Submartingale doob decomposition has increasing compensator

The Doob formula and conditional linearity give $A_n-A_{n-1}=E[X_n-X_{n-1}\mid {\mathcal{F}}_{n-1}]=E[X_n\mid {\mathcal{F}}_{n-1}]-X_{n-1}$ a.s., since $X_{n-1}$ is known and integrable. If $X$ is a submartingale the right side is nonnegative for every $n\ge 1$, hence so is the compensator increment. Conversely nonnegative compensator increments imply $E[X_n\mid {\mathcal{F}}_{n-1}]\ge X_{n-1}$ for every $n\ge 1$, which is the submartingale definition Martingale submartingale and supermartingale. AC is inherited from the Doob construction and its conditional classes.

If every increment is nonnegative a.s., the measurable sets $N_n=\{A_n<A_{n-1}\}$ are null. Their union $N={\bigcup }_{n\ge 1}{N}_n$ is measurable and null. For $\omega \mathrm{\notin }N$ every successive inequality holds, so finite chaining gives $A_m(\omega )\le A_n(\omega )$ whenever $m\le n$. Conversely, if all paths off a measurable null $N$ are nondecreasing, each $N_n$ is contained in $N$, and hence has probability zero. This proves the path formulation for any chosen measurable versions.