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.
 A: $\alpha$ 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.
