European Contracts
It's a really important question and as @noob2 commented, the FTAP is normally applied to European-style derivatives, even if they are (strongly) path-dependent, including barrier options and Asian options. The idea is always the same, $V_t=B_t\mathbb{E}^\mathbb{Q}\left[\frac{\xi_T}{B_T}\Big|\mathcal{F}_t\right]$, that is the derivative's price process is the conditional risk-neutral expectation of the future discounted payoff, $\xi_T$ (which may depend on hitting barrier levels etc). It essentially follows from the fact that for any integrable random variable $X$, the process $\mathbb{E}[X|\mathcal{F}_t]$ is a martingale. If you add the price process $V_t$ to a market where discounted asset prices are martingales, then you don't introduce new arbitrage (by the FTAP) and thus, $V_t$ is a no-arbitrage compatible price for trading the payoff $\xi$. So why is the early exercise such a problem for the martingale property?
Buyer's Price for American contracts
The price of a payoff that can be exercised at any time is much more related to something like $$U_t=\sup_{\tau\in \mathcal S_{t,T}}\left\{\mathbb{E}^\mathbb{Q}\left[\frac{\xi_\tau}{B_\tau}\bigg|\mathcal{F}_t\right]\right\},$$
where the supremum is taken over the set of all stopping times (exercise strategies) with values in $\{t,...,T\}$. Of course, $U_T=\frac{\xi_T}{B_T}$. This process $U$ relates to Snell's Envelope. A stopping time $\tau$ is optimal if $U_t=\mathbb{E}^\mathbb{Q}\left[\frac{\xi_\tau}{B_\tau}\Big|\mathcal{F}_t\right]$. The option price would be $B_tU_t$.
Two important properties:
- $U$ dominates the payoff $\xi$: we know this, an American option is always worth at least its immediate payoff (by no-arbitrage)
- $U$ is a supermartingale: that causes the problem with the FTAP (see below)
Optimal Exercise
Let's (recursively) define the following stopping time, $\tau_t^*$ via $\tau_T^*=T$ and for $t<T$ as
\begin{align*}
\tau^*_t=\begin{cases}
t & \text{if } \frac{\xi_t}{B_t}\geq \mathbb{E}^\mathbb{Q}\left[\frac{\xi_{\tau_{t+1}^*}}{B_{\tau_{t+1}^*}}\bigg|\mathcal F_t\right], \\\\
\tau_{t+1}^* & \text{if }\frac{\xi_t}{B_t}< \mathbb{E}^\mathbb{Q}\left[\frac{\xi_{\tau_{t+1}^*}}{B_{\tau_{t+1}^*}}\bigg|\mathcal F_t\right].
\end{cases}
\end{align*}
So what does $\tau^*_t$ mean economically? If the immediate payoff $\xi_t$ is bigger than the continuation value, $B_t\mathbb{E}^\mathbb{Q}\left[\frac{\xi_{\tau_{t+1}^*}}{B_{\tau_{t+1}^*}}\bigg|\mathcal F_t\right]$, then exercise the option ($\tau_t^*=t$) and otherwise, keep holding the option.
Two properties related to this stopping time
- $U_t=\mathbb{E}^\mathbb{Q}\left[\frac{\xi_{\tau_t^*}}{B_{\tau_t^*}}\bigg|\mathcal{F}_t\right]$, i.e. $\tau_t^*$ is optimal
- $U_t=\max\{\frac{\xi_t}{B_t},\mathbb{E}^\mathbb{Q}[U_{t+1}|\mathcal{F}_t]\}$ starting with $U_T=\frac{\xi_T}{B_T}$. This property is also used to define Snell's envelope and captures the entire idea of binomial trees: start at maturity and work backwards, comparing every time whether exercise is optimal (the payoff $\frac{\xi_t}{B_t}$ is larger) or the continuation value of keeping the option for another period. This representation also immediately tells you that $U$ is a supermartingale: $$U_t=\max\left\{\frac{\xi_t}{B_t},\mathbb{E}^\mathbb{Q}[U_{t+1}|\mathcal{F}_t]\right\}\geq \mathbb{E}^\mathbb{Q}[U_{t+1}|\mathcal{F}_t]$$
Summary
Because you can exercise at any time, your option value is a supremum over all exercise strategies (stopping times). The FTAP and martingale pricing would simply take the payoff and construct the corresponding price process by discounting and conditioning but for American options you have to think about the optimal stopping time.
A few notes
- The notes above are kind of from a buyer's perspective. You can take a hedger's perspective and show that a seller has the same price if the buyer behaves optimally.
- As always, if markets are incomplete, $\mathbb Q$ is not unique and infinitely many fair prices may exist.
- All the statements above are proven via backward induction: show that it holds for $t=T$ (normally trivially by construction) and show that if it holds for $t+1$, then it also holds for $t$.
