Stopped Feynman-Kac Functional

Let D be a domain in ℝd, and for a function q ∈ βd, let $$ e_q (\tau _\mathcal{D} ) = \exp (\int_0^{\tau _\mathcal{D} } {q(X_t )dt} ), $$ where {X t} is the Brownian motion in ℝd, and τD is the exit time from D defined in Section 1.5. The random variable in (1) is well defined if and only if $$ \int_0^{\tau _\mathcal{D} } q (X_t )dt $$ is well defined, almost surely. This is trivially the case if q ∈ 0, or if q is bounded and τD < ∞. To see that this is also the case when q ∈ J and τD < ∞, we need the Corollary to Proposition 3.8 which implies that for each t > 0, $$ \int_0^t {\left| q \right.(X_s )\left| {ds < \infty } \right.} $$ a.s.; thus, the same is true when t is replaced by τD provided the latter is finite a.s. As before, ‘a.s.’ will be omitted in what follows when the context is obvious. Under these circumstances we have 0 < e_q (τD) < ∞; and in fact for each x ∈ ℝd , if P_x{τD < ∞} > 0, then $$ 0 < E^x \left\{ {\tau _\mathcal{D} < \infty ;e_q (\tau _\mathcal{D} )} \right\} \leqslant \infty . $$