Inspired by Process with Markov property but not strong Markov property.
I'm having trouble rigorously verifying that this is a Markov process but not a strong Markov process. From the definition, I want to show that $$X_t(\omega) = \max \{t-T(\omega), 0\}, \text{ where } T\sim \text{Exp}(1)$$ satisfies $$P(X_t \in B \mid X_s) = P(X_t \in B \mid \mathcal F_s)$$ for all $s<t$ and $B\in \mathcal B$ (a Borel set). Now from here, I believe we can simplify to only check that $$P(X_t = x \mid X_s) = P(X_t = x \mid \mathcal F_s)$$ for all $x$, though I'm not sure why we can do this simplification.
Q1: Can we make this simplification and why?
I got that $$P(X_t = x \mid X_s) = P(X_t = x \mid X_s = 0) \cdot 1_{[X_s = 0]} + P(X_t = x \mid X_s > 0) \cdot 1_{[X_s > 0]}$$ where $P(X_t = x \mid X_s = 0)$ is $P(T \geq x)$ if $x=0$ and $P(T = t-x)$ if $x>0$; and $P(X_t = x \mid X_s > 0)$ is $1$ if $x=t-s+X_s$ and $0$ otherwise.
Then, we decide that $\mathcal F_s = \sigma[\bigcup_{i \in [0,s]} X_i^{-1}(\mathcal B)]$. However, I am stuck here. I don't know how to understand $P(X_t \in B \mid \mathcal F_s)$.
Q2: Is everything I have so far correct? How do I continue? Is there an easier way to prove Markov-ness than this?
and lastly
Q3: I haven't even gotten to disproving strong Markov, so that would be my third question. However, it is possible that after getting answers to Q1 and Q2 I'll be able to handle this on my own.
$\def\F{\mathscr{F}}$For the first question, since $X_s = \max(X_t + s - t, 0)$ for $0 \leqslant s < t$, then $σ(X_s) \subseteq σ(X_t)$, which implies that$$ \F^X_t = σ(\{X_s \mid 0 \leqslant s \leqslant t\}) = σ(X_t), $$ thus $P(X_t \in B \mid X_s) = P(X_t \in B \mid \F^X_s)$ trivially holds for any $0 \leqslant s < t$ and $B \in \mathscr{B}(\mathbb{R})$.
For the second question, the deconposition of $P(X_t = x \mid X_s)$ is correct, but$$ P(X_t = x \mid X_s = 0) = \frac{P(X_t = x,\ X_s = 0)}{P(X_s = 0)} = \begin{cases} \dfrac{P(T \geqslant t)}{P(T \geqslant s)}; & x = 0\\ \dfrac{P(T = t - x)}{P(T \geqslant s)} = 0; & x > 0 \end{cases}. $$ In fact, the issue is that $P(X_t \leqslant x \mid X_s)$ should be considered instead of $P(X_t = x \mid X_s)$ because $σ(X_s)$ can be generated by $\{\{X_t \leqslant x\} \mid x \in \mathbb{R}\}$ but not by $\{\{X_t = x\} \mid x \in \mathbb{R}\}$.