n}^{-1}(C_n)$ is of the form $\pi_A^{-1}(C)$ where $A = \bigcup_n A_n$ is countable). Moreover, $\mu$ is a probability measure and, under $\mu$, the $\pi_a$ are iid $U(0,1)$ random variables. So $(\Omega, \mathcal{F}, \mu)$ satisfies the hypotheses, taking $\xi_a = \pi_a$ and $u = \pi_2$.
We then define $\alpha$ as you say, via $\alpha(\omega) = \pi_{\pi_2(\omega)}(\omega)$. Then $\alpha$ is certainly not a random variable. If it were, then $\alpha^{-1}([0,1/2])$ would be of the form $\pi_A^{-1}(C)$ for $A$ countable and $C \subset [0,1]^A$. Let $b \in [0,1] \setminus A$. Define $\omega$ via $\omega(2)=b$, $\omega(b)=1$, and $\omega(a) = 0$ for all $a \in [0,1] \setminus \{b\}$. Define $\omega'$ similarly but with $\omega'(b)=0$. Then $\alpha^{-1}([0,1/2])$ contains $\omega'$ but not $\omega$, whereas $\pi_A^{-1}(C)$ must contain both of $\omega,\omega'$ or neither.
This doesn't rule out the possibility of being able to choose some more exotic $(\Omega, \mathcal{F}, P)$ (which should perhaps be left to the set theorists). But even if you could, I agree with fedja that no good can come of it. For example, working formally, you might observe that for each finite $A \subset [0,1]$, $\alpha$ is independent of $\{\xi_a : a \in A\}$ (since almost surely $u \notin A$), whence $\alpha$ is independent of $\{\xi_a : a \in [0,1]\}$, which appears to be absurd.