Confusion about chiral anomaly (Fujikawa's method)

I am reading Fujikawa’s method for calculating chiral anomaly, see this wiki page.

The method can be described as follows.
It starts with the path integral
begin{equation}
Z=intmathcal{D}psimathcal{D}bar{psi}e^{iint mathcal{L}}.
end{equation}
First we calculate the shift of Lagrangian $mathcal{L}(psi,partial psi)$ due to the local chiral transformation
begin{equation}
psi’=e^{ialpha(x)gamma^5}psi.
end{equation}
we get
begin{equation}
mathcal{L}(psi’,partial psi’)=mathcal{L}(psi,partial psi)+alpha partial_{mu}(bar{psi}gamma^{mu}gamma^5psi).
end{equation}
On the other hand, we also calculate the path integral measure due to above chiral transformation, we have
begin{equation}
mathcal{D}psi’mathcal{D}bar{psi}’=mathcal{D}psimathcal{D}bar{psi}e^{2iint alpha(x)A(x)}.
end{equation}
We require $Z$ should be invariant under local chiral transformation, then we have
begin{equation}
partial_{mu}(bar{psi}gamma^{mu}gamma^5psi)=-2A(x).
end{equation}

Now here is my confusion: why do we require $Z$ is invariant under local transformation $psi’=e^{ialpha(x)gamma^5}psi$? If we don’t, then there is no reason to expect $bar{psi}gamma^{mu}gamma^5psi$ to be conserved. (It is called "anomaly" because we expect it is conserved but it is not.)