Let $(B_t)_t$ be a Brownian motion, then I am given a stopping time $\tau_s:=\min(\inf\{t \ge 0; B_t=a\}, \inf\{t \ge s; B_t=b\}; \inf \{t \ge 0;B_t=c\}),$ where $a<0<b<c.$
Now, I want to show that $P(B_{\tau_s}=c)$ is a continuous function in $s.$
If anything is unclear, please let me know, I am happy to provide additional information.
Briefly, let $L^b_t$ be the local time of Brownian motion in state $b$, normalized so that $E[L^b_u]=\int_0^u p_t(0,b)\,dt$. Then, because $L^b_t$ is constant on the interval $[s,T_b(s)]$, $$ \int_s^\infty e^{-\alpha t} p_t(0,b)\,dt =E[\int_s^\infty e^{-\alpha t}\,d_tL^b_t]=E[\int_{T_b(s)}^\infty e^{-\alpha t} d_tL^b_t]. $$ Now by the strong Markov property applied at time $T_b(s)$, the latter expectation is equal to $$ \eqalign{ E[e^{-\alpha T_b(s)}\int_0^\infty e^{-\alpha u} d_uL^b_{T_b(s)+u}] &=E\left[[\exp(-\alpha T_b(s))]\cdot E^b[\int_0^\infty e^{-\alpha u}d_uL^b_u]\right],\cr &=E\left[\exp(-\alpha T_b(s))\right]\cdot \int_0^\infty e^{-\alpha u}p_u(b,b)\,du,\cr } $$ because $L^b_\cdot$ is additive ($L_{u+v}=L^b_v+L^b_u\circ\theta_v$, where $\theta_v$ is the shift-time-by-$v$-units operator). (Here $E^b$ is expectation for Brownian motion started in state $b$.)