Let $b\in C^1(\mathbb R)$, $\sigma\in C^2(\mathbb R)$, $$Lf:=bf'+\frac12\sigma^2f''\;\;\;\text{for }f\in C^2(\mathbb R)$$ and $$L^\ast g:=\frac12(\sigma^2g)''-(bg)'\;\;\;\text{for }g\in C^2(\mathbb R).$$ Moreover, let $\varrho\in C^2(\mathbb R)$ with $$\int\varrho(x)\:{\rm d}x=1\tag1$$ and $$L^\ast\varrho=0\tag2$$ Now, let $\mu$ denote the measure with density $\varrho$ with respect to the Lebesgue measure $\lambda$.
Are we able to show that $$\int f(Lg)\:{\rm d}\mu=\int g(Lf)\:{\rm d}\mu\tag3$$ for all $f,g\in C_c^2(\mathbb R)$?
Clearly, $$\int(Lf)g\:{\rm d}\lambda=\int f(L^\ast g)\:{\rm d}\mu\tag4$$ for all $f,g\in C^2(\mathbb R)$ such that $f$ or $g$ is compactly supported. So, if $f\in C^2(\mathbb R)$ and $g\in C_c^2(\mathbb R)$, then $\varrho g\in C_c^2(\mathbb R)$ and $$\int(Lf)g\:{\rm d}\mu=\int fL^\ast(\varrho g)\:{\rm d}\lambda\tag5.$$ Now, $$L^\ast(\varrho g)=\underbrace{(L^\ast\varrho)}_{=\:0}g+(\sigma^2\varrho)'g'+\varrho\underbrace{\left(\frac12\sigma^2g''-bg'\right)}_{=\:L^\ast g\:+\:b'g}\tag6.$$ However, I don't know how we need to proceed from here.