I'm trying to understand the proof of optional stopping theorem for continuous Martingales. They begin by recalling the equivalent fact for a Martingales that take values on a discrete set:
Recall the optional sampling theorem for stopping times $\alpha, \beta: \Omega \rightarrow {t_j}_{j \geq 1}$ ̨with values in a discrete set: $$(X(t_j),F_{t_j})_{j \geq 1} \text{ submartingale, } \alpha \leq \beta \leq C \implies X(\alpha) \leq E[X(\beta)|F_{\alpha}]$$
They proceed to use the statement in the proof of the continuous case:
Using the discrete version of optional sampling theorem with $\alpha = \sigma_j \wedge k$ and $\beta = \tau_j \wedge k$ reveals $X(\sigma_j \wedge k), X(\tau_j \wedge k) \in L^1(P)$ and $X(\sigma_j \wedge k) \leq E[X(\tau_j \wedge k)|F_{\sigma_j \wedge k}]$.
I see where the inequality comes from, but what about the assertion that $X(\sigma_j \wedge k), X(\tau_j \wedge k) \in L^1(P)$?
Here is a direct way to see integrability using the definition of $\tau_j$. The definition of $\tau_j$ given in your book (included here for completeness) is $$\tau_j(\omega) = \begin{cases} \tau(\omega) \qquad \text{ if } \tau(\omega) = \infty, \\n2^{-j} \qquad \text{ if }(n-1)2^{-j} \leq \tau(\omega) < n2^{-j} \end{cases}$$ where $\tau$ is some stopping time (the case of $\sigma_j$ will be similar so I just deal with $\tau_j$ here).
In particular, $\tau_j \wedge k \in \{ 2^{-j}n : n \leq 2^j k\}$. Hence $$|X(\tau_j \wedge k)| \leq \sum_{n=1}^{2^jk} |X(2^{-j}n)| \in L^1(P)$$ since, by assumption, $X(2^{-j}n) \in L^1(P)$ for each $n$. So $X(\tau_j \wedge k) \in L^1(P)$.