I am closely following this answer. The purpose of an effective theory is to write down a Lagrangian that describes physics at scale $\mu$ and with some cutoff $\Lambda$.
In renormalized perturbation theory we split our Lagrangian as $$\mathcal{L}_{\text{bare}}(\Lambda) = \mathcal{L}_{\text{physical}} + \mathcal{L}_{\text{counter}}(\Lambda).$$ The linked answer suggests that the (Wilsonian) effective Lagrangian $\mathcal{L}_{\text{eff}}$ obtained by integrating out momenta between $\mu$ and $\Lambda$ is equal to $\mathcal{L}_{\text{physical}}$.
But if we take $\mathcal{L}_{\text{eff}}$ to be the Lagrangian with physical parameters, $\mathcal{L}_{\text{physical}}$, wouldn't we be exactly where one starts in bare perturbation theory? Wouldn't $\mathcal{L}_{\text{eff}}$ still have divergent parameters?
As a concrete example suppose one wants to compute the magnetic moment of the electron using an effective QED Lagrangian, which from above, we take to be $\mathcal{L}_{\text{physical}}$. Then once one computes the necessary form factors wouldn't we find things diverge?
