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.
If a submartingale $X=\left(X_{n}\right)_{n \in \mathbb{Z}_{+}}$ is "uniformly integrable", then $X_{n} \xrightarrow[a.e.]{L^{1}}X_{\infty}$ and $\left(X_{n}\right)_{n \in \overline{\mathbb{Z}}_{+}}$ becomes a closed submartingale.
To prove the proposition, we need some facts:
- (Criterion of $L^{1}$-convergence) Let $X_{n} \in L^{1}$; then $X_{n} \xrightarrow{L^{1}} X$ if and only if $X_{n}$ are uniformly integrable and
$X_{n} \rightarrow_\mathbb{P} X$.
- (Vitali's convergence theorem) Suppose that $f_{n} \rightarrow f$ a.e. $(\mathbb{P})$. If the $f_{n}$ are uniformly integrable, then $f$ is integrable and such that
$$
\int f_{n} d \mathbb{P} \rightarrow \int f d \mathbb{P}.
$$
By the definition of uniform integrability, $(X_n)_{n\in \mathbb{Z}_+}$ is bounded in $L^1$ and, moreover, $\left(X_{n} 1_{A}\right)_{n \in \mathbb{Z}_{+}}$ is also uniformly integrable for any $A \in \mathscr{F}_{\infty}$. Thus, by Doob's submartingale convergence theorem, we have
$$
X_{n} \stackrel{\text { a.e. }}{\longrightarrow} X_{\infty} \in L^{1}\left(\Omega, \mathscr{F}_{\infty}, \mathbb{P}\right) \Rightarrow X_{n} 1_{A} \stackrel{\text { a.e. }}{\longrightarrow} X_{\infty} 1_{A} \text { and } X_{n} 1_{A} \rightarrow_\mathbb{P} X_{\infty} 1_{A}.
$$
Then by fact 1, we see as $n \rightarrow \infty, X_{n} \rightarrow X_{\infty}$ in $L^{1}\left(\Omega, \mathscr{F}_{\infty}, \mathbb{P}\right)$. Moreover, by fact 2, it follows that as $m \rightarrow \infty$,
$$
\text{E}\left[X_{m} 1_{A}\right] \rightarrow \text{E}\left[X_{\infty} 1_{A}\right] ,\quad \forall A \in \mathscr{F}_{\infty}.
$$
Finally, to show $\left(X_{n}\right)_{n \in \overline{\mathbb{Z}}_{+}}$ is a closed submartingale, we only need to prove $\text{E}\left[X_{\infty} \mid \mathscr{F}_{n}\right] \geq X_{n}$
a.e. for all $n \geq 0$. For this, by $\text{E}\left[X_{m} \mid \mathscr{F}_{n}\right] \geq X_{n}$ a.e. for all $m>n$, it follows that for any $A \in \mathscr{F}_{n}$ we have
$$
\text{E}\left[X_{n} 1_{A}\right] \leq \text{E}\left[X_{m} 1_{A}\right] \rightarrow \text{E}\left[X_{\infty} 1_{A}\right] \text { as } m \rightarrow+\infty.
$$
This completes the proof.
See Proof of Fact 1,
Proof of Fact 2.
Best Answer
If $S_{n}=\sum_{i=1}^{n} a_{i}$, you have to $\mathbb{E}[Y_{n}|\mathcal{F}_{n-1}]\le Y_{n-1}$, where $Y_{n}=X_{n}-S_{n}$, then $Y_{n}$ is supermartingale.