= Solution
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 $n\geq0$ with $2^{-n}\leq\delta<2^{1-n}$. Let $s_j=2^{-j}\lfloor2^js\rfloor$ and $t_j=2^{-j}\lfloor2^jt\rfloor$. 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 $s,t$. Thus
$$
|\xi_t-\xi_s|\leq2\sum_{j\geq n}A_j
\leq2^{-n\alpha}K_\alpha\leq K_\alpha|t-s|^\alpha.
$$
The <dyadic rationals> are dense in $[0,1]$, so this uniform bound gives a unique continuous extension $X$ to the entire interval, with
$$
\boxed{X_t=\xi_t\quad(t\in D),\qquad
|X_t-X_s|\leq K_\alpha|t-s|^\alpha.}
$$
All 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>.
Back to article page
