I'm trying to prove this proposition about the stopping time. Could you please verify whether my proof looks fine or contains logical mistakes? Thank you so much!
Let $\left(\Omega, \mathcal{F},(\mathcal{F}_{n})_{n \in \mathbb{N}}, \mathbb P\right)$ be a filtered probability space, $(X_{n})_{n \in \mathbb N}$ a sub-martingale, and $T$ a stopping time. Then $(X_{n \wedge T})_{n \in \mathbb N}$ is also a sub-martingale.
My attempt:
- $X_{n \wedge T}$ is $\mathcal F_n$-measurable
We have $X_{n \wedge T}$ is $\mathcal F_{n \wedge T}$-measurable and $\mathcal F_{n \wedge T} \subseteq \mathcal F_n$. The claim then follows.
- $X_{n \wedge T}$ is integrable
We first observe that the maximum and the minimum of finite number of integrable functions are integrable. We have $\min \{X_0 , \ldots, X_n\} \le X_{n \wedge T} \le \max \{X_0 , \ldots, X_n\}$, i.e., $X_{n \wedge T}$ is bounded by two integrable functions. Moreover, $X_{n \wedge T}$ is measurable. Hence $X_{n \wedge T}$ is integrable.
- $\mathbb E [ X_{(n+1) \wedge T} | \mathcal F_n ] \le X_{n \wedge T}$
We have $$\begin{aligned} \mathbb E [ X_{(n+1) \wedge T} | \mathcal F_n ] &= \mathbb E [ X_{(n+1) \wedge T} \mathbf{1}_{\{T \le n\}} + X_{(n+1) \wedge T} \mathbf{1}_{\{T > n\}} | \mathcal F_n ] \\ &= \mathbb E [ X_{(n+1) \wedge T} \mathbf{1}_{\{T \le n\}} | \mathcal F_n ] +\mathbb E [ X_{(n+1) \wedge T} \mathbf{1}_{\{T > n\}} | \mathcal F_n ] \\ &= \mathbb E [ X_{n \wedge T} \mathbf{1}_{\{T \le n\}} | \mathcal F_n ] + \mathbb E [ X_{n+1} \mathbf{1}_{\{T > n\}} | \mathcal F_n ] \\ \end{aligned}$$
It follows from $X_{n \wedge T}$ is $\mathcal F_n$-measurable and $\{T \le n\} \in \mathcal F_n$ that $X_{n \wedge T} \mathbf{1}_{\{T \le n\}}$ is $\mathcal F_n$-measurable. Hence $ \mathbb E [ X_{n \wedge T} \mathbf{1}_{\{T \le n\}} | \mathcal F_n ] = X_{n \wedge T} \mathbf{1}_{\{T \le n\}}$.
It follows from $\{T > n\} \in \mathcal F_n$ that $\mathbf{1}_{\{T > n\}}$ is $\mathcal F_n$-measurable. Hence $\mathbb E [ X_{n+1} \mathbf{1}_{\{T > n\}} | \mathcal F_n ] = \mathbf{1}_{\{T > n\}} \mathbb E [ X_{n+1} | \mathcal F_n ]$. On the other hand, $(X_n)_{n \in \mathbb N}$ is a sub-martingale. So $\mathbb E [ X_{n+1} | \mathcal F_n ] \le X_n$.
As such, we get $$\begin{aligned} \mathbb E [ X_{(n+1) \wedge T} | \mathcal F_n ] &\le X_{n \wedge T} \mathbf{1}_{\{T \le n\}} + \mathbf{1}_{\{T > n\}} X_n \\ &= X_{n \wedge T} \mathbf{1}_{\{T \le n\}} + \mathbf{1}_{\{T > n\}} X_{n \wedge T} \\ &= X_{n \wedge T} \end{aligned}$$