 Solution

Lorentz invariance requires $A^{\mu }$ and $J^{\mu }$ to transform as four-vectors, $\chi$ as a scalar, $m$ and ${\mu }_0$ as scalars, and the integration domain and boundary conditions to respect the transformation. Then every contracted term and $d^{4}x$ are Lorentz invariant.

Under $A_{\mu }\mapsto A_{\mu }+{\partial }_{\mu }\chi$, the field strength is unchanged. The source term changes by a boundary term pluswhich vanishes for a conserved current and suitable boundary behaviour of $\chi$. The mass term is not gauge invariant. Thus the action has the usual local gauge invariance whenwith the gauge function or its boundary contribution suitably controlled. For $m>0$, the mass term breaks this gauge symmetry.