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 
$$
\frac{d}{dt} \langle A(t)\rangle= \frac{d}{dt} \psi(t),\langle A\psi(t)\rangle =\langle -i H \psi(t), A\psi(t)\rangle+ \langle\psi(t), A (-iH)\psi(t)\rangle\\ =i(\langle H \psi(t), A\psi(t)\rangle - \langle\psi(t), A H\psi(t)\rangle) \quad \ast.
$$
Quantum Hamiltonian $H$ is densely defined on $\mathcal{H}$ with domain $D(H)$ and if $A$ does not preserved $D(H)$ the first term $\langle H \psi(t), A\psi(t)\rangle$ is not equal to $\langle\psi(t), HA\psi(t)\rangle$ and $\frac{d}{dt}\langle A(t)\rangle \neq 0$. 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.