A cadlag Feller process for $\mathcal F$ is Markov w.r.t $\mathcal F_+$ (Th. 46, Chap. 1, Stochastic Integration - Protter)

Question

In page 35 of the book Stochastic Integration by P. Protter, he defines a Feller process as follows:

Then he states the following theorem.

In the proof, he used the following strategy:

Next, he argues as follows:

There are two points that I don't understand. How is $Y= (Y_s)_{0\leq s \leq t}$ is cadlag? Clearly, we cannot state

$$ \lim_{n \rightarrow +\infty)} P_{t-(s+1/n)}f(X_{s+1/n})= P_{t-s}f(X_s) $$ just based on the Feller property. Also, in the next sentence, he claims that every cadlag $\mathcal F$-martingale is $\mathcal F_+$-martingale, which maybe wrong. I don't even understand why $Y$ is cadlag immediately...

Answer 1

I also don't understand his argument, but here is another approach to the theorem. By the Feller property and cadlag property of $X$, we have (every equality is a.s.)

$$ Y_s = E[f(X_t)|\mathcal F_s] = P_{t-s}f(X_s) = \lim_{n \rightarrow \infty} P_{t-s}f(X_{s+\frac{1}{n}} )= \lim_{n \rightarrow \infty} E[f(X_{t+\frac{1}{n}})| \mathcal F_{s+\frac{1}{n}}] $$

On the other hand, we can prove that $Z_n := E[f(X_{t+\frac{1}{n}})| \mathcal F_{s+\frac{1}{n}}]$ converge in $L^1$ to $E[f(X_{t})| \mathcal F_{s+}]$. To see this, we first observe

$$ E\Big[\Big|Z_n - E[f(X_t)|\mathcal F_{s+\frac{1}{n}}]\Big|\Big] \leq E[|f(X_{t+\frac{1}{n}})-f(X_t)|] \rightarrow 0, $$

by bounded convergence since $f$ is continuous and bounded, $X$ is cadlag. Also, there is a result, known as Levy's downwards theorem (see e.g. wiki), which states that

$$ E[f(X_t)|F_{s+\frac{1}{n}}] \overset{a.s.+L^1}{\longrightarrow} E[f(X_t)|F_{s+}], $$ and so $Z_n$ converge in $L^1$ to $E[f(X_{t})| \mathcal F_{s+}]$ and a.s. to $Y_s$. So, we must have

$$ E[f(X_{t})| \mathcal F_{s+}] = E[f(X_{t})| \mathcal F_{s}] = P_{t-s}f(X_s), $$

and $X$ remains Markov w.r.t filtration $\mathcal F_+$. The left limit of $Y$ exists a.s. again by Levy's upwards theorem, but I think it's irrelevant here.