Brownian Motion infinitesimal generator $\int Lf(x) \mu(dx) = 0$

Question

Take $B(t)$ as a standard Brownian Motion.

Consider a real-valued Markov process with infinitesimal generator L.
Recall that µ is an invariant measure if $$\int E[f(B(t))] \mu(dx) =
\int f(x) \mu(dx)$$ for any bounded measurable function $f$ and all $t
> 0$.

Prove that $$\int Lf(x) \mu(dx) = 0$$ for functions $f$ in the domain
of $L$.

I need some help with this problem. I know that $E[B(t)]=0$ but it seems like unless $f$ is linear, I cannot make the jump to $E[f(B(t))]=0$.

Answer 1

By $\mu$-invariance we have $\int (E^x f(B(h))-f(x))\mu(dx) = 0$ for all $h>0$ hence using dominated convergence $$ \int L f(x) \mu(dx) = \lim_{h\to 0} h^{-1}\int (E^x f(B(h))-f(x))\mu(dx) = 0. $$