For positive supermartingale, if $X_k(\omega)=0$ for some $k$, then $X_n(\omega)=0$ for all $n\ge k$.

Question

Suppose that the supermartingale $(X_n)_{n \in \mathbb{N}}$ is positive, meaning that for each $n\in \mathbb{N}$, $X_n \ge 0$ almost surely. Prove that the following is true of almost every $\omega \in \Omega:$ If $X_k(\omega)=0$ for some $k\in \mathbb{N}$, then $X_n(\omega)=0$ for all $n\ge k$.

I'm not sure how this problem can be solved. First I need to identify the almost sure set for which the statement holds. I think the candidate is the set for which $X_n\ge 0$ for each $n\in N$. But then choosing a specific point $\omega$ in the set, I can't use expectation properties such as $E[X_k]\ge E[X_n]$ for all $n\ge k$. The problem does not seem very complex but I can't figure out a way to solve this. I would greatly appreciate any help.

Answer 1

Hint: Since $(X_n)_{n \in \mathbb{N}}$ is a supermartingale, we have

$$\int_F X_n \, d\mathbb{P} \leq \int_F X_k \, d\mathbb{P}$$

for all $F \in \mathcal{F}_k := \sigma(X_1,\ldots,X_k)$ and $k \leq n$. For $F := \{X_k=0\}$ this shows, by the non-negativity, that

$$\int_F X_n \, d\mathbb{P}=0.$$

Conclude.