In Cheng&Li's book "Gauge theory of elementary particle physics" section 16.3, the $U(1)$ problem is presented. Here is the story:
Suppose we have a theoy which contains only two quarks $q_1=u$ and $q_2=d$. The symmetry group is given by $SU(2)_L\times SU(2)_R\times U(1)_V\times U(1)_A$.
$U(1)_A$ is axial symmery, i.e., $q_i\rightarrow e^{i\beta \gamma_5}q_i,i=1,2$. In the limit $m_{u,d} \rightarrow 0$, the chiral current \begin{equation} J^5_{\mu}=\bar{u}\gamma_{\mu}\gamma_5u+\bar{d}\gamma_{\mu}\gamma_5d \end{equation} does not seem to correspond any observed symmetry in hardon spectra. If the $U(1)_A$ is spontaneously broken, then a massles Goldstone boson will appear. But in experiment, no such particle is observed.
My confusion is: why do we expect a Goldstone boson? This $U(1)_A$ symmetry is based on the assumption "$m_{u,d} \rightarrow 0$". But in the real world, both $m_u$ and $m_d$ are nonzero, i.e., we might have mass term like $\bar{u}m_uu+\bar{d}m_dd$. Therefore the $U(1)_A$ is not a symmetry. What is the logic and whole story here? I think I misunderstood something.