# $||\phi-\phi_\epsilon||_{L^1(\mu)}<\epsilon$ and $||\phi-\tilde{\phi}_\epsilon ||_{L^1(v\mu)}<\epsilon$, then $\tilde{\phi}_\epsilon= \phi_\epsilon$?

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 $$\mu$$ ?