Let
- $E$ be a locally compact separable metric space
- $(T(t))_{t\ge0}$ be a strongly continuous contraction semigroup on $C_0(E)$ with generator $(\mathcal D(A),A)$
- $(T_n(t))_{t\ge0}$ be a uniformly continuous contraction semigroup on $B(E)$ (bounded Borel measurable functions $E\to\mathbb R$ equipped with the supremum norm) for $n\in\mathbb N$
- $(\Omega,\mathcal A,\operatorname P)$ be a probability space
- $(X_t)_{t\ge0}$ be an $E$-valued càdlàg process on $(\Omega,\mathcal A,\operatorname P)$ with $$\operatorname E\left[f(X_t)\mid(X_r)_{r\le s}\right]=(T(t-s)f)(X_s)\;\;\;\text{almost surely}\tag1$$ for all $f\in C_0(E)$ and $t\ge s\ge0$
- $(X^n_t)_{t\ge0}$ be an $E$-valued càdlàg process on $(\Omega,\mathcal A,\operatorname P)$ with $$\operatorname E\left[f(X^n_t)\mid(X^n_r)_{r\le s}\right]=(T_n(t-s)f)(X_s)\;\;\;\text{almost surely}\tag2$$ for all $f\in C_b(E)$ and $t\ge s\ge 0$
Assume $$X^n_0\xrightarrow{n\to\infty}X_0\tag3$$ weakly and that for all $f\in\mathcal D(A)$ and $t\ge0$, there is a $(B_n)_{n\in\mathbb N}\subseteq\mathcal B(E)$ and a $(f_n)_{n\in\mathbb N}\subseteq B(E)$ with $$\operatorname P\left[\forall s\in[0,t]:X^n_s\in B_n\right]\xrightarrow{n\to\infty}1\tag4,$$ $$\sup_{n\in\mathbb N}\left\|f_n\right\|_\infty<\infty\tag5$$ and $$\left\|f_n-f\right\|_{B_n}+\left\|A_nf_n-Af\right\|_{B_n}\xrightarrow{n\to\infty}0\tag6$$ (where $\left\|g\right\|_{B_n}:=\sup_{x\in B_n}|g(x)|$).
Fix $f\in\mathcal D(A)$ and $t\ge 0$. Are we able to conclude $\operatorname E\left[f(X^n_t)\right]\xrightarrow{n\to\infty}\operatorname E\left[f(X_t)\right]$?
My attempt is to write $$\operatorname E\left[f(X^n_t)\right]=\operatorname E\left[(T_n(t)f)(X^n_0)\right]=\operatorname E\left[f(X^n_0)\right]+\operatorname E\left[\int_0^t(A_nf)(X^n_r)\:{\rm d}r\right]\tag7$$ and analogously $$\operatorname E\left[f(X_t)\right]=\operatorname E\left[(T(t)f)(X_0)\right]=\operatorname E\left[f(X_0)\right]+\operatorname E\left[\int_0^t(Af)(X_r)\:{\rm d}r\right]\tag8.$$ With $(7)$ and $(8)$, we obtain $$\left|\operatorname E\left[f(X^n_t)\right]-\operatorname E\left[f(X_t)\right]\right|\le\left|\operatorname E\left[f(X^n_0)\right]-\operatorname E\left[f(X_0)\right]\right|+\operatorname E\left[\int_0^t\left|(A_nf)(X^n_r)-(Af)(X_r)\right|\:{\rm d}r\right]\tag9,$$ but I'm unable to proceed from here.
EDIT: On the other hand, the claim would follow from $$\operatorname E\left[f(X^n_t)-(T(t)f)(X^n_0)\right]\xrightarrow{n\to\infty}0.\tag{10}$$ We know that there are $B_n$ and $f_n$ as above. Let $U^n:=f_n(X^n)$ and $V^n:=(A_nf_n)(X^n)$. Then we may somehow use that $U^n-\int_0^{\;\cdot\;}V^n_s\:{\rm d}s$ is a martingale. Maybe we need to stop this martingale at $$\tau_n:=\inf\left\{t>0:\left(\int_0^t\left|(A_nf_n)(X^n_s)\right|^2\:{\rm d}s\right)^2>\sqrt t(\left\|Af\right\|_\infty+1)\right\},$$ noting that $\operatorname P\left[\tau_n\le t\right]\xrightarrow{n\to\infty}0$. The idea is that we may insert this martingale in $(10)$ and somehow show the convergence.
Remark: There is a superior result in section 8 of chapter 4 in the book of Ethier and Kurtz, but I would like to find a direct proof in this special situation.