Update: So it turns out that the question I was given is actually wrong! $X_t$ should be defined with $\{T>t\}$ not $\{T<t\}$.
My question is the following. It looks really easy, and so I 'just did it'... getting a completely difference answer. I'm clearly doing something rather stupid, but I can't for the life of me see what!
Let $T \sim \text{Exp}(1)$. Set $X_t = e^t 1_{\{T<t\}}$. Show that $$E(X_t1_{\{T>r\}}) = E(X_s1_{\{T>r\}}) \; \text{for} \ r \le s \le t.$$
Here's what I've done:
$$E(X_t1_{T>r}) = E(e^t1_{r<T<t}) = e^tP(r<T<t) = e^t(e^{-r}-e^{-t}) = e^{t-r}-1,$$ $$E(X_s1_{T>r}) = E(e^s1_{r<T<s}) = e^sP(r<T<s) = e^s(e^{-r}-e^{-s}) = e^{s-r}-1.$$ Thus the two aren't equal...
Any advice on this would be most appreciated!
Just before this, the question asks to describe the natural filtration of $X$. It then goes on after to ask to deduce that $(X_t)_{t \ge 0}$ is a cadlag martingale. That should be fine: just use the fact that the indicator functions will form a $\pi$ system containing the $\sigma$-algebra and apply Dynkin's lemma (or something like that) -- that's not the important bit of this SE question.
I think your calculation is correct (there is a typo in the 2nd line, 2nd "=", it should read $e^s$ instead of $e^t$). It implies that $(X_t)_{t \geq 0}$ is a submartingale. In fact, that's the best we can expect as $\mathbb{E}X_t$ is (strictly) increasing. (Note that $\mathbb{E}X_0=0$ and $\mathbb{E}(X_t)= e^t \mathbb{P}(T<t) \to \infty$ as $t \to \infty$.)