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The supermartingale property and linearity of conditional expectation give, for $n\ge 1$,Each $G_n$ is integrable and ${\mathcal{F}}_{n-1}$-measurable, while $G_0=0$. Thus $Z_n$ is integrable andMoreover, all terms $G_j$, $j\le n+1$, are ${\mathcal{F}}_n$-measurable, so $Z_{n+1}\in m{\mathcal{F}}_n$. This is predictability of the decreasing component in the Doob decomposition theorem.

Choose versionsThe stochastic process $(A_n)$ is increasing, and its $k$th increment is ${\mathcal{F}}_{k-1}$-measurable, so it is predictable. The Doob decomposition theorem therefore givesTo check the martingale property directly, each summand is integrable and has zero conditional expectation given ${\mathcal{F}}_{k-1}$. The sums are an adapted process for the given filtration. Thus $\mathbb{E}[M_n\mid {\mathcal{F}}_{n-1}]=M_{n-1}$. Its bounded increments satisfy $|M_n-M_{n-1}|\le 1$.