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Henry.L
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Feller condition guarantees the right continuitySorry I probably missed "$(P_t)_{t \ge 0}$ of the joint measurea Hunt process" when I first composed this answer and that is why confusion arise. In your argument $\begin{align*} \lim_{t \to 0}P_{t}f(x)\overset{\Delta}{=}\lim_{t \to 0}E_{x}[f(X_t)]=E_{x}[f(X_0)]=E_{x}[f(x)]=f(x) \end{align*}$

The second equality will fail without Feller property becauseI think it is most beneficial if we start from and stick to definition.

$$\lim_{t \to 0}E_{x}[f(X_t)]=lim_{t\rightarrow 0}\int_\Omega f(X_t(\omega))dP_x(\omega) = \int_\Omega lim_{t\rightarrow 0}f(X_t(\omega))dP_x(\omega) \overset{?}{=} \int_\Omega f(lim_{t\rightarrow 0} X_t(\omega))dP_x(\omega)=E_{x}[f(X_0)]$$ evenHunt process is a strong Markov process with dominancesome regularity on integrand sequence $\{f\circ X_t\}$sample functions, you can only proceed the second equalityusually càdlàg sample functions but not necessarily the thirdsubject to some alternations. Suppose $X$ is on a (locally) compact space $E$ and there is a one-to-one correspondence between Hunt process and Feller-Dynkin semigroups.

The paper you cited provided (You cannot argue by continuity1)(2) as a set of equivalent conditions that allows you to regard the transition semigroup $f$ because$(P_t)_{t \ge 0}$ as a Feller-Dynkin semigroup as well. A Hunt process may admits a transition semigroup satisfying $X_t(\omega)$ is not only(2) of course but it also needs one sequence(1), which is guaranteed by Feller to make the operations within the transition semingroup closed.

I hope I made myself clear now.

Feller condition guarantees the right continuity of the joint measure. In your argument $\begin{align*} \lim_{t \to 0}P_{t}f(x)\overset{\Delta}{=}\lim_{t \to 0}E_{x}[f(X_t)]=E_{x}[f(X_0)]=E_{x}[f(x)]=f(x) \end{align*}$

The second equality will fail without Feller property because

$$\lim_{t \to 0}E_{x}[f(X_t)]=lim_{t\rightarrow 0}\int_\Omega f(X_t(\omega))dP_x(\omega) = \int_\Omega lim_{t\rightarrow 0}f(X_t(\omega))dP_x(\omega) \overset{?}{=} \int_\Omega f(lim_{t\rightarrow 0} X_t(\omega))dP_x(\omega)=E_{x}[f(X_0)]$$ even with dominance on integrand sequence $\{f\circ X_t\}$, you can only proceed the second equality but not necessarily the third one(You cannot argue by continuity of $f$ because $X_t(\omega)$ is not only one sequence), which is guaranteed by Feller.

Sorry I probably missed "$(P_t)_{t \ge 0}$ of a Hunt process" when I first composed this answer and that is why confusion arise.

I think it is most beneficial if we start from and stick to definition.

Hunt process is a strong Markov process with some regularity on sample functions, usually càdlàg sample functions but subject to some alternations. Suppose $X$ is on a (locally) compact space $E$ and there is a one-to-one correspondence between Hunt process and Feller-Dynkin semigroups.

The paper you cited provided (1)(2) as a set of equivalent conditions that allows you to regard the transition semigroup $(P_t)_{t \ge 0}$ as a Feller-Dynkin semigroup as well. A Hunt process may admits a transition semigroup satisfying (2) of course but it also needs (1) to make the operations within the transition semingroup closed.

I hope I made myself clear now.

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Henry.L
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Feller condition guarantees the right continuity of the joint measure. In your argument $\begin{align*} \lim_{t \to 0}P_{t}f(x)\overset{\Delta}{=}\lim_{t \to 0}E_{x}[f(X_t)]=E_{x}[f(X_0)]=E_{x}[f(x)]=f(x) \end{align*}$

The second equality will fail without Feller property because

$$\lim_{t \to 0}E_{x}[f(X_t)]=lim_{t\rightarrow 0}\int_\Omega f(X_t(\omega))dP_x(\omega) = \int_\Omega lim_{t\rightarrow 0}f(X_t(\omega))dP_x(\omega) \overset{?}{=} \int_\Omega f(lim_{t\rightarrow 0} X_t(\omega))dP_x(\omega)=E_{x}[f(X_0)]$$ even with dominance on integrand sequence $\{f\circ X_t\}$, you can only proceed the second equality but not necessarily the third one(You cannot argue by continuity of $f$ because $X_t(\omega)$ is not only one sequence), which is guaranteed by Feller.