Let
- $(\Omega,\mathcal A,\operatorname P)$ be a probability space
- $(\mathcal F_t)_{t\ge0}$ be a filtration of $\mathcal F$
- $\mathcal D$ be the set of product measurable processes $X$ on $(\Omega,\mathcal A,\mathcal F,\operatorname P)$ such that $(X_\tau:\tau\text{ is a finite stopping time on }(\Omega,\mathcal A,\mathcal F,\operatorname P))$ is uniformly integrable
How can we show that if $X$ is a product measurable process on $(\Omega,\mathcal A,\mathcal F,\operatorname P)$ with $$|X|\le|Y|\;\;\;\text{almost surely}\tag1$$ for some $Y\in L^1(\operatorname P)$, then $X\in\mathcal D$?
I've read this claim in Stochastic Integration Theorey by Péter Medvegyev, Corollary 1.145. However, I don't see why the claim is true.
If $Y$ would belong to $L^p(\operatorname P)$ for some $p>1$, then it would be clear to me, since $(1)$ yields the $L^p$-boundedness of $(X_\tau:\tau\text{ is a finite stopping time on }(\Omega,\mathcal A,\mathcal F,\operatorname P))$.
If you have any collection $\mathcal A$ of random variables that is dominated by an integrable random variable $Y$, then $\mathcal A$ is uniformly integrable. This is because for any $\epsilon>0$, there exists some $\delta>0$ such that $$ P(A) < \delta \implies E[|Y|1_A]<\epsilon \implies \sup_{X \in \mathcal A} E[|X|1_A]< \epsilon$$
Now note that if $|X(t, \omega)| \leq |Y(\omega)|$, then for any a.s. finite stopping time $\tau$ we also have that $|X(\tau(\omega),\omega)| \leq |Y(\omega)|$; hence the collection $\mathcal A:=\{X_{\tau}:\tau$ a finite stopping time$\}$ is dominated by $Y$.