Feller continuity of the stochastic kernel Given a metric space $X$ with a Borel sigma-algebra, the stochastic kernel $K(x,B)$ is such that $x\mapsto K(x,B)$ is a measurable function and a $B\mapsto K(x,B)$ is a probability measure on $X$ for each $x$
Let $f:X\to \mathbb R$. We say that $f\in \mathcal C(B)$ if $f$ is continuous and bounded on $B$. 
Weak Feller continuity of $K$ means that if $f\in\mathcal C(X)$ then $F\in\mathcal C(X)$ where
$$
F(x):=\int\limits_X f(y)K(x,dy).
$$
I wonder if it implies that if $g\in \mathcal C(B)$ then 
$$
G(x):=\int\limits_Bg(y)K(x,dy)
$$
also belongs to $\mathcal C(B)$?

To make it clearer, you can re-write $ G(x) = \int\limits_X g(y)\mathbf{1}_B(y) K(x,{\rm d} y)$. Now in general $g{\mathbf 1}_B$ is not continuous anymore so as Nate pointed out you should not expect $G$ to be such. 
However, if you take a continuous $g $
with $\overline{{\rm support}(g)} \subsetneq B$ then (I let you :) show this) $g{\mathbf 1}_B$ is still continuous  and so is $G$ then.
