Let $E$ be a locally compact polish space, $(P_t)_{t \geq 0}$ be a Chapman-Kolmogorov family on $E$ and $B(E,\mathbb{R})$ be the bounded measurable functions from $E$ to $\mathbb{R}$ . Define the linear operator
$$ \bar{P}_t:B(E,\mathbb{R}) \rightarrow B(E,\mathbb{R}), P_tf(x)=\int_\mathbb{E}f(y)P_t(x,dy). $$
Question: Is $(P_t)$ already uniquely determined by $\bar{P}_t\mid_{C_0(E,\mathbb{R})}$?
I know that it's uniquely determined by $(\bar{P}_t)$ as $P_t(x,A)=(\bar{P}_t\mathbb{1_A})(x)$ for $x \in E, A \in \mathcal{B}(E)$, but I am not sure about the restriction.
Two probability measures on a Polish space $E$ coincide on $\mathcal B(E)$ once their integrals assign the same values to each function in $C_b(E,\Bbb R)$, for example. Thus for each $x\in E$ and each $t\ge 0$, the probability measure $P_t(x,\cdot)$ is determined by $\bar P_t|_{C_b(E,\Bbb R)}$.
And $C_0(E,\Bbb R)$ is enough in the locally compact case.