# Simple Question on Convergence of Martingales

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.

(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.