Show submartingale property. Let $\tau$ be a stopping time. Let $X_k$ be iid random variables such that $E[X_i] = m < \infty$. Also, $m>0$.Show that $\sum_{k=1}^{\min(\tau,n)} X_k$ is a submartingale. 
We need to show that:
$E[\sum_{k=1}^{\min(\tau,n)} X_k | F_{n-1}] \geq \sum_{k=1}^{\min(\tau,n-1)} X_k$
My attempt: We can easily show that the process $(\sum_{k=1}^{n}X_k)_{n\in \mathbb{N}}$ is a supermartingale. It follows by Doob's optional sampling theorem, that $E[X_{\tau}] \geq E[x_0]$. Now write:
$$E[\sum_{k=1}^{\min(\tau,n)} X_k|F_{n-1}] = E[\mathbb{1}_{\tau<n}\sum_{k=1}^{\tau} X_k+\mathbb{1}_{\tau \geq n}\sum_{k=1}^{n} X_k | F_{n-1}] = E[\mathbb{1}_{\tau<n}(\sum_{k=1}^{\tau-1}  X_k+X_{\tau})+\mathbb{1}_{\tau \geq n}\sum_{k=1}^{n} X_k | F_{n-1}]$$
From here, I think some of the terms are $F_{n-1}$ measurable. But the $\tau<n$ indicator is confusing me because I don't know how to deal with it.  
 A: Here is a proof based on direct calculations; it does not use the optional stopping theorem.
For brevity of notation set $S_n := \sum_{k=1}^n X_k$. Obviously,
$$\begin{align*} \sum_{k=1}^{\min\{\tau,n\}} X_k = \left( 1_{\{\tau < n\}} + 1_{\{\tau \geq n\}} ´\right) \sum_{k=1}^n X_k = \underbrace{1_{\{\tau < n\}}S_{\tau}}_{=:Y_1} + \underbrace{1_{\{\tau \geq n\}} S_n}_{Y_2}. \end{align*}$$
As $\tau$ is a stopping time, we have $$\{\tau \geq n\} = \{\tau < n\}^c \in \mathcal{F}_{n-1},$$ and therefore, using pull out,
$$\mathbb{E}(Y_2 \mid \mathcal{F}_{n-1}) = 1_{\{\tau \geq n\}} \mathbb{E}(S_n \mid \mathcal{F}_{n-1}).$$
Since $(S_n)_{n \in \mathbb{N}}$ is a submartingale (check it!), we get
$$\mathbb{E}(Y_2 \mid \mathcal{F}_{n-1}) \geq 1_{\{\tau \geq n\}} S_{n-1}.$$
On the other hand,
$$\{\tau<n\} = \bigcup_{j=0}^{n-1} \{\tau=j\} \tag{1}$$
and so
$$\begin{align*} \mathbb{E}(Y_1 \mid \mathcal{F}_{n-1}) &= \sum_{j=0}^{n-1} \mathbb{E}(1_{\{\tau=j\}} S_{\tau} \mid \mathcal{F}_{n-1}) = \sum_{j=0}^{n-1} \mathbb{E}(1_{\{\tau=j\}} S_j \mid \mathcal{F}_{n-1}). \end{align*}$$
As $\{\tau=j\} \in \mathcal{F}_j$, we can use the pull out to conclude
$$\begin{align*} \mathbb{E}(Y_1 \mid \mathcal{F}_{n-1}) = \sum_{j=0}^{n-1} 1_{\{\tau=j\}} \mathbb{E}(S_j \mid \mathcal{F}_{n-1})  &= \sum_{j=0}^{n-1} 1_{\{\tau=j\}} S_j \stackrel{(1)}{=} 1_{\{\tau<n\}} S_{\tau}. \end{align*}$$
Adding all up gives
$$\mathbb{E} \left( \sum_{k=1}^{\min\{\tau,n\}} X_k\mid \mathcal{F}_{n-1} \right) \geq 1_{\{\tau<n\}} S_{\tau} + 1_{\{\tau \geq n\}} S_{n-1} = S_{\min\{\tau \wedge (n-1)\}}.$$
