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](https://math.stackexchange.com/questions/851197/uncountable-family-of-random-variables). 


*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.



Uncountable family of random variables
$\alpha$ 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.


