Let $X_n : \Omega \to \{ -1, 1 \}$ be a random variable with $P(X = -1) = P(X = 1) = 1/2$, like tossing a coin, and $M_n = \sum_{i=1}^n X_n$. Also let $\tau_m : \Omega \to \mathbb N \cup \{ \infty\}$ be the stopping time $\tau_m(\omega) := \inf\{ n : M_n(\omega) = m \}$ and $S_n = 4 \cdot 2^{M_n}$. I want to evaluate for $m > 0$ $$ E\left[ \left( \frac{1}{1+r}\right)^{\tau_{-m}} (K - S_{\tau_{-m}})\right] $$ where $K \in \mathbb R$ and $r > 0$ are some numbers. In my textbook we have $$ E\left[ \left( \frac{1}{1+r}\right)^{\tau_{-m}} (K - S_{\tau_{-m}})\right] = 4(1 - 2^{-m}) E\left[ \left( \frac{1}{1+r}\right)^{\tau_{-m}} \right]. $$ I understand that this holds if $\tau_{-m}(\omega) < \infty$, but what if $\tau_{-m}(\omega) = \infty$, i.e. we never have $M_n(\omega) = m$? Then this expression does not make any sense at all?
Remark: For context, this is from the book Stochastic Calculus for Finance I, I linked the relevant part from Google books.