Multistep martingale characterization

For $n=m$, adaptedness and integrability give $E[X_m\mid {\mathcal{F}}_m]=X_m$. For the submartingale case fix $m$ and induct on $n-m$. If $n>m$, nesting and the tower identity give $E[X_n\mid {\mathcal{F}}_m]=E[E[X_n\mid {\mathcal{F}}_{n-1}]\mid {\mathcal{F}}_m]\ge E[X_{n-1}\mid {\mathcal{F}}_m]\ge X_m$. The first inequality uses the one-step hypothesis and conditional order; the second is the induction hypothesis. All conditioned variables are integrable.

For the supermartingale case the identical tower identity has both inequalities reversed, since conditional order preserves the relation $\le$. For the martingale case the inner conditional expectation equals $X_{n-1}$, so induction gives equality at every pair. Each induction uses only finitely many almost-sure identities. AC here is inherited from the existence of the conditional classes; no representatives at all pairs are selected.

Conversely, each all-pairs condition evaluated at $(m,n)=(k,k+1)$ is its defining one-step condition. This includes $k=0$, and proves each equivalence.