I'm tryin to verify that the following stopped process is a martingale, i.e.,
Theorem Let $(M_n, n \in \mathbb{N})$ be a martingale and $T$ a stopping time with respect to a filtration $(\mathcal{F}_n, n \in \mathbb{N})$. Then the stopped process $(M'_n, n \in \mathbb{N})$ with $M'_n := M_{T \wedge n}$ where $T \wedge n := \min \{T, n\}$ is also a martingale with respect to $(\mathcal{F}_n, n \in \mathbb{N})$.
Could you have a check on my below attempt?
Proof We have $M'_n$ is $\mathcal F_{T \wedge n}$-measurable. We have $T \wedge n \le n$, so $\mathcal F_{T \wedge n} \subset \mathcal F_n$. It follows that $M'_n$ is $\mathcal F_n$-measurable. We have $M'_n$ is integrable because $$ \mathbb E [|M'_n|] = \mathbb E \left [ \sum_{k=0}^{n-1} |M_k| 1_{\{T=k\}} + |M_n| 1_{\{T\ge n\}} \right] \le \sum_{k=0}^n \mathbb E [|M_k|] < \infty. $$
Notice that $\{T \ge n+1\} \in \mathcal F_n$. We have $$ \begin{align} \mathbb E [M'_{n+1} | \mathcal F_{n}] &= \mathbb E [ M_{T \wedge (n+1)} | \mathcal F_{n}] \\ &= \mathbb E \left [ \sum_{k=0}^{n} M_k 1_{\{T=k\}} + M_{n+1} 1_{\{T\ge n+1\}} \bigg \lvert \mathcal F_n \right] \\ &= \sum_{k=0}^{n} \mathbb E \left [ M_k 1_{\{T=k\}} \bigg \lvert \mathcal F_n \right] + \mathbb E [M_{n+1} 1_{\{T\ge n+1\}} | \mathcal F_n] \\ &= \sum_{k=0}^{n} M_k 1_{\{T=k\}} + \mathbb E [M_{n+1} | \mathcal F_n] 1_{\{T\ge n+1\}} \\ &= \sum_{k=0}^{n} M_k 1_{\{T=k\}} + M_{n} 1_{\{T\ge n+1\}} \\ &= M_{T \wedge n}. \end{align} $$
This completes the proof.