Symmetry protects against symmetry-breaking counterterms in renormalization My lecturer said that when we renormalize a theory (any theory, not necessarily a renormalizable one) we can do so by adding counterterms to the original Lagrangian $\mathscr{L}_B$, turning it into $\mathscr{L}_R + \mathscr{L}_C $. At any order $n$, in the counterterms part of the Lagrangian we can expect all possible scalar combinations of fields, with coupling constants with dimensions (in unit of mass, or length) up to some function of $n$. The only counterterms that are prohibited "from the start" are the ones that explicitly break the symmetries of $\mathscr{L}_B$.
For example, if in $\mathscr{L}_B$ we have massless fermions, chiral symmetry protects against counterterms such as $-\delta m \bar{\psi} \psi$, which would give mass to fermions and break chiral symmetry. I don't have clear why it is so. Why do symmetries protect against explicitly symmetry-breaking counterterms?
 A: The "renormalized plus counterterm" decomposition is a rewriting of the original Lagrangian. Any symmetry present in one notation needs to be present in the other.
Sometimes people describe this approach carelessly and claim that the counterterm Lagrangian is a "new piece" that gets added to the original Lagrangian when you want to consider loops. This is not true. It just looks that way because
\begin{equation}
\mathcal{L} = \frac{1}{2} \partial_\mu \phi_0 \partial_\mu \phi_0 + \frac{1}{2}m_0^2 \phi_0^2 + \frac{\lambda_0}{4!}\phi_0^4
\end{equation}
takes the same form as the renormalized Lagrangian when you drop the zero subscripts. But
\begin{equation}
\mathcal{L}_{ren} = \frac{1}{2} \partial_\mu \phi \partial_\mu \phi + \frac{1}{2} m^2 \phi^2 + \frac{\lambda}{4!} \phi^4
\end{equation}
is a different Lagrangian. Do $\mathcal{L}$ and $\mathcal{L}_{ren}$ still give the same results at tree level? Not necessarily because that depends on what scheme you use.
In practice, it is common to use minimal subtraction so that the difference between bare and renormalized quantities (though infinite) is higher order in $\lambda$. One can therefore calculate poles in $\mathcal{L}_{ren}$ loop diagrams to disentangle the counterterms that were already present from the rest of $\mathcal{L}$. But it is important to remember that you're not changing the theory. Merely splitting it in a way that allows for better control.
