If $A$ is a classical symmetry, it is possible that after quantization $A$ is no longer a symmetry. One of the ways to see this in the operator formulation of quantum mechanics is the following.

If $\psi \in \mathcal{H}$ is a state, then $$ \frac{d}{dt} <A(t)>= \frac{d}{dt} <\psi(t),A\psi(t)> = <-i H \psi(t), A\psi(t)>+ <\psi(t), A (-iH)\psi(t)>\\ =i(<H \psi(t), A\psi(t)>- <\psi(t), A H\psi(t)>) \quad \ast. $$ Quantum Hamiltonian $H$ is densely defined on $\mathcal{H}$ with domain $D(H)$ and if $A$ does not preserved $D(H)$ the first term $<H \psi(t), A\psi(t)>$ is not equal to $<\psi(t), HA\psi(t)>$ and $\frac{d}{dt} <A(t)> \neq 0$. This is one of the mechanisms of quantum anomaly in quantum mechanics.

My question is how to derive formula $\ast$ using path integral formulation? Somehow the fact that Feynman measure is not invariant under symmetry $A$ has to play a role in such computation, but I don't see how to do it.