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It is well known one cannot construct uncountable many independent random variables on $([0, 1], \mathcal{B}[0, 1], \lambda)$. ($\lambda$ Lebesgue measure.)

Also, one can clearly construct infinitely many independent random variables of the same distribution given a Brownian motion. Say $B_1, B_2 - B_1, B_3 - B_2, \ldots$

But can one construct uncountably many such variables from a Brownian motion?

I am curious because if we could it would be a simple proof that we cannot construct a Brownian motion on $([0, 1], \mathcal{B}[0, 1], \lambda)$.

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    $\begingroup$ Of course one can construct a BM on $[0,1]$ with Borel $\sigma$-algebra: simply construct first $B_t$ for rational $t$, and then extend by continuity. $\endgroup$
    – Mateusz Kwaśnicki
    Commented Mar 25, 2020 at 16:24
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    $\begingroup$ More generally, one cannot construct uncountably many independent random variables on any standard Borel space. And $C([0,1])$ with its Borel $\sigma$-algebra is a standard Borel space (since $C([0,1])$ with its uniform norm is a complete separable metric space). $\endgroup$
    – Nate Eldredge
    Commented Mar 25, 2020 at 17:14
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    $\begingroup$ Practically speaking, since basically all of everyday probability theory is done on standard Borel spaces, this means that you will never encounter uncountably many independent random variables, so you might as well quit thinking about such horrors :-) $\endgroup$
    – Nate Eldredge
    Commented Mar 25, 2020 at 17:15

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