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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?

BCLC
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Sinusx
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    why does uncountability mean it is not a r.v.? – Lost1 Jun 29 '14 at 12:45
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    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
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    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
  • The first function maps w onto u(w), the second maps constant $a$ onto $\xi(a)$, how is this not composition of two functions. For measure theory to work, any sort of intuive things, such as this example, should all be measurable, else it is a crap theory. – Lost1 Jun 29 '14 at 13:00
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    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
  • @Lost1 Could you give a link? – Sinusx Jun 29 '14 at 13:19
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    http://math.stackexchange.com/questions/315144/two-questions-about-brownian-motion the first question and see the karatzas and shreve reference. this question is different though... – Lost1 Jun 29 '14 at 13:26
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    A counter-example was given by Nate Eldredge at Math Overflow. – Davide Giraudo Jul 26 '14 at 10:23
  • I’m voting to close this question because this question was answered off-site – Jack M Oct 27 '21 at 00:53

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