Past exam of the mathematics course of the University of Cambridge / 2005 / iii / Paper 34 / 2

The almost sure submartingale convergence theorem states that a discrete-time submartingale $Y_n$ satisfying ${\sup }_n\mathbb{E}{Y}_n^{+}<\infty$ converges almost surely to a finite integrable limit. In particular, a martingale with ${\sup }_n\mathbb{E}|M_n|<\infty$ converges almost surely to an integrable $M_{\infty }$, and $\mathbb{E}|M_{\infty }|\le {\sup }_n\mathbb{E}|M_n|$ by Fatou's lemma. This is the martingale convergence theorem. Uniform integrability gives the stronger conclusion of L1 convergence; it is not implied by bounded first moments alone.

Take $\Omega =[0,1)$ with its Borel sigma-algebra and Lebesgue measure of total mass one. Let ${\mathcal{F}}_n$ be the finite sigma-algebra generated by the half-open dyadic intervals of length $2^{-n}$. Set $X_0=f(1)-f(0)$ if the indexing begins at one. Each $X_n$ is adapted to this filtration, and if $L$ is a Lipschitz constant of $f$, then $|X_n|\le L$, so every coordinate is integrable.

On a parent interval $[u,u+2^{-n})$, conditional expectation averages the two equally likely child values. Their average isThus $\mathbb{E}[X_{n+1}\mid {\mathcal{F}}_n]=X_n$, establishing the dyadic slope martingale. One can equivalently include $1$ as a separate null atom without changing any probability law.

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.