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?

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    $\begingroup$ why does uncountability mean it is not a r.v.? $\endgroup$ – Lost1 Jun 29 '14 at 12:45
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    $\begingroup$ it appears to me this is just a composition of two measurable function (aka random variables), so it is measurable (aka a random variable) $\endgroup$ – Lost1 Jun 29 '14 at 12:48
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    $\begingroup$ 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. $\endgroup$ – Sinusx Jun 29 '14 at 12:57
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    $\begingroup$ 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. $\endgroup$ – Lost1 Jun 29 '14 at 13:09
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    $\begingroup$ A counter-example was given by Nate Eldredge at [Math Overflow][1]. [1]: mathoverflow.net/questions/173037/… $\endgroup$ – Davide Giraudo Jul 26 '14 at 10:23

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