Concerning the following result in Karatzas and Shreve (Problem 3.19):
Proposition The following three conditions are equivalent for a non-negative right-continuous submartingale $\{X_t,0\leq t < \infty\}$:
It is uniformly integrable.
It converges in $L^1$ as $t\rightarrow \infty$.
It converges $\mathbb{P}$-a.s. as $t\rightarrow \infty$ to an integrable random variable $X_{\infty}$ such that $\{X_t,0\leq t \leq \infty\}$ is a submartingale.
The book provides a solution to this problem. I am interested in the implication $(iii)\implies(i)$.
To establish "$(3)\implies(1)$" the authors argue the following:
For $0\leq t<\infty$ and $\lambda>0$ we have $$\int_{\{|X_t|\geq\lambda\}}X_t d\mathbb{P} \leq \int_{\{|X_t|\geq\lambda\}}X_{\infty}d\mathbb{P}$$ which converges uniformly in $t$ to $0$ since $\mathbb{P}[|X_t|\geq\lambda]\leq(1/\lambda)\mathbb{E}[X_t]\leq(1/\lambda)\mathbb{E}[X_{\infty}]$.
Question: Why does the inequality $\int_{\{|X_t|\geq\lambda\}}X_t d\mathbb{P} \leq \int_{\{|X_t|\geq\lambda\}}X_{\infty}d\mathbb{P}$ hold? Does it really hold for any $\lambda, t$?
A question about the same problem was answered a year ago here. The credits of the template for this question also go to the author of this question. Also there is a quite an alternative approach to prove "$(3)\implies(1)$" by spalein here.
Best Answer
Since $(X_t)_{0 \leq t \leq \infty}$ is a submartingale, we have $$\mathbb{E}(X_{\infty} \mid \mathcal{F}_t) \geq X_t \tag{1}$$ for all $t \geq 0$. This implies $$\int_F X_{\infty} \, d\mathbb{P} \geq \int_F X_t \, d\mathbb{P} \quad \text{for all $F \in \mathcal{F}_t$.}\tag{2}$$ To prove this, recall that by the very definition of the conditional expectation $$\int_F X_{\infty} \, d\mathbb{P} = \int_F \mathbb{E}(X_{\infty} \mid \mathcal{F}_t) \, d\mathbb{P} \quad \text{for all $F \in \mathcal{F}_t$.}$$ Combining this with $(1)$, we get $$\int_F X_{\infty} \, d\mathbb{P} = \int_F \mathbb{E}(X_{\infty} \mid \mathcal{F}_t) \, d\mathbb{P} \geq \int_F X_t \, d\mathbb{P}, \qquad F \in \mathcal{F}_t,$$
which proves $(2)$. Choosing $F := \{|X_t| \geq \lambda\} \in \mathcal{F}_t$ in $(2)$ gives the desired inequality.