Let $(\tau_n)$ be a <localizing sequence> for $M$. For $s\leq t$, the <optional sampling theorem> gives
$$
\mathbb E[M_{t\wedge\tau_n}\mid\mathcal F_s]=M_{s\wedge\tau_n}.
$$
Both sides converge almost surely to $M_t$ and $M_s$, and $|M_{u\wedge\tau_n}|\leq Z$. Conditional dominated convergence therefore yields $\mathbb E[M_t\mid\mathcal F_s]=M_s$, so $M$ is a <martingale>. Moreover, the family $(M_t)_{t\geq0}$ is dominated by the integrable random variable $Z$, hence is <uniformly integrable>. Thus \b[$M$ is a uniformly integrable martingale].

Solution

Past exam of the mathematics course of the University of Cambridge / 2019 / iii / Paper 202 / 1 / 1 / Solution

A localizing sequence for a <local martingale> $M$ is an increasing sequence of <stopping times> $T_n\uparrow\infty$ almost surely such that every stopped process $M^{T_n}$ is a true martingale. One can often choose the times so that the stopped processes are bounded.

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