= 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 $|\dot f|\leq L$ everywhere.
If $a,b$ are <dyadic rationals>, then for every sufficiently fine grid the integral telescopes:
$$
\int_a^b X_n(x)\,dx=\sum_{[u,v)\subset[a,b)}\bigl(f(v)-f(u)\bigr)=f(b)-f(a).
$$
Pass to the limit by <L1 convergence>. For arbitrary endpoints choose dyadic $a_j\to a$, $b_j\to b$ with $a_j\leq b_j$. 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. Therefore
$$
\boxed{\int_a^b\dot f(x)\,dx=f(b)-f(a)\quad(0\leq a\leq b\leq1),\qquad\|\dot f\|_\infty\leq L.}
$$
This constructs the bounded integral density directly, without assuming in advance that a <Lipschitz function> has an <almost everywhere> derivative.
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