 Solution

The dyadic slope martingale is uniformly bounded by $L$. The martingale convergence theorem gives an almost-sure limit $\dot{f}$, and dominated convergence gives L1 convergence. Define $\dot{f}=0$ on the null exceptional set and at the endpoint. This gives a measurable representative with $|\dot{f}|\le L$ everywhere.

If $a,b$ are dyadic rationals, then for every sufficiently fine grid the integral telescopes:Pass to the limit by L1 convergence. For arbitrary endpoints choose dyadic $a_j\to a$, $b_j\to b$ with $a_j\le b_j$. The error in the integrals is at most $L(|a_j-a|+|b_j-b|)$, and Lipschitz continuity controls the error in the endpoint values by the same expression. ThereforeThis constructs the bounded integral density directly, without assuming in advance that a Lipschitz function has an almost everywhere derivative.