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Here are two theorems that I've come across which explain when a martingale converges almost surely:

  1. If $\{X_n\}$ is a martingale bounded above or bounded below, then $\lim_{n \to \infty}X_n = X$ a.s
  2. If $\{X_n\}$ is a supermartingale bounded below or a submartingale bounded above, then it converges almost surely.

I've learned that a martingale is both a submartingale and supermartingale. Therefore, I believe theorem 2. could be seen as a corollary to theorem 1.

However, if that is the case, then shouldn't it also be true that if we have submartingale bounded below or a supermartingale bounded above then this sequence also converges almost surely? I'm hesitant to conclude this - because intuitively that does not make a lot of sense.

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You are missing the implication order.

First notice that all of those results are only one:

(1) If $X_n$ is a submartingale bounded above, then exists $X$ such that $X_n \to X$.

(2) The case $X_n$ is a supermartingale bounded below follows since $-X_n$ will be a submartingale bounded above.

(3) The case $X_n$ is a martingale bounded (above/below) follows since every martingale is also a (sub/super)martingale and then we use (1/2).

But now if $X_n$ is a supermartingale bounded above, it don't need to be a martingale. Take the following supermartingale with filtration $\mathcal{F_n} = \{\Omega, \emptyset\}$: $$ X_n = -n $$ We have: $$ E(X_{n+1}| \mathcal{F_n}) = -n-1 = X_n - 1 < X_n $$ It is bounded above by zero, but $X_n \to \infty$. And clearly, it is not a martingale.

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