question about Fatou lemma in stochastic process.

Question

$X^k_t(\omega)\ge 0$ are stochastic process which satisfy: $$\mathbb E(\sup_{0\le s\le t}X_s^k)\le M$$

and $X^k_t\to X$ in $[0,t]$ uniformly, can we conclude: $$\mathbb E(\sup_{0\le s\le t}X_s)\le M$$ ?

by Fatou's lemma: $$\mathbb E(\sup_{0\le s\le t}X_s)=\mathbb E(\sup_{0\le s\le t}\liminf_{k\to\infty} X_s^k)\stackrel{??}{\leq} \mathbb E(\liminf_{k\to\infty}\sup_{0\le s\le t} X_s^k)\le \liminf_{k\to\infty}\mathbb E(\sup_{0\le s\le t} X_s^k)\le M$$

what can we say about $??$ above?

Answer 1

Fix $\omega \in \Omega$, $\epsilon>0$ and set

$$m := \sup_{0 \leq s \leq t} X_s(\omega).$$

By the definition of "sup" we can choose $s \in [0,t]$ such that

$$m-\epsilon \leq X_s(\omega) \leq m.$$

Since $X_s^k(\omega)$ converges to $X_s(\omega)$, there exists $k_0 \in \mathbb{N}$ such that

$$|X_s^k(\omega)-X_s(\omega)| \leq \epsilon$$

for all $k \geq k_0$; hence,

$$\sup_{0 \leq r \leq t} X_r^k(\omega) \geq X_s^k(\omega) \geq m-2\epsilon = \sup_{0 \leq s \leq t} X_s(\omega)-2\epsilon.$$

This implies

$$\liminf_{k \to \infty} \sup_{0 \leq r \leq t} X_r^k(\omega) \geq \sup_{0 \leq s \leq t} X_s(\omega)-2\epsilon.$$

Letting $\epsilon \to 0$ gives

$$\liminf_{k \to \infty} \sup_{0 \leq s \leq t} X_s^k(\omega) \geq \sup_{0 \leq s \leq t} X_s(\omega)$$

... and that's exactly the inequality you need in order to apply Fatou's lemma.