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For a function on the dyadic rationals in $[0,1]$, let $A_n$ be its largest absolute adjacent increment on the level-$n$ grid. If $K_{\alpha }<\infty$ for $\alpha >0$, the function extends uniquely to a Hölder continuous function withChoose $n$ with $2^{-n}\le |t-s|<2^{1-n}$. The level-$n$ left approximations differ by at most two grid steps, and each finer approximation adds at most one level increment. Thus the difference is bounded by $2{\sum }_{j\ge n}{A}_j\le 2^{-n\alpha }{K}_{\alpha }$. For a stochastic process with $\Vert {\xi }_t-{\xi }_s{\Vert }_p\le C|t-s{|}^{\beta }$, the bound $\Vert A_n{\Vert }_p\le C{2}^{-n(\beta -1/p)}$ makes this series converge in the Lp norm for $0<\alpha <\beta -1/p$.

Fix an exponent in the range from (a), and work on the single probability-one event where $K_{\alpha }<\infty$ and all dyadic values are finite. We give the dyadic increment chaining argument. For dyadic $s<t$, put $\delta =t-s$ and choose $n\ge 0$ with $2^{-n}\le \delta <2^{1-n}$. Let $s_j=2^{-j}\lfloor 2^{j}s\rfloor$ and $t_j=2^{-j}\lfloor 2^{j}t\rfloor$. The level-$n$ approximations are at most two grid steps apart, so their difference is bounded by $2A_n$. At each subsequent level an approximation either stays fixed or moves by one adjacent level increment. Because $s,t$ are dyadic, these approximations eventually equal $s,t$. ThusThe dyadic rationals are dense in $[0,1]$, so this uniform bound gives a unique continuous extension $X$ to the entire interval, withAll these equalities hold on that one event, not merely one event per dyadic time. Define $X$ to be the zero path on its null complement. Each $X_t$ is measurable as the limit of the random variables at deterministic left dyadic approximations. Alternatively, their measurable linear interpolations converge uniformly to $X$, which also shows measurability as a random element of $C([0,1])$. This is the continuous extension version of a continuous modification.

This is an infinite binary expansion with digits in $\{0,1\}$. If the usual expansion of $\alpha$ has infinitely many $1$ digits, list their positions as $b_1<b_2<\cdots$. If it terminates with a final term $2^{-k}$, replace that term bythis also covers every dyadic rational. For $\alpha =1$, use $b_n=n$. Thus every $\alpha \in (0,1]$ has the required form.

The representation by an infinite strictly increasing sequence is unique. If two sequences first differ at exponent $k$, one sum contains $2^{-k}$ and the other does not. The latter's entire possible tail is at most ${\sum }_{j>k}{2}^{-j}=2^{-k}$, while the former has $2^{-k}$ plus its own nonempty tail, so the sums cannot agree.