In my study of strong Markov property of an RCLL canonical Markov process I encounter the following definition:
Suppose $Y_t:\omega\rightarrow \omega(t)$ is canonical Markov process with respect to its raw filtration $\mathbb{F}^0$, with transition probability kernel $(K_t)_t$, where $\omega$ is RCLL function of $t$ taking values in polish space S, then $Y$ has strong Markov property wrt to the RC filtration $\mathbb{F}$ if for all $\mathbb{F}$-stopping time $\tau$,
$$E[f(Y_{\cdot+\tau})\mid\mathcal{F}_\tau]=E_{Y_\tau}[f(Y)]$$ on $\{\tau<\infty\}$, where $E_\mu$ refer to taking expectation conditional on initial value $\mu$ and the same transition probability, and $f(\omega)$ bounded measurable real positive RV of the sample path.
Then there is this claim that the above condition is equivalent to for all $\mathbb{F}$-stopping time $\tau$,
$$E[f(Y_{\cdot+\tau}){\bf 1}_{\tau<\infty}]=E[E_{Y_\tau}[f(Y)]{\bf 1}_{\tau<\infty}],$$
i.e., the first equation only need to hold in integrated form. I tried to prove this claim but was not able to make any progresses. Any suggestions? Thank you.
Let $\tau$ be a bounded $\mathbb{F}$-stopping time and $F \in \mathcal{F}_{\tau}$. We define a new stopping time by setting
$$\varrho(\omega) := \begin{cases} \tau(\omega), & \omega \in F, \\ \infty, &\text{otherwise} \end{cases}.$$
Then, by assumption,
$$\mathbb{E}(f(Y_{\cdot+\varrho}) 1_{\{\varrho<\infty\}}) = \mathbb{E}(\mathbb{E}_{Y_{\varrho}}(f(Y)) 1_{\{\varrho<\infty\}}),$$
i.e.
$$\mathbb{E}(f(Y_{\cdot+\tau}) 1_F) = \mathbb{E}(\mathbb{E}_{Y_{\tau}}(f(Y)) 1_F).$$
Since this holds for any $F \in \mathcal{F}_{\tau}$, we conclude
$$\mathbb{E}(f(Y_{\cdot+\tau}) \mid \mathcal{F}_{\tau}) = \mathbb{E}_{Y_{\tau}}(f(Y)).$$
Remark: A very similar result holds true for martingales. In fact, an adapted integrable process $(M_t,\mathcal{F}_t)_{t \geq 0}$ is a martingale if and only if
$$\mathbb{E}(M_{\varrho}) = \mathbb{E}(M_{\tau})$$
for all bounded ($\mathcal{F}_t)$-stopping times $\varrho$ and $\tau$.