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An almost sure event is a measurable event of probability one. Countable intersections of such events still have probability one, allowing a conclusion to hold simultaneously for countably many parameter choices. Further deterministic reasoning may then extend it to uncountably many parameters, as in Brownian Hölder regularity or nowhere monotonicity of Brownian motion. An uncountable intersection need not preserve probability one without such additional reasoning.

Here is the one-parameter Kolmogorov continuity theorem. Suppose a real-valued stochastic process on $[0,1]$ satisfies, for some $p>0$, $q>0$ and finite $C$,Then it has a continuous modification of a stochastic process whose paths are Hölder continuous of every positive exponent $\gamma <q/p$. A modification means equality almost surely at each fixed time, not an a priori simultaneous equality at all times. Other compact time intervals follow by rescaling; the vector-valued version uses a norm in place of absolute value.

For the proof fix $0<\gamma <q/p$ and use the dyadic grids $D_n=\{j{2}^{-n}:0\le j\le 2^n\}$. Let $D_n^{\ast }$ be the largest adjacent increment on this grid. The union bound and Markov inequality giveThe right-hand side is summable. The Borel-Cantelli first lemma shows that, outside a null set, an almost-surely finite random constant $K$ can absorb the finitely many early grids so that

We now use dyadic increment chaining. For two distinct dyadic points $s,t$, choose $n$ with $2^{-(n+1)}<|t-s|\le 2^{-n}$. Approximate them from below by $s_j,t_j\in D_j$, for $j\ge n$. Each refinement moves by zero or one grid step, while $s_n,t_n$ are at most two grid steps apart. The sequences eventually equal the original dyadic points. Telescoping giveswhere a deterministic finite $C_{\gamma }$ absorbs the geometric series and $2^{-n}<2|t-s|$. Thus the sample function on the dense union of dyadic grids has a unique Hölder-continuous extension to $[0,1]$. Define this extension to be $\tilde{X}$ on the probability-one event, and set it to zero on the exceptional null set.

For a fixed $t$, take dyadic $t_j\to t$. The moment condition gives $X_{t_j}\to X_t$ in probability, while the constructed extension gives $X_{t_j}\to {\tilde{X}}_t$ almost surely. Uniqueness of a probability limit yields ${\tilde{X}}_t=X_t$ almost surely. This proves the required modification property. Performing the argument for a countable increasing sequence of exponents tending to $q/p$ gives one version with all smaller positive Hölder exponents, since its extensions agree on the dyadic points.

For Brownian motion, Gaussian increments give, for any integer $r\ge 2$,Use $p=2r$ and $q=r-1$. The theorem gives every exponent $\alpha <(r-1)/(2r)=1/2-1/(2r)$. For any $0<\alpha <1/2$, choose $r$ large enough. Intersect probability-one events over integer $r$ and compact intervals $[0,N]$ to obtainThe usual continuous Brownian version has the same property: its continuous paths agree with the constructed version on a countable dense set and hence everywhere, outside one null set. The Hölder constant depends on the compact interval. This is the precise local meaning of Brownian Hölder regularity; it is not a uniform Hölder bound on the entire unbounded time axis.

First construct a centred Gaussian process on the countable dyadic set with covariance function $\mathbb{E}(B_s{B}_t)=\min (s,t)$. These consistent finite-dimensional normal distributions exist because, for coefficients $a_j$,the Kolmogorov extension theorem realizes them on one probability space. Every increment has the normal distribution $N(0,|t-s|)$, so for every finite $p\ge 1$,Use the preceding estimate with $\beta =1/2$, $p>2$ and $0<\alpha <1/2-1/p$. It yields an almost surely Hölder continuous extension from the dense dyadic set to $[0,1]$. Approximate arbitrary times by dyadic times; continuity of the paths and of the Gaussian covariance function preserves all finite-dimensional laws. Disjoint increments are jointly Gaussian with zero cross-covariances, hence independent, with the required variances. Together with $B_0=0$ and continuity, this constructs Brownian motion.

For any $\alpha <1/2$, choose $p$ large enough. Taking a countable increasing sequence of exponents tending to $1/2$ gives simultaneous Hölder regularity of every order $\alpha <1/2$ on a common probability-one set; the same argument on each compact time interval gives local Brownian Hölder regularity. This method does not give the endpoint exponent. In fact that endpoint fails: if a path had a finite $1/2$-Hölder seminorm $K$, all its $2^n$ adjacent normalized increments at level $n$ would have magnitude at most $K$. These are independent standard normal random variables, so for a fixed integer $m$ the probability that their maximum is at most $m$ is $\mathbb{P}(|Z|\le m{)}^{2^n}\to 0$. The event of a finite $K$ is contained in the union over $m$ of events with this bound at every level, each of probability zero. Thus Brownian paths are not $1/2$-Hölder continuous on $[0,1]$ almost surely.

Use the Kolmogorov continuity theorem in its one-parameter form: if on a compact interval a process satisfies $\mathbb{E}|X_t-X_s{|}^p\le C|t-s{|}^{1+\beta }$ for some $p,\beta >0$, it has a modification whose paths are Hölder continuous of every order $\gamma <\beta /p$ on that interval.

For Brownian motion, normal increments give, for every $p>0$,Every such normal moment is finite. Taking $p>2$ yields $\beta =p/2-1$, hence any order below $1/2-1/p$. For a prescribed $0<\alpha <1/2$, choose $p$ with $1/2-1/p>\alpha$.

The continuous modification and the given continuous Brownian motion agree at all rational times on one almost sure event; continuity makes them agree everywhere on the interval. To obtain all exponents and all compact intervals simultaneously, apply the theorem to integer intervals $[0,N]$ and a countable sequence of positive exponents increasing to $1/2$, then intersect these almost sure events. A bound at exponent $\gamma >\alpha$ implies a bound at $\alpha$ on a compact interval. Consequently the Brownian Hölder regularity conclusion iswith finite random constants on one common event of probability one. This uses the usual positive-exponent meaning of Hölder continuity.

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.

The Gaussian process assumption and the covariance formula give a centered Gaussian increment with varianceFor every finite $p>1$, a standard normal variable $Z$ therefore givesFor any $0<\alpha <1/2$, choose $p>\max (2,(1/2-\alpha {)}^{-1})$. Then $\alpha <1/2-1/p$, so parts (a) and (b) give a Hölder continuous function of exponent $\alpha$ as an extension. To obtain one extension with every required exponent, take a countable sequence ${\alpha }_j\uparrow 1/2$ and apply the construction with suitable $p_j$. Intersect the probability-one events. Their continuous extensions agree on the dense set $D$ and hence agree everywhere. For each smaller exponent choose $j$ with $\alpha <{\alpha }_j$; since $|t-s|\le 1$, the bound with ${\alpha }_j$ also implies the bound with $\alpha$. ThusThe exponent constants may depend on $\alpha$ and the sample path. This proves Brownian Hölder regularity for the continuous Gaussian extension without making an uncountable intersection of unrelated full-probability events.

False for a general continuous $f$. For example, take $f(t)=t^{1/3}$. The Brownian Hölder regularity gives $B_t/t^{1/3}\to 0$ almost surely as $t\downarrow 0$, whilealmost surely. The original and translated path laws therefore concentrate on disjoint measurable events and are mutually singular measures.

More generally, the Cameron-Martin theorem gives the exact criterion: translation by $f$ is absolutely continuous precisely when $f$ is an absolutely continuous function, $f(0)=0$, and $f^{\prime }\in L^2[0,1]$.