Proof of the Submartingale Convergence Theorem
Theorem
Given a probability space $( \Omega , \mathcal{F} , P)$ and a submartingale $\left\{ ( X_{n} , \mathcal{F}_{n} ) \right\}$, if we assume $\displaystyle \sup_{n \in \mathbb{N}} E X_{n}^{+} < \infty$, then $X_{n}$ converges almost surely to some random variable $X_{\infty}: \Omega \to \mathbb{R}$. $$E X_{\infty} < E X_{\infty}^{+} < \infty$$
Proof
Strategy: Use the properties of limit supremum and limit infimum. $$ X^{\ast}:= \limsup_{n \in \mathbb{N}} X_{n} \\ X_{\ast}:= \liminf_{n \in \mathbb{N}} X_{n} $$ Then, $$ \left( X^{\ast} > X_{\ast} \right) = \bigcup_{a < b \\ a, b \in \mathbb{Q}} \left( X^{\ast} > b > a > X_{\ast} \right) $$ since $a,b \in \mathbb{Q}$, we can divide $(X^{\ast} > X_{\ast})$ into $P$ countably. Therefore, for all rational numbers $a$, $b$ between $X^{\ast}$ and $X_{\ast}$, if $P \left( X^{\ast} >b >a> X_{\ast} \right) = 0$ then $P \left( X^{\ast} > X_{\ast} \right) = 0$, meaning that $X_{n}$ converges almost surely to $X_{\infty}$.
Part 1. $P \left( X^{\ast} > b > a > X_{\ast} \right) \le P \left( \beta_{\infty} (a,b) = \infty \right) $
Let’s denote the number of upcrossings within the closed interval formed by two rational numbers $a, b \in \mathbb{Q}$ and $[a,b]$ as $\beta_{N} (a,b) $, and its limit as $\displaystyle \beta_{\infty} (a,b):= \lim_{N \to \infty} \beta_{N} (a,b)$. If $X^{\ast} > b > a > X_{\ast}$, it implies that $X_{n}$ has descended below $a$ and ascended above $b$ for infinitely many $n \in \mathbb{N}$. Hence, $\beta_{\infty} (a,b) = \infty$, and in proposition, $$ X^{\ast} > b > a > X_{\ast} \implies \beta_{\infty} (a,b) = \infty $$ In set notation, $$ \left( X^{\ast} > b > a > X_{\ast} \right) \subset \left( \beta_{\infty} (a,b) = \infty \right) $$ Taking the probability of $P$, $$ P \left( X^{\ast} > b > a > X_{\ast} \right) \le P \left( \beta_{\infty} (a,b) = \infty \right) $$
Part 2. $P \left( \beta_{\infty} (a,b) < \infty \right) = 1$
Since $$ P \left( |X| = \infty \right) \ne 0 \implies E \left( |X| \right) = \infty $$ by the contrapositive, almost surely $$ E \beta_{\infty} (a,b) < \infty \implies \beta_{\infty} (a,b) < \infty $$
Upcrossing expected value upper bound: $\displaystyle E \beta_{N} (a,b) \le {{ E X_{N}^{+} + |a| } \over { b-a }}$
Monotone convergence theorem: Let the sequence of non-negative measurable functions $\left\{ f_{n} \right\}$ satisfy $f_{n} \nearrow f$. Then, $$ \lim_{n \to \infty} \int_{E} f_{n} dm = \int_{E} f dm $$
By assumption, $$ E \beta_{N} (a,b) \le {{ E X_{N}^{+} + |a| } \over { b-a }} \le {{ \sup_{N \in \mathbb{N}} E X_{N}^{+} + |a| } \over { b-a }} < \infty $$ since by definition of $\beta_{N} (a,b)$, we have $\beta_{N} (a,b) \nearrow \beta_{\infty} (a,b)$, by the monotone convergence theorem, $$ \begin{align*} \infty &>& \lim_{N \to \infty} E \beta_{N} (a,b) \\ =& E \lim_{N \to \infty} \beta_{N} (a,b) \\ =& E \beta_{\infty} (a,b) \end{align*} $$ Summing up, since $E \beta_{\infty} (a,b) < \infty$, almost surely $\beta_{\infty} (a,b) < \infty$, meaning $P \left( \beta_{\infty} (a,b) < \infty \right) = 1$ holds.
Part 3. $\displaystyle P \left( X^{\ast} = X_{\ast} \right) = 1$
According to Part 1~2, for all $a, b \in \mathbb{Q}$ where $X^{\ast} > b > a > X_{\ast}$, $$ P \left( X^{\ast} > b > a > X_{\ast} \right) \le P \left( \beta_{\infty} (a,b) = \infty \right) = 0 $$ Since probability $P$ is a measure, $$ \begin{align*} P \left( X^{\ast} > X_{\ast} \right) =& P \left[ \bigcup_{a < b \\ a, b \in \mathbb{Q}} \left( X^{\ast} > b > a > X_{\ast} \right) \right] \\ =& \sum_{a < b \\ a, b \in \mathbb{Q}} P \left( X^{\ast} > b > a > X_{\ast} \right) \\ \le & \sum_{a < b \\ a, b \in \mathbb{Q}} 0 \\ =& 0 \end{align*} $$ Summing up, since $P \left( X^{\ast} \le X_{\ast} \right) = 1$, the limit $X_{\infty}$ of $X_{n}$ almost surely exists.
Part 4. $E X_{\infty} < \infty$
By the decomposition of absolute values, $$ |X_{n}| = X_{n}^{+} + X_{n}^{-} = 2 X_{n}^{+} - X_{n} $$ Since $\left\{ ( X_{n} , \mathcal{F}_{n} ) \right\}$ is a submartingale, hence $E X_{n} \ge E X_{1}$, $$ E |X_{n}| = 2 E X_{n}^{+} - E X_{n} \le 2 E X_{n}^{+} - E X_{1} $$ Given that we assumed $\displaystyle \sup_{n \in \mathbb{N}} E X_{n}^{+} < \infty$, $$ \sup_{n \in \mathbb{N}} E | X_{n} | \le 2 \sup_{n \in \mathbb{N}} E X_{n}^{+} - E X_{1} < \infty $$
Fatou’s lemma: For a sequence of non-negative measurable functions $\left\{ f_{n} \right\}$, $$\displaystyle \int_{E} \left( \liminf_{n \to \infty} f_{n} \right) dm \le \liminf_{n \to \infty} \int_{E} f_{n} dm $$
$\sup_{n \in \mathbb{N}} E | X_{n} | < \infty$ and by Fatou’s lemma, $$ \begin{align*} \infty &>& \sup_{n \in \mathbb{N}} E | X_{n} | \\ \ge& \liminf_{n \to \infty} E | X_{n} | \\ =& \liminf_{n \to \infty} \int_{\Omega} | X_{n} | d P \\ =& \int_{\Omega} \liminf_{n \to \infty} | X_{n} | d P \\ =& \int_{\Omega} | X_{\infty} | d P \\ =& E |X_{\infty}| \end{align*} $$ Therefore, $E | X_{\infty} |$ also exists.
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Corollary
Particularly, in the proof process of Part 4., if $X_{n} < 0$, then from $X_{n} = X_{n}^{+} - X_{n}^{-}$ to $X_{n}^{+} = 0$, implying that even the condition $\displaystyle \sup_{n \in \mathbb{N}} E X_{n}^{+} < \infty$ becomes unnecessary.