Past exam of the mathematics course of the University of Cambridge / 2016 / iii / Paper 201 / 4 / d

Let $\mu$ be the given Brownian motion law on $C[0,1]$. Choose $\alpha <\beta <1/2$ and an integer $p$ with $p(1/2-\beta )>1$. For the largest adjacent level-$n$ increment $A_n$, Brownian scaling and the Markov inequality givewhere $c_p=\mathbb{E}|\mathcal{N}(0,1){|}^p<\infty$. These probabilities are summable, so the Borel-Cantelli lemma gives a finite random $K$ such that $A_n\le K{2}^{-n\beta }$ for every $n$, after absorbing the finitely many exceptional levels.

For $s<t$, choose $m$ with $2^{-m}\le t-s<2^{1-m}$. Their left dyadic approximations at level $m$ differ by at most two grid steps; each finer approximation adds at most one adjacent increment. Continuity and dyadic increment chaining therefore giveThus paths are $\beta$-Hölder and hence $\alpha$-Hölder on $[0,1]$. This direct proof supplies the relevant Brownian Hölder regularity, also obtainable from the Kolmogorov continuity theorem.

The full-measure subset $E_{\alpha }=C^{\alpha }[0,1]$ is measurable in $C[0,1]$, because continuity identifies its seminorm with a countable supremum:Its finiteness event is a countable union of countable intersections of coordinate-measurable events. The sigma-algebra specified on $E_{\alpha }$ is precisely the trace of the coordinate sigma-algebra on $C[0,1]$. Define $\nu (A)=\mu (A)$ for trace-measurable $A\subseteq E_{\alpha }$; equivalently restrict the Brownian law to its full-measure subset. Since $\mu (E_{\alpha })=1$, no renormalization is needed, and all finite-dimensional coordinate laws are unchanged. This is the required probability measure:The construction uses the requested coordinate sigma-algebra and makes no assumption about a separable Hölder-norm topology.