I'm really stuck on this bit, maybe someone can help be along:
Let $X$ be a canonical, right-continuous Markov process with values in a Polish state space $E$, equipped with Borel $\sigma$-algebra $\mathcal{E}$. Assume $t \mapsto \mathbb{E}_{X_t}f(X_s)$ right-continuous everywhere for each bounded continuous function $f : E \mapsto R$. For $x \in E$ consider the stopping time $\sigma_x = \inf\{t > 0 \,|\, X_t \neq x\}$.
Now I want to show using the strong markov property that
$$ \mathbb{P}_x\{X_{\sigma_x}=x,\, \sigma_x < \infty\} = \mathbb{P}_x\{X_{\sigma_x}=x,\, \sigma_x < \infty\}\mathbb{P}_x\{\sigma_x = 0\}.$$
What I've got so far is by the strong markov property for canonical process is $$ \mathbb{1}_{\{\sigma_x < \infty\}}E_x(Z\circ\theta_{\sigma_x}|\mathcal{F}_{\sigma_x}) = \mathbb{1}_{\{\sigma_x < \infty\}}E_{X_{\sigma_x}}(Z)$$ for some $\mathcal{F}_\infty$-measurable random variable $Z$. This is then equal to $$ \mathbb{1}_{\{\sigma_x < \infty\}}E_x (\mathbb{1}_{A}Z\circ\theta_{\sigma_x}) = \mathbb{1}_{\{\sigma_x < \infty\}}E_x(\mathbb{1}_AE_{X_{\sigma_x}}Z)$$ for some $A \in \mathcal{F}_{\sigma_x}$. My problem is how to choose $Z$ and $A$ to get the desired identity. My first thought to take $A = \{\sigma_x = 0\}$ and $Z = \mathbb{1}_{\{X_{\sigma_x}=x\}}$ gives \begin{align*} \mathbb{1}_{\{\sigma_x < \infty\}}E_x (\mathbb{1}_{\{\sigma_x = 0\}}\mathbb{1}_{\{X_{\sigma_x}=x\}}\circ\theta_{\sigma_x}) &= \mathbb{1}_{\{\sigma_x < \infty\}}E_x(\mathbb{1}_AE_{X_{\sigma_x}}Z) \\ &= P_x\{X_{\sigma_x}=x,\,\sigma_x < \infty\}\} \end{align*} but I can't figure out what the left side becomes, and wether it gives the right expression. Any ideas?
