A USEFUL RESULT (Doob's Upcrossing Inequality)
Let $(X_n)_{\geq0}$ be a submartingale, let $a<b$ and let $U_n$ be the number of upcrossings of $[a,b]$ before time $n$. Then
$$
\mathbb{E}\{U_n\}\leq\frac{1}{b-a}\mathbb{E}\{(X_n-a)^{+}\}
$$
THEOREM (Backwards Martingale Convergence Theorem)
Given a probability space $(\Omega$, $\mathcal{F}$, $\mathbb{P})$, let $(X_{-n}, \mathcal{F}_{-n})$ be a backwards martingale (hence $\mathbb{E}\{X_{-n}\mid\mathcal{F}_{-m}\}=X_{-m}$ a.s., with $0\leq n<m$) and $\mathcal{F}_{\infty}=\bigcap\limits_{n=0}^{\infty}\mathcal{F}_{-n}$. Then, the sequence $(X_{-n})$ converges a.s. and in $\mathcal{L}^1$ to a limit $X$ as $n\rightarrow\infty$ (in particular, $X$ is a.s. finite and is integrable)
A PART OF THE PROOF OF Backwards Martingale Convergence Theorem
Let $(U_{-n})$ be the number of upcrossings of $(X_{-n})_{n\geq0}$ of $[a,b]$ between time $-n$ and $0$. $(U_{-n})$ is increasing as $n$ goes up and so $U(a,b)=\lim\limits_{n\rightarrow\infty} U_{-n}$. By Monotone Convergence Theorem (equality below) and by Doob's Upcrossing Inequality (red inequality below) \begin{align*} \mathbb{E}\{U(a,b)\}&=\lim\limits_{n\rightarrow\infty}\mathbb{E}\{U_{-n}\}\\ &\color{red}{\leq}\frac{1}{b-a}\mathbb{E}\{(X_0-a)^+\}<\infty \end{align*} hence $\mathbb{P}\{U(a,b)<\infty\}=1$. With some work one can also show that $X=\lim\limits_{n\rightarrow\infty}X_{-n}$ exists a.s..
At this point, let $\phi(x)=x^+$ which is convex and increasing and $\color{red}{obviously}$ $\color{red}{\phi(X_{-n})}$ $\color{red}{is}$ $\color{red}{integrable}$, all $n$.
Then, $\color{red}{Jensen's}$ $\color{red}{inequality}$ $\color{red}{and}$ $\color{red}{\mathbb{E}\{X_{-n}\mid\mathcal{F}_{-m}\}=X_{-m}}$ $\color{red}{imply}$ $\color{red}{that}$ $\color{red}{X_{-n}^{+}\leq\mathbb{E}\{X_0^+\mid\mathcal{F}_{-n}\}}$, hence $\mathbb{E}\{X_{-n}^+\}\leq\mathbb{E}\{X_0^+\}$.
Finally, Fatou's lemma (first inequality below) and the fact that $X_{-n}^+\geq0$ and $X_{-n}^+\rightarrow X^+$ a.s. yield $$ \mathbb{E}\{\limsup\limits_{n\rightarrow\infty}X_{-n}^+\}=\mathbb{E}\{\liminf\limits_{n\rightarrow\infty}X_{-n}^+\}=\mathbb{E}\{X^+\}\leq\liminf\limits_{n\rightarrow\infty}\mathbb{E}\{X_{-n}^+\}\color{red}{\leq}\mathbb{E}\{X_0^+\}<\infty $$
I cannot really understand the four red-colored parts in the proof above.
$\color{red}{1)}$ Why does $\lim\limits_{n\rightarrow\infty}\mathbb{E}\{U_{-n}\}
\leq\dfrac{1}{b-a}\mathbb{E}\{(X_0-a)^+\}$ hold true after applying Doob's Upcrossing Inequality? What I would expect is that $\lim\limits_{n\rightarrow\infty}\mathbb{E}\{U_{-n}\}
\leq\dfrac{1}{b-a}\lim\limits_{n\rightarrow\infty}\mathbb{E}\{(X_n-a)^+\}$. So, how is $\dfrac{1}{b-a}\lim\limits_{n\rightarrow\infty}\mathbb{E}\{(X_n-a)^+\}$ related to $\dfrac{1}{b-a}\mathbb{E}\{(X_0-a)^+\}$?
$\color{red}{2)}$ Why can one obviously state that $\phi(X_{-n})=(X_{-n})^+$ is integrable (all $n$) (i.e. $\mathbb{E}\{X_{-n}^+\}<\infty$)?
$\color{red}{3)}$ By Jensen's inequality and $\mathbb{E}\{X_{-n}\mid\mathcal{F}_{-m}\}=X_{-m}$ I would expect that $(\mathbb{E}\{X_{-n}\})^+\leq\mathbb{E}\{X_{-n}^+\}=\mathbb{E}\{X_0^+\mid\mathcal{F}_{-n}\}$ and not that $X_{-n}^+\leq\mathbb{E}\{X_{-n}^+\}=\mathbb{E}\{X_0^+\mid\mathcal{F}_{-n}\}$. Hence, how is $(\mathbb{E}\{X_{-n}\})^+$ related to $X_{-n}^+$?
$\color{red}{4)}$ I would expect that, by definition, $\liminf\limits_{n\rightarrow\infty}\mathbb{E}\{X_{-n}^+\}\leq\lim\limits_{n\rightarrow\infty}\mathbb{E}\{X_{-n}^+\}$ and not that $\liminf\limits_{n\rightarrow\infty}\mathbb{E}\{X_{-n}^+\}\leq\mathbb{E}\{X_{-0}^+\}$. So, how is $\lim\limits_{n\rightarrow\infty}\mathbb{E}\{X_{-n}^+\}$ related to $\mathbb{E}\{X_{-0}^+\}$?
$\def\F{\mathscr{F}}$1) There seems to be a typo, i.e. it should be $E((X_0 - a)^+)$ instead of $E((-X_0 - a)^+)$. Note that for any fixed $n \geqslant 1$, $\{X_m \mid -n \leqslant m \leqslant 0\}$ is a submartingale with respect to $\{\F_m \mid -n \leqslant m \leqslant 0\}$, so $X_0$ is the ending term on the right in the submartingale, and Doob's upcrossing inequality yields$$ E(U_{-n}) \leqslant \frac{1}{b - a}E((X_0 - a)^+), $$ which implies $\lim\limits_{n → ∞} E(U_{-n}) \leqslant \dfrac{1}{b - a}E((X_0 - a)^+) < +∞$.
2) The existence of the conditional expectation $E(X_{-n} \mid \F_{-m})$ implicitly requires that $X_{-n}$ be integrable, so $E(X_{-n}^+) \leqslant E(|X_{-n}|) < +∞$.
3) This is Jensen's inequality for conditional expectation:$$ E(X_0^+ \mid \F_{-n}) = E(φ(X_0) \mid \F_{-n}) \geqslant φ(E(X_0 \mid \F_{-n})) = φ(X_{-n}) = X_{-n}^+. $$ 4) Since the last line implies that $E(X_{-n}^+) \leqslant E(X_0^+)$, then $\inf\limits_{m \geqslant n} E(X_{-m}^+) \leqslant E(X_0^+)$ for any $n \geqslant 0$ and$$ \varliminf_{n → ∞} E(X_{-n}^+) = \lim_{n → ∞} \inf_{m \geqslant n} E(X_{-m}^+) \leqslant E(X_0^+). $$