Calculate Radon-Nikodym derivative For the laws of two pure-jump Markov processes $\mu_1$ and $\mu_2$ on $\mathbb R^n$, which generators are
$H_1f(x)=\int h(x,dy) (f(y)-f(x))$
and $H_2f(x)=\int e^{-g(x,y)} h(x,dy) (f(y)-f(x))$ (corresponding), where $\int c(x,dy)<\infty$ $\forall x$, $g$ is continuous and bounded. The Radon-Nikodym derivative is equal to $$\frac{d \mu_1}{d\mu_2}|_{F_t}(X)=
\exp(-\sum_{s\le t} g(X_s^-,X_s)-\int_0^t \int h(X_s,dy)(e^{-g(X_s,y)}-1)  ).$$ (This is my concern, but I'm not sure that this is the exact form of derivative, as I have some problems with computations).
Please, help me prove it.
 A: Since $X_t$ is a pure-jump process it can be reconstructed from the initial value, jump times, and value after each jump. Let $\tau_0=0$ and $\tau_j$ be the $j$'th jump time. Consider the discrete-time process $Y_j=(\tau_j,X_{\tau_j})$, $j=0,1,...$, which is a homogeneous Markov process with transition kernel
$$ k_i(t,x,ds,dy)=\exp\left[\bar h_i(x)(t-s)\right]h_i(x,dy)ds, \quad s>t,$$
and zero otherwise, where $\bar h_i(x)=\int h_i(x,dy)$.
Inserting the initial distribution of $X_0$ and the probability of having no jump between $\tau_n$ and $t$ will give an exponential from $0$ to $t$ from collecting the exponentials when multiplying all the transition kernels for $\tau_n<t$. Taking the ratio, we find that your formula is correct up to a different sign. Denote the set of jump times in $[0,t]$ by $J_t=\{\tau_j|\tau_j\leq t\wedge j>0\}$. Then 
$$\frac{d\mu_1}{d\mu_2}\Bigg|_{\mathcal{F}_t}=\exp\left[\int_0^t \left(\bar h_2(X_s)-\bar h_1(X_s)\right)ds\right]\prod_{s\in J_t}\frac{dh_1(X_{s^-},\cdot)}{dh_2(X_{s^-},\cdot)}(X_s),$$
where $\bar h_i(x)=\int h_i(x,dy)$.
When $h_1=h$ and $h_2=e^{-g}h$, we have $dh_1/dh_2=e^g$ and we can rewrite the above as
$$\frac{d\mu_1}{d\mu_2}\Bigg|_{\mathcal{F}_t}=\exp\left[\int_0^t h(X_s,dy)\left(e^{-g(X_s,y)}-1\right)ds+\sum_{s\in J_t}g(X_{s^-},X_s)\right].$$
