Why $P^\mu[X_{S+t} \in \Gamma \mid \mathcal{F}_{S+}]=P^\mu[X_{S+t} \in \Gamma \mid X_{S}]=0 \text{ ,} P^\mu\text{-a.s. on } \{S=\infty\}$ For any progressively measurable process $X$ and any optional time $S$ of $\{\mathcal{F}_t\}$ why do we have that
$$P^\mu[X_{S+t} \in \Gamma \mid \mathcal{F}_{S+}]=P^\mu[X_{S+t} \in \Gamma \mid X_{S}]=0 \text{     ,} P^\mu\text{-a.s. on } \{S=\infty\}$$
This is remark $2.6.4$ in Section 2.6 of Karatzas and Shreve's Brownian Motion and stochastic Calculus
The author mentions at the beginning of the remark that $\{X_{S+t} \in \Gamma \}:= \{X_{S+t} \in \Gamma,S<\infty \}$. 

Now since $S$ is an $\{\mathcal{F}_t\}$ optional time, $S$ is $ \mathcal{F}_{S+}$ measurable
So we can write
\begin{equation}
\begin{split}
P^\mu[X_{S+t} \in \Gamma \mid \mathcal{F}_{S+}] 1_{\{S=\infty\}}=E^\mu[1_{\{X_{S+t} \in \Gamma\}}  1_{\{S=\infty\}} \mid \mathcal{F}_{S+}]=E^\mu[1_{\{X_{S+t} \in \Gamma,S<\infty\}}\}  1_{\{S=\infty\}} \mid \mathcal{F}_{S+}]\\=0
\end{split}
\end{equation}
since the product of the two indicators functions is zero everywhere
Is this reasoning correct? If not , could you tell me what is problem. Furthermore how could I show that $P^\mu[X_{S+t} \in \Gamma \mid X_{S}]=0$ on $\{S=\infty \}$. Where is the progressive measurability of $X$ being used?
 A: *

*I think your reasoning of $P^\mu[X_{S+t}\in\Gamma|\mathcal F_{S+}]=0$ on $\{S=\infty\}$ is correct.

*According to the definition of $\sigma(X_S)$ given in Problem 1.1.17, $\{S=\infty\}\in\sigma(X_S)$. So I think you can apply the same reasoning as above to verify that $P^\mu[X_{S+t}\in\Gamma|X_S]=0$ on $\{S=\infty\}$.

*To me it seems, therefore, that in asserting 
$$
 P^\mu[X_{S+t}\in\Gamma|\mathcal F_{S+}]
=P^\mu[X_{S+t}\in\Gamma|X_S]
=0\text{  on }\{S=\infty\},
$$ 
we don't need the progressive measurability of $X$. I guess that the authors say "for any progressively measurable process $X$" because immediately preceding the remark, they've defined Markov processes as progressively measurable processes satisfying the Markov property (Definitions 2.6.2 and 2.6.3). That is, "for any progressively measurable process $X$," it seems to me, is to say "for any progressively measurable process $X$ (whether it's a Markov process or not)."
$\quad\quad$Markov processes are defined among progressively measurable processes probably in view of Proposition 1.2.18, i.e., to make sure $X_S\in\mathcal F_{S+}$ (the Markov property is an assertion that a conditional expectation given $\mathcal F_{S+}$ be equal to that given $X_S$, so among other things $X_S$ must be measurable with respect to $\mathcal F_{S+}$).
