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(Note: It is very possible I misunderstood the questions.) By $X$ dominates $Y$ up to an additive constant, do you mean $X,Y$ are supermartingales with bounded increments, and $X(S)>Y(S)-C$ for some …

This is for martingales index by natural numbers. Also, I call a martingale which converges to 0 "singular". I have also seen them called "potentials". … Then there are two nonnegative martingales $(P_k)$ and $(N_k)$ such that such that $M_{k}=P_k-N_k$ a.e. for all $k$, and $\left\Vert M\right\Vert =\left\Vert P\right\Vert +\left\Vert N\right\Vert = \ …

My motivation for thinking it might be true is that something similar is true for martingales, namely $\displaystyle P\{ \max_{n \leq k \leq m} |M_k - M_n| > \epsilon\} \leq \frac{||M_m - M_n||_1}{ … , and I know there are many similarities between backward martingales and ergodic averages. …

We have been using martingales as long as probability theory (going back to work of von Mises). … However, since algorithmic randomness is often about sequences of $0$s and $1$s, the martingales are usually dyadic martingales. Also they are usually nonnegative. …

Equations (A) and (B) and de Finetti's theorem can all be proved using reverse martingales. … Similarly, for which types of pointwise ergodic theorems and ergodic decompositions is there a proof using reverse martingales? Pointers to any relevant references would be helpful. …

To answer the second question, it seems that such ergodic averages can be represented as reverse martingales when the $H_n$ are finite subgroups of $G$. … Then on can use the geometry of $G$ to turn a theorem about reverse martingales (for example a maximal theorem) into a theorem about ergodic averages. …

Assume $M_t$ is a submartingale adapted to $\mathcal{F}_t$ such that $t \mapsto \mathbb{E}M_t$ is right-continuous (which is always true of martingales on right-continuous filtrations). … I think I have a proof for martingales (it involves algorithmic randomness, so it is not at all standard), but since I cannot find this written anywhere, I am worried I might be mistaken. …