# 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?

• why does uncountability mean it is not a r.v.? – Lost1 Jun 29 '14 at 12:45
• it appears to me this is just a composition of two measurable function (aka random variables), so it is measurable (aka a random variable) – Lost1 Jun 29 '14 at 12:48
• Thank you for your comment. If you consider $\{ \xi _a \}_{a \in [0;1]}$ as a function of $a$, it does not have to be measurable. – Sinusx Jun 29 '14 at 12:57
• As it stands, you have not specified your sample space. I see your concern and it is not as easy as i thought. I actually asked a similar question before about a bm. – Lost1 Jun 29 '14 at 13:09
• A counter-example was given by Nate Eldredge at [Math Overflow][1]. [1]: mathoverflow.net/questions/173037/… – Davide Giraudo Jul 26 '14 at 10:23