We have a sigma-algebra $\mathcal{F}=\mathcal{F}_{\infty}$, a stopping time $T$ and an integrable random variable $X$ and define a martingale by $X_t = \mathbb{E}[X \mid \mathcal{F}_t], t\in[0,\infty]$, for some unspecified underlying filtration $\{\mathcal{F}_t\}$. We need to show that $X^T_t = X_{T \wedge t} = \mathbb{E}\left[X_T\mid \mathcal{F}_t\right]$. The topic on Optional Sampling has not been covered yet.
First of all, I don't see how we can prove measurability (of both $X^T_t$ and $X_T$); is it progressively measurable or cadlag? Integrability is an issue as well, but not my main question: that is about the equality in the title.
I wanted to show that for all $A \in \mathcal{F}_t$, it holds that $\mathbb{E}\left[X_{T \wedge t}\textbf{1}_A\right]=\mathbb{E}\left[X_T\textbf{1}_A\right]$. I thought it would be clever to do the following: \begin{align*} \mathbb{E}\left[X_{T \wedge t}\textbf{1}_A\right]&=\mathbb{E}\left[X_{T \wedge t}\textbf{1}_A\textbf{1}_{\Omega}\right] = \mathbb{E}\left[X_{T \wedge t}\textbf{1}_A\textbf{1}_{\left(\{T\leq t\} \cup \{ T > t\}\right)}\right]\\ &= \mathbb{E}\left[X_{T \wedge t}\textbf{1}_A\textbf{1}_{\{T\leq t\}}\right]+ \mathbb{E}\left[X_{T \wedge t}\textbf{1}_A\textbf{1}_{\{T> t\}}\right]\\ &=\mathbb{E}\left[X_{T}\textbf{1}_A\textbf{1}_{\{T\leq t\}}\right]+ \mathbb{E}\left[X_{T \wedge t}\textbf{1}_A\textbf{1}_{\{T> t\}}\right], \end{align*} but this did not yield anything useful. Can anyone of you shed some light on this? Many thanks in advance!