I have a question about Dynkin **Hunt** formula.


Last day, I found a formula in this paper [enter link description here](https://pdfs.semanticscholar.org/06bb/40bf01bee9cecaea7aa2e9e7cda7cb5c35e9.pdf).
The formula is the equation (2.5) in this paper, which is called Dynkin **Hunt** formula. I know Dynkin formula. Dynkin formula states
\begin{align\*}
R\_{\alpha}f(x)=R\_{\alpha}^{U}f(x)+E\_{x}[e^{-\alpha \tau\_{U}}R\_{\alpha}f(X\_{\tau\_{U}})],
\end{align\*}
where $\{X\_t,P\_x\}$ is a Markov(Hunt) process defined on a topological space $E$ and $R\_{\alpha}$ is its resolvent. $R\_{\alpha}^{U}f(x)$ is defined by
\begin{equation\*}
R\_{\alpha}^{U}f(x)=E\_{x}[\int\_{0}^{\tau\_{U}}e^{-\alpha t}f(X\_t)\,dt],
\end{equation\*}
where $\tau\_{U}$ is the first exit time from an open set $U$. In other words, $R\_{\alpha}^{U}$ is the resolvent of subprocess $(X\_{t}^{U},P\_x)$ of $(X\_t,P\_x)$ on $U$.


You can see this formula in *Dirichlet Forms and Symmetric Markov Processes* by M. Fukushima, Y. Oshima, and M. Takeda. The equation (4.1.6) in this book is Dynkin formula.


**Question**


Assume $\{X\_t\}$ has transition probability density function $p(t,x,y)$ and transition density function of $\{X\_{t}^{U}\}$ is denoted by $p^{U}(t,x,y)$.


Can we show the Dynkin-**Hunt** formula
\begin{equation\*}
p(t,x,y)=p^{U}(t,x,y)+E\_{x}[1\_{\{\tau\_{U} \le t\}}p(t-\tau\_{U},x,y)]
\end{equation\*}
for every $t>0$, $x,y \in E$?


**My attempt**


Assume resolvent density $R\_{\alpha}(x,y)$ is continuous in $y$. Then, 
from Dynkin formula, we see
\begin{equation\*}
R\_{\alpha}(x,y)=R\_{\alpha}^{U}(x,y)+E\_{x}[e^{-\alpha \tau\_{U}}r\_{\alpha}(X\_{\tau\_{Y}},y)]
\end{equation\*}
for every $\alpha>0,x,y \in E$. Therefore,
\begin{align\*}
\int\_{0}^{\infty}e^{-\alpha t}p(t,x,y)\,dt
&=\int\_{0}^{\infty}e^{-\alpha t}p^{U}(t,x,y)\,dt \\
&+\int\_{0}^{\infty}e^{-\alpha t}E\_{x}[1\_{\{\tau\_{U} \le t\}}p(t-\tau\_{U},X\_{\tau\_{U}},y)]\,dt
\end{align\*}
Therefore, for every $x,y \in E$,
\begin{equation\*}
p(t,x,y)=p^{U}(t,x,y)+E\_{x}[1\_{\{\tau\_{U} \le t\}}p(t-\tau\_{U},x,y)]
\end{equation\*}
holds for a.e. $t$. Can we refine this equation? That is, can we show 
\begin{equation\*}
p(t,x,y)=p^{U}(t,x,y)+E\_{x}[1\_{\{\tau\_{U} \le t\}}p(t-\tau\_{U},x,y)]
\end{equation\*}
for every $t>0$, $x,y \in E$?


**If you know related works about Dynkin-Hunt formula, please let me know.**



Dynkin Hunt formula
The authors of the paper you cite mis-state the matter slightly. Dynkin's formula is indeed a direct consequence of the strong Markov property. Your Laplace inversion argument (with no assumption of continuity) shows that for each fixed $x$,
$$
p(t,x,y)=p^{U}(t,x,y)+E\_{x}[1\_{\{\tau\_{U} \le t\}}p(t-\tau\_{U},x,y)]
$$
for $m\otimes\lambda$-a.e. $(y,t)$, where $m$ is the symmetry measure of $X$ and $\lambda$ is Lebesgue measure on $(0,\infty)$. The key observation now is that the two sides of this identity, as functions of $(y,t)$, are *finely continuous* with respect to the space-time process $(X\_t, r-t)\_{0\le t<r}$, for which $m\otimes\lambda$ is a reference measure. As such, the two sides of the identity agree *for all* $(y,t)$. A nice discussion of such methods can be found in the paper "Excursions of dual processes" by R.K. Getoor and M.J. Sharpe [*Advances in Mathematics* **45** (1982) 259–309]. A detailed discussion of the "Dynkin-Hunt" identity, in the context of Brownian motion, appears in the book *From Brownian motion to Schrödinger's equation* of K.L. Chung and Z.X. Zhao. In the paper you cite, the proces $X$ is symmetric, so arguments based on *quasi-continuity* come into play; perhaps this is what the authors had in mind.


