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

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 $$\mathcal{L}^1$$

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 $$$$\frac{1}{b-a} \lim\limits_{n\rightarrow\infty}\mathbb{E}\{(X_n-a)^{+}\} \leq \frac{1}{b-a} \sup\limits_{n}\mathbb{E}\{(X_n-a)^{+}\}$$$$ ?

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, $$$$\limsup\limits_{n\rightarrow\infty}\mathbb{E}\{(X_n-a)^{+}\} = \lim_{m \to \infty} \sup_{n \geq m} \mathbb{E}\{(X_n-a)^{+}\} \leq \sup_{n \geq 0} \mathbb{E}\{(X_n-a)^{+}\}.$$$$