. There is a second part to Theorem 8.2.4 (optional sampling), which says the following.
Theorem 8.8.1 (Optional sampling - Part II). Let $X(t), t \geq 0$, be $a$ submartingale, and let $\tau$ be a stopping time. Then $\mathbb{E} X(t \wedge \tau) \leq \mathbb{E} X(t)$. If $X(t)$ is a supermartingale, then $\mathbb{E} X(t \wedge \tau) \geq \mathbb{E} X(t)$. If $X(t)$ is a martingale, then $\mathbb{E} X(t \wedge \tau)=\mathbf{E} X(t)$.
The proof is technical and is omitted. The idea behind the statement about submartingales is the following. Submartingales tend to go up. Since $t \wedge \tau \leq t$, we would expect this upward trend to result in the inequality $\mathbb{E} X(t \wedge \tau) \leq$ $\mathbb{E} X(t)$. When $\tau$ is a stopping time, this intuition is correct. Once we have Theorem 8.8.1 for submartingales, we easily obtain it for supermartingales by using the fact that the negative of a supermartingale is a submartingale. Since a martingale is both a submartingale and a supermartingale, we obtain the equality $\mathbb{E} X(t \wedge \tau)=\mathbb{E} X(t)$ for martingales.
Use Theorem 8.8.1 and Lemma 8.5.1 to show in the context of Subsection 8.5.1 that
$$
\widetilde{\mathbb{E}}\left[e^{-r T}(S(T)-K)^{+}\right]=\max _{\tau \in \mathcal{T}_{0, T}} \widetilde{\mathbb{E}}\left[e^{-r \tau}(S(\tau)-K)^{+}\right],
$$
where as usual we interpret $e^{-r \tau}(S(\tau)-K)^{+}$to be zero if $\tau=\infty$. The righthand side is the American call price analogous to Definition 8.4.1 for the American put price. The left-hand side is the European call price.