let $P_x$ a probability measure under which $(B_t)_{t\geq 0}$ is a brownian motion startetd at $x>0$ and let $\tau_0 := \inf\lbrace B_t = 0\rbrace$ the first hitting time of $0$. For positive $t$, $y$ and $\varepsilon$, I want to show:
$$P_x(B_t \in (y,y+\varepsilon),\tau_0< t) = P_x(B_t \in (-y-\varepsilon,-y))$$
I want to use the strong Markov property at the stopping time $\tau_0$. Do I have to integrate with respect to $P_x(\tau_0 \in ds)$?
EDIT: We can safely write $$P_x(B_t \in (y,y+\varepsilon),\tau_0< t) = E_x[1_{\lbrace\tau_0 < t\rbrace} E_x[1_{\lbrace B_t\in (y,y+\varepsilon) \rbrace}|\mathcal{F}_{\tau_0}]].$$ From my understanding this equals $$\int\limits_{\Omega} 1_{\lbrace\tau_0(\omega) < t\rbrace} \left(E_x[1_{\lbrace B_t\in (y,y+\varepsilon) \rbrace}|\mathcal{F}_{\tau_0}]\right)(\omega)P_x(d\omega).$$
Now, is it true, that for any $\omega$ with $\tau_0(\omega) < t$ we can apply the strong Markov property so that $$ \left(E_x[1_{\lbrace B_t\in (y,y+\varepsilon) \rbrace}|\mathcal{F}_{\tau_0}]\right)(\omega) = E_{B_{\tau_0(\omega)}} [1_{\lbrace B_t-\tau_0(\omega)\in (y,y+\varepsilon) \rbrace}] $$ a.s. ?
Is the random variable $\left(E_x[1_{\lbrace B_t\in (y,y+\varepsilon) \rbrace}|\mathcal{F}_{\tau_0}]\right)(\omega)$ a.s. the same as $\left(E_x[1_{\lbrace B_t\in (y,y+\varepsilon) \rbrace}|\mathcal{F}_{\tau_0(\omega)}]\right)(\omega)$ ?