I have a hard time understanding the proof of the following result in Karatzas and Shreve (Problem 3.19, page 18).
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, but I have hard time understanding the argument used to establish that $(ii)\implies(iii)$ and $(iii)\implies(i)$.
To establish $(ii)\implies(iii)$ the authors's argument is following:
Let $X_{\infty}$ be the $L^1$ limit of $X_t$. For any $A_s\in\mathcal{F}_s$ we have
$$ \int_A X_s d\mathbb{P}\leq \int_A X_td\mathbb{P}$$ Letting $t\rightarrow \infty$ we obtain that $$\int_A X_s d\mathbb{P}\leq \int_A X_{\infty} d\mathbb{P}$$ for all $0\leq s < \infty$.
Question: How do we arrive at the last inequality. Why can we pass the limit through the integral?
To establish $(iii)\implies(i)$ 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: How does uniform limit of $\mathbb{P}[|X_t|\geq\lambda]$ as $\lambda\rightarrow \infty$ establishes uniform integrability?
Best Answer
$(ii) \implies (iii):$ You've assumed that $X_t \to X_\infty$ in $L^1$ as $t \to \infty$. We can then write $$\bigg |\int_A X_t d \mathbb{P} - \int_A X_\infty d \mathbb{P} \bigg |\leq \int_A |X_t - X_\infty| d \mathbb{P} \leq \|X_t - X_\infty\|_{L^1} \to 0$$ to see that $\int_A X_t d\mathbb{P} \to \int_A X_\infty d\mathbb{P}$.
$(iii) \implies (i):$ To show uniform integrability, we want to show that for $\varepsilon > 0$ there is a $\lambda$ such that $$\sup_t \int_{\{|X_t| \geq \lambda\}} |X_t| d \mathbb{P} \leq \varepsilon.$$ The authors do this using the standard fact that for $Y \in L^1$, $\int_A |Y| d \mathbb{P} \to 0$ as $\mathbb{P}(A) \to 0$ in the sense that for $\varepsilon > 0$, there is a $\delta>0$ such that $\mathbb{P}(A) < \delta$ implies that $\int_A |Y| d \mathbb{P} < \varepsilon$.
Here is a proof of that fact in your specific case. We have \begin{align} \int_{\{|X_t| \geq \lambda\}} |X_\infty| d\mathbb{P} =& \int_{\{|X_t| \geq \lambda\}\cap \{|X|_\infty < K\}} |X_\infty| d\mathbb{P} + \int_{\{|X_t| \geq \lambda\} \cap \{|X|_\infty \geq K\}} |X_\infty| d\mathbb{P} \\ \leq & K \mathbb{P}(|X_t| \geq \lambda) + \int_{\{|X|_\infty \geq K\}} |X_\infty| d\mathbb{P}. \end{align} We can pick $K$ such that $\int_{\{|X|_\infty \geq K\}} |X_\infty| d\mathbb{P} < \frac{\varepsilon}{2}$. For this $K$ we have shown that $$\sup_t \int_{\{|X_t| \geq \lambda\}} |X_t| d \mathbb{P} \leq K \sup_t \mathbb{P}(|X_t| \geq \lambda) + \frac{\varepsilon}{2}.$$ Since $\mathbb{P}(|X_t| \geq \lambda) \to 0$ uniformly in $t$ as $\lambda \to \infty$, the right hand side of this is less than $\varepsilon$ for $\lambda$ large enough.