The bare Lagrangian of the $\phi^4$-theory can be written in terms of bare parameters as $$\mathcal{L}=\frac{1}{2}(\partial_\mu\phi_0)^2-\frac{1}{2}m_0^2\phi_0^2+\frac{\lambda_0}{4!}\phi_0^4\tag{1}.$$ The same bare Lagrangian, in terms of renormalized parameters and counterterms as $$ \mathcal{L}=\frac{1}{2}(\partial_\mu\phi_r)^2-\frac{1}{2}m^2\phi_r^2+\frac{\lambda}{4!}\phi_r^4+\frac{\delta_Z}{2}(\partial_\mu\phi_r)^2-\frac{\delta_m}{2}\phi_r^2+\frac{\delta_\lambda}{4!}\phi_r^4 $$ $$=\mathcal{L}_{renorm}+\mathcal{L}_{counterterm}\tag{2}. $$
I'm interested in evaluating the 1PI effective action for this theory. For this, I have to compute the integral $$Z[j]=e^{iW[j]}=\int D\phi \exp[{i\int d^4x( \mathcal{L} +j\phi)}]$$
I think, it is the total Lagrangian $\mathcal{L}$, in either forms (1) or (2), can be used to evaluate $Z$.
But if I understand it correct, A. Zee's book on Quantum Field Theory in Nutshell, calculates Z using $\mathcal{L}_{renorm}$ part of $\mathcal{L}$ and not using full $\mathcal{L}$. See Chapter IV.3 Eqn. (1), (11). He uses, $A,B,C$ for counterterms.
Why is it that only the renormalized part of the Lagrangian used for the calculation of effective action? Am I missing something?