If $X$ is a right-continuous, discrete Markov process, then $\displaystyle\lim_{t\downarrow 0}\operatorname P_x\left[X_t=x\right]=1$ Let


*

*$E$ be an at most countable Polish space and $\mathcal E$ be the discrete topology on $E$

*$X=(X_t)_{t\ge 0}$ be a discrete Markov process with values $(E,\mathcal E)$ and distributions $(\operatorname P_x)_{x\in E}$



Claim$\;\;\;$Let $$\operatorname p_t(x,y):=\operatorname P_x\left[X_t=y\right]\;\;\;\text{for }x,y\in E\text{ and }t\ge 0\;.$$ If $X$ is right-continuous, then $$\lim_{t\downarrow 0}p_t(x,x)=1\;.\tag 1$$

How can we prove the claim? I've tried the following: Let $$\tau:=\inf\left\{s>0:X_s\ne X_0\right\}$$ and $x\in E$. Clearly, since $\operatorname P_x$-almost surely $X_0=x$, $$\operatorname P_x\left[\tau>t\right]=\operatorname P_x\left[X_s=X_0\text{ for all }s\in (0,t]\right]\le p_t(x,x)\;\;\;\text{for all }t>0\;.\tag 2$$ Two questions:


*

*How exactly does the right-continuity of $X$ imply $$\tau>0\;\;\;\operatorname P_x\text{-almost surely}\tag 3$$ for all $x\in E$? 

*Why can we conclude $(1)$ from $(2)$ and $(3)$?

 A: Here is a proof by contradiction:
Suppose that $p_t(x,x)$ does not converge to $1$ as $t\downarrow 0$ . Then there exist a sequence of times $t_n \downarrow 0$ and an $\epsilon >0$ such that $p_{t_n}(x,x)\le 1- \epsilon$ for all $n$. 
Moreover, the latter implies, by taking complements, that $P_x(X_{t_n}\ne x) \ge \epsilon$ for all $n$. As the space is discrete, I'll assume the discrete topology, so that $X_{t_n}\ne x$ implies that $X_{t_n}$ lies outside the fixed neighborhood $V=\{x\}$ of $x$.
Now, by the reverse Fatou lemma, $$ \epsilon \le \limsup_n P_x(X_{t_n}\ne x) \le P_x(\limsup_n\{X_{t_n}\ne x\}) = P_x\{X_{t_n}\not \in V \text{ for infinitely many }t_n\},$$
and this implies that, with probability at least $\epsilon$, $X_{t_n}\overset{n}{\not \to} x$, contradicting the right continuity claim.
A: I don't think your Claim is correct in the generality you have stated it in.
Consider a Markov chain on the state space $E=\{1, 1/2, 1/3,\ldots\}\cup\{0\}$ with the topology it inherits as a subspace of the real line. The process $X_t$ is to be a sort of "pure birth" process: If started in state $1/n$, $X_t$ holds there for an exponential random time with mean value $n^{-2}$ and then jumps to state $1/(n-1)$ (here $n=2,3,\ldots$), etc. The state $1$ is absorbing. Starting from the state $0$ the process move (very rapidly at first!) through the non-zero states in increasing order. Think of the law $\Bbb P_0$ as the limit $\lim_{x\downarrow 0}\Bbb P_x$. The sample paths of this chain are right continuous. But  $\Bbb P_0[X_t\not=0,\forall t>0]=1$. In particular, $p_t(0,0)=0$ for all $t>0$, but $\Bbb P_x[\tau=0]=1$.
As requested by Did: Let $\xi_k$, $k=2,3,\ldots$ be independent exponential random variables, with $\Bbb E \xi_k=k^{-2}$ ($\xi_k$ represents the holding time in state $1/k$). Define $T_n=\sum_{k>n} \xi_k$. This is the time it takes $X_t$ to reach state $1/n$ if started in state $0$. We then have $p_t(0,1/n)=\Bbb P[T_n\le t]-\Bbb P[T_{n-1}\le t]$. One can easily write down an expression for the Laplace tansform $\int_0^\infty e^{-\alpha t}p_t(0,1/n)\,dt$. I'm not sure how easy it would be to express $p_t(0,1/n)$ in a closed form expresion.
