I am trying to understand the proof of Theorem 2.2.2(Optional Stopping Theorem) in Fleming and Harrington's Counting Processes and Survival Analysis. Let $\{X(t):0\leq t<\infty \}$ be a right-continuous $\mathcal{F}_t$-martingale, and let $\tau$ be an $\mathcal{F}_t$-stopping time. The author mentioned that $X(t\wedge \tau)$ is $\mathcal{F}_t$-measurable. How do we prove this? There're proofs for discrete case, but I cannot find the proof for a continuous case.
I guess maybe we can write $X(t\wedge \tau)$ as $\int_0^t I(\tau\geq t)dX_t + X_0$. To make this formula meaningful, we need $X(\cdot)$ to be of bounded variation on $[0,t]$, is it true for an arbitrary martingale?
By the way there is another thing I am not sure about. When they proved $E\{X(t\wedge \tau)|\mathcal{F}_s\}=X(s\wedge \tau) a.s.$, they wrote $X(t\wedge \tau)=I_{\{\tau<s\}}X(t\wedge \tau) + I_{\{ \tau \geq s \}}X(t\wedge\tau)$. Is there any reason to use $\{\tau <s\}$ instead of $\{\tau \leq s\}$? I cannot see there is any difference here.
It is a standard fact that $X_{\tau}$ is $\mathcal F_{\tau}$ measurable. You can find a proof in any book that discusses stopping times. Once you know this you can just use the identity $X_{\tau \wedge t} ^{-1} (A)= ((\tau \leq t)\cap X_\tau^{-1}(A)) \cup [(\tau >t)\cap (X_t^{-1}(A)]$. Use definition of $\mathcal F_{\tau}$ for the first part.