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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.