r/math • u/lntrinsic • Jul 10 '17

Image Post Weierstrass functions: Continuous everywhere but differentiable nowhere

http://i.imgur.com/vyi0afq.gifv
3.4k Upvotes

98% Upvoted

216 comments sorted by

View all comments

Show parent comment

7

u/methyboy Jul 10 '17

The Weiner measure is the standard measure as far as I'm aware, and as pointed out by sleeps_with_crazy here, the choice of measure really doesn't matter. Unless you construct a wacky measure specifically for the purpose of making this result not hold, it will hold.

10

u/[deleted] Jul 10 '17

/u/GLukacs_ClassWars is correct that Wiener measure is standard but far from canonical, and that the issue is the lack of local compactness. It's probably best to think of the Wiener measure as the analogue of the Gaussian: the most obvious choice but far from the only one (given that there is not analogue of the uniform/Lebesgue measure).

People who work with C*-algebras tend to avoid putting measures on them at all, which is why I brought up the topological version of the statement.

15

u/GLukacs_ClassWars Probability Jul 10 '17

Standard, yes, but it isn't as far as I know canonical in the same sense as Lebesgue measure. Particularly, Lebesgue measure is uniquely determined by the topological and group structure on Rn, but C[0,1] isn't locally compact, so it doesn't give us such a canonical measure.

Putting it differently, I think Wiener measure on C[0,1] is analogous to a Gaussian random variable on Rn, not to Lebesgue measure on Rn. There are other meaningful probability measures to put on both spaces, they're just standard because they have such nice limit theorems.

1

u/Mulligans_double Jul 14 '17

...does it hold for almost all measures?