# Show if M is a positive local martingale with $E(M_0)<\infty$ then $M$ is a supermartingale.

A similar question was asked here but I have a question about one of the steps.

To show that $$M_t \in M := \{M_t : t \in [0,\infty)$$ is integrable we consider $$M_{t \wedge T_n}$$ where $$\{T_n\}$$ are stopping times almost surey increasing to $$\infty$$. Then $$M_{t \wedge T_n} \to M_t$$ almost surely. We can now apply Fatou's Lemma to get that $$E(M_t) \le \liminf\limits _{n \to \infty}E(M_{t \wedge T_n}) = E(M_0)$$

But why does $$\liminf\limits _{n \to \infty}E(M_{t \wedge T_n}) = E(M_0)$$?

$$M_{t\wedge T_n}: t \geq 0$$ is a martingale for each $$n$$. So $$EM_{t\wedge T_n}=EM_{0\wedge T_n}=EM_0$$.
• Maybe there is a property of martingales that I am not familiar with but why does $M_{t\wedge T_n}$ being a martingale for $t \ge 0$ and for every $n$ imply that $EM_{t\wedge T_n}=EM_{0\wedge T_n}$? Commented Jun 5, 2019 at 9:57
• @alpastor If $\{X_t\}$ is a martingale then $EX_t$ is in dependent of $t$. To see this just take expectation in the definition of martingale. Commented Jun 5, 2019 at 9:59
• You mean independent right? $$E(X_s)=E\left[E(X_s)\right]=E\left[E(X_t|\mathcal F_s)\right] = E(X_t)$$. Thank you, for the explanation Commented Jun 5, 2019 at 10:06