Number of d.o.f. in massless and massive Maxwell field? Consider the Maxwell kinetic Lagrangian
$$L_{kin}=-\frac{1}{4}(F_{\mu\nu})^2$$
where the field strength is $F_{\mu\nu}=\partial_\mu A_\nu-\partial_\nu A_\mu$. A vector quantity can contain at most 4 degrees of freedom (d.o.f.). Just from the definition of $F_{\mu\nu}$ it is clear that the longitudinal degree of freedom $\partial_\mu\chi$ in the vector
$$A_\mu=A_\mu^\perp+\partial_\mu \chi$$
cancels out of $L_{kin}$. Additionally, one can use the gauge invariance $A_\mu\to A_\mu+\partial_\mu\xi$ to e.g. eliminate the spatially longitudinal component in $A_i=A_i^T+\partial_i A^L$. This shows that the massless Maxwell field only propagates 2 degrees of freedom.
However, if we add a mass term to the Lagrangian:
$$L_m=-\frac{1}{2}m^2A_\mu A^\mu$$
This reintroduces a kinetic term for the longitudinal component $\partial_\mu\chi$ above. Additionally, the gauge invariance is broken.
My question is:

Without gauge invariance, how can we show in this case that the massive field propagates 3 degrees of freedom and not 4?

 A: The reason why there are only 3 physical degrees of freedom,  can be deduced by simply computing the Euler-Lagrange equation for the field $V^{\mu}$.
The Lagrangian reads:
$$L_{V} = -\frac{1}{4} V_{\mu\nu}V^{\mu\nu} + \frac{1}{2} m^{2}V^2 $$
and the E-L equations are:
$$(\Box + m^{2})V^{\sigma} - \partial^{\sigma} \partial_{\rho}V^{\rho}=0$$
Up to now, there are 4 physical degrees of freedom (i.e. the 4 components of $V^{\mu}$), but if you derive both members of the E-L equations by $\partial_{\sigma}$ you will find:
$$ \Box(\partial_{\sigma} V^{\sigma}) + m^{2}(\partial_{\sigma} V^{\sigma}) -\Box(\partial_{\rho} V^{\rho}) =0 \rightarrow m^{2} (\partial_{\sigma} V^{\sigma})=0$$
If you suppose that you are describing a massive field theory (i.e. $m$ different from 0), the only way to satisy  the above constrain is that 
$\partial_{\sigma}V^{\sigma}=0$.
At this point you can rewrite the equation of motion of your field theory as:
$$(\Box + m^{2})V^{\mu}=0$$
$$ \partial_{\mu}V^{\mu}=0$$
That is a Klein-Gordon equation plus a constrain. The presence of this constrain, clearly means that you have 4-1=3 physical degrees of freedom of your theory. The latters are the 3 spatial components of the vector field $V^{\mu}$.
