Past exam of the probability, measure and martingales course of the University of Oxford / 2023 / 2 / c

Each $Z_n$ is a bounded Borel measurable function of $X_n$, including the specified value at $X_n=1$. Therefore $(Z_n)$ is an adapted process for $({\mathcal{F}}_n)$ and consists of integrable random variables.

Fix an atom of a sigma-algebra in ${\mathcal{F}}_n$ with positive probability, corresponding to the dyadic interval $[u,u+\delta )$, with $\delta =2^{-n}$. By the uniform distribution, each half has conditional probability $1/2$. The conditional expectation of $Z_{n+1}$ on that atom is thereforeThe middle values cancel. On the zero-probability atom the conditional expectation can be chosen as zero, agreeing with our version of $Z_n$. Henceso $(Z_n)$ is a martingale. This is the dyadic slope martingale associated with the Lipschitz function $f$.

The uniformly integrable martingale convergence theorem states that a uniformly integrable martingale $(U_n)$ has an integrable terminal random variable $U_{\infty }$ such that $U_n\to U_{\infty }$ almost surely and in the L1 norm, andThe last identity follows by passing to the limit of a sequence in $\mathbb{E}[U_m{1}_A]=\mathbb{E}[U_n{1}_A]$ for $A\in {\mathcal{F}}_n$ and $m\ge n$, using convergence in L1.

Part (b)(iii) verifies uniform integrability for $(Z_n)$, soMoreover $|Z_{\infty }|\le L$ almost surely. In this particular problem the common bound also yields convergence in Lp for every finite $p\ge 1$, including convergence in L2: the dominated convergence theorem applies to $|Z_n-Z_{\infty }{|}^p\le (2L{)}^p$. General uniform integrability alone does not assert these higher-power conclusions.

Choose $Z_{\infty }$ as the pointwise limit of a sequence of $(Z_n)$ where it exists, and as zero otherwise. The convergence event and every $Z_n$ are $\sigma (X)$-measurable, so this is an exactly $\sigma (X)$-measurable version, bounded by $L$. Part (a)(ii), the Doob-Dynkin lemma, supplies a Borel measurable function $g$ on $[0,1]$ such that $Z_{\infty }=g(X)$. Applying scalar clipping to $g$ with bounds $[-L,L]$ preserves that equality and ensures $|g(u)|\le L$ everywhere.

Part (c)(ii) and the conditional expectation formula from part (b)(iv) now giveSince the indices $n$ form a countable family, these identities may be taken on one common full-probability event.

For fixed $n,k$, both the slope and the displayed average are constant on the atom $A_{n,k}$, whose probability is $2^{-n}>0$. An equality holding almost surely of these constants forces the deterministic equalitySum over $0\le k<j$. The telescoping sum givesThus the desired identity holds at every dyadic endpoint, including $1$, without evaluating the undefined slope at $X=1$.

Finally, given $x\in [0,1]$, take dyadic endpoints $x_n\to x$. The Lipschitz condition implies $|f(x_n)-f(x)|\le L|x_n-x|$, and the bound on the Borel measurable function $g$ givesPassing to the limit of a sequence provesThis is a martingale proof of the integral representation of a Lipschitz function. It constructs a bounded Lebesgue integrable function representing the increments of $f$, and in particular proves absolute continuity of $f$. No prior assumption that $f$ has a derivative is needed.