Suppose that $(X_n,F_n)$ be a martingale and uniformly integrable. Show that if $T$ is a stopping time then $(X_{n \wedge T},F_{n \wedge T})$ is also a uniformly integrable martingale. The uniformly integrable part is trivial.
My attempt:
First note that $X_{min\{n+1,T\}}=X_{n+1}-(X_{n+1}-X_{T})I_{\{n<T\}}$. Since $n \wedge T \leq n$, we have $E(X_{n+1 \wedge T} | F_n)=E(X_{n+1}|F_n)-E(X_{n+1}|F_n)I_{n<T}+E(X_{T} | F_n)I_{n<T}$. Now if $n<T$ then $X_{n \wedge T}=X_n$ and $E(X_{n+1 \wedge T} | F_n)=X_n=X_{n \wedge T}$. Now if $T \leq n$ then then the two terms on the last two terms on the LHS are zero and we get $E(X_{n+1 \wedge T} | F_n)=X_n \neq X_{n \wedge T}$. I'm not sure where the problem is. Any help would be appreciated.
First off, your inequality in the indicator function is in the wrong direction, it should be $$X_{min\{n+1,T\}}=X_{n+1}-(X_{n+1}-X_{T})I_\color{red}{{\{n>T\}}}$$ Second, I don't think that distinguishing between the random events $n>T$ and $n\le T$ is a proper way to proceed, you should just stick with the indicator functions all the way.
As for your problem, first notice that $$ X_{n+1 \wedge T} = X_{n \wedge T} + (X_{n+1} - X_n)\mathbf 1_{T\ge n+1}$$ (i.e., the process $(X_{\boldsymbol\cdot \wedge T})$ can only be incremented at time $n+1$ if $T$ has not yet been reached).
You then have $$\begin{align}\mathbb E[X_{n+1 \wedge T} \mid F_n] &= \mathbb E[X_{n \wedge T} + (X_{n+1} - X_n)\mathbf 1_{T\ge n+1}\mid F_n] \\ &= \mathbb E[X_{n \wedge T}\mid F_n] + \mathbf 1_{T\ge n+1}\mathbb E[(X_{n+1} - X_n)\mid F_n]\end{align} $$ Finally notice that $X_{n\wedge T}$ is $F_n$-measurable and use martingale property of $X$ to conclude.