Let $\Omega\subset \mathbb{R}^n$ be an open and bounded set. Let $\mu:\mathcal{B}(\Omega)\to [0,+\infty)$ a bounded Radon measure and let $\varphi, \, v \in L^1(\Omega,\,\mu)$, $v\geq 0$. Then $v\mu$ is a bounded Radon measure too (and it is absolutely continuous with respect to $\mu$), where $$v\mu(B):=\int_Bv\,d\mu.$$
It is well is well known that $\forall \varepsilon>0$ there exists $\varphi_\varepsilon$ smooth (for my purpose $C^0$ is enough) s.t. $||\varphi - \varphi_\varepsilon ||_{L^1(\mu)}<\varepsilon$ and $\tilde{\varphi}_\varepsilon$ smooth s.t. $||\varphi - \tilde{\varphi}_\varepsilon ||_{L^1(v\mu)}<\varepsilon$.
Can I suppose that $\tilde{\varphi}_\varepsilon= \varphi_\varepsilon$ ? Perhaps it is sufficient to use the fact that $v\mu$ is AC with respect to $\mu$ ?