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I'm trying to read a proof in the book or Revuz and Yor (prop 1.5 Chapt 13, p.516-517, "continuous martingales and Brownian Motion" 3rd edition), but I'm struggling with a detail. I think it lies between functional analysis and probability theory. The detail is to show that some probability measures are tight under some assumptions, but the difficulty comes from the fact that these measures are not defined on $\mathbb{R}$ but on the Wiener space which is defined as $ C(\mathbb{R}_+,\mathbb{R}) $ with the topology of the compact convergence.

So here is my problem: If $P$ is a probability measure on this Wiener space (let's call it $W$), then we want to show that $\forall \epsilon\in (0,1)\ \ \exists K_{\epsilon}\subset W$ such that $P(K_{\epsilon})>1-\epsilon$

i) the set $\{\ w(0),\ w\in K_{\epsilon} \}$ is bounded in $\mathbb{R}$

ii) for all $N \in \mathbb{N}$,

$\underset{\delta \to 0}{\lim} \underset{w\in K_{\epsilon}}{\sup}V^N(w,\delta)=0$

where

$V^N(w,\delta)=\sup\{ |w(t)-w(t')|; \ |t-t'|\leq \delta\ \text{ and } t,t'\leq N\}$

A little additional information

1) conditions i) and ii) above ensure that $K_{\epsilon}$ is compact for the given topology (by a version of Arzela Ascoli).

2) The topology of compact convergence used here is obtained by using the following metric:

$d(w,w')=\sum_{n=1}^{\infty}2^{-n}\frac{\sup_{t\leq n} |w(t)-w'(t)|}{1+\sup_{t\leq n} |w(t)-w'(t)|}$

I tried to proceed by contradiction taking $\bar{B}(0,1-\delta)$ (ball around the null function for the above metric) as a candidate for $K_\epsilon$. If it is possible to show that any sequence $w_n$ in such a $K$ has a convergent subsequence, then I think I can show the result. But I'm not sure that we necessarily have such a subsequence.

Thank you for your help.

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  • $\begingroup$ Do you know that Brownian paths are lip(a) for all a < 1/2 ? These are actually a very smooth subset of all continuous functions, and picking a group with a uniform but large bound on the lip(a) value should give a precompact set of large probability. $\endgroup$
    – user83457
    Jun 17, 2016 at 12:49
  • $\begingroup$ @micheal Thank you for your answer michael. But I'm a bit confused: in this context the probability measure I have does not necessarily induce a brownian motion. $\endgroup$
    – GWa
    Jun 17, 2016 at 13:24
  • $\begingroup$ Is it assumed that the probability measure has to be absolutely continuous with respect to Wiener Measure? (the law of Brownian motion) $\endgroup$ Jun 17, 2016 at 19:56
  • $\begingroup$ @D.Kelleher, a priori no. The theorem starts with "A sequence $(P_n)$ of probability measures on $W$ is weakly relatively compact if and only if..." But thanks for the suggestion. $\endgroup$
    – GWa
    Jun 18, 2016 at 18:46

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