A doubt on the proof of Martingale Convergence Theorem on Jacod-Protter

Question

Theorem: Let $(X_n)_{n\geq1}$ be a submartingale such that $\sup\limits_{n}\mathbb{E}\{X_n^{+}\}<\infty$. Then, $\lim\limits_{n\rightarrow\infty} X_n = X$ exists a.s. (and is finite a.s.). Moreover, $X$ is in $\mathcal{L}^1$

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Proof: Let $U_n$ be the number of upcrossings of $[a,b]$ (where $a$, $b \in \mathbb{Q}$) before time $n$. By definition, $U_n$ is non-decreasing, hence $U(a,b)=\lim\limits_{n\rightarrow\infty} U_n$ exists. By Monotone Convergence Theorem, \begin{equation*} \begin{split} \mathbb{E}\{U(a,b)\} &= \lim\limits_{n\rightarrow \infty} \mathbb{E}\{U_n\} \\ &\leq \frac{1}{b-a} \sup\limits_{n}\mathbb{E}\{(X_n-a)^{+}\} \end{split} \end{equation*} where the inequality follows from the Doob's upcrossing inequality according to which $\mathbb{E}\{U_n\}\leq\frac{1}{b-a}\mathbb{E}\{(X_n-a)^{+}\}$

The point I cannot fully get is the one related to the inequality. That is, starting from $\mathbb{E}\{U_n\}\leq\frac{1}{b-a}\mathbb{E}\{(X_n-a)^{+}\}$, considering that by Monotone Convergence Theorem $\lim\limits_{n\rightarrow \infty} \mathbb{E}\{U_n\} = \mathbb{E}\{U(a,b)\}$, applying limit on both sides of the inquality I would get \begin{equation*} \lim\limits_{n\rightarrow \infty} \mathbb{E}\{U_n\} = \mathbb{E}\{U(a,b)\} \leq \frac{1}{b-a} \lim\limits_{n\rightarrow\infty}\mathbb{E}\{(X_n-a)^{+}\} \end{equation*}

Now, according to what is quoted above, why can I state that \begin{equation} \frac{1}{b-a} \lim\limits_{n\rightarrow\infty}\mathbb{E}\{(X_n-a)^{+}\} \leq \frac{1}{b-a} \sup\limits_{n}\mathbb{E}\{(X_n-a)^{+}\} \end{equation} ?

Answer 1

The limit is equal to the $\limsup$, because in this case the limit exists. However, the $\sup$ is always greater than or equal to the $\limsup$, because the $\limsup$ tells you the extreme large values taken at the tail, whereas the $\sup$ records these values from time $0$ onwards. So in particular, \begin{equation} \limsup\limits_{n\rightarrow\infty}\mathbb{E}\{(X_n-a)^{+}\} = \lim_{m \to \infty} \sup_{n \geq m} \mathbb{E}\{(X_n-a)^{+}\} \leq \sup_{n \geq 0} \mathbb{E}\{(X_n-a)^{+}\}. \end{equation}