If $T_k$ localizes $X$, then so does $S_k \wedge T_k$?

Question

There is a fact mentioned in the answer of this question that I can't seem to show. It's probably simple but I'm not seeing it. Essentially it says:

If $X$ local martingale with localizing sequence $T_k$, and given $(S_k) \rightarrow \infty$, a sequence of stopping times then $S_k \wedge T_k$ is also a localizing sequence for X.

Answer 1

By Doob's Optional Stopping Theorem, stopped martingales are again martingales. So, for all $k$, we stop the martingale $(X_{t \wedge T_k})_t$ with the stopping time $S_k$ to obtain a martingale $(X_{t \wedge S_k \wedge T_k})_t$.

As $T_k, S_k \to \infty$, also $S_k \wedge T_k \to \infty$ is a localizing sequence for $X$.