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
\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}
?
Best Answer
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}