Vanishing action integral for Gravitation Field

For a Gravitation Field Action Integral looks like:

begin{equation}label{1}
S_{gravity} = frac{c^3}{16pi G}int Rsqrt{-g} d^4x.
end{equation}

А Least Action Principle says the $delta S_{gravity} + delta S_{matter} = 0$.

But we know the $delta S_{gravity}$ itself is zero:
begin{equation}
delta S_{gravity} = frac{c^3}{16pi G} int G_{munu} sqrt{-g} delta g^{munu}d^4x = frac{c^3}{16pi G} int sqrt{-g} dx^4 G_{munu}^{,, ;nu}xi^{mu} = 0.
end{equation}
Where $G_{munu}$ — Einstein tensor, $xi^{mu}$ — Killing vector.

Also, for matter

begin{equation}label{2}
delta S_{matter} = int T_{munu}sqrt{-g} delta g^{munu}d^4x = int sqrt{-g}T^{munu}_{,, ;nu}xi_{mu}d^4x = 0.
end{equation}

So, the equation $delta S_{gravity} + delta S_{matter} = 0$ looks like summation of zeroth $0 + 0 =0$ in contrast, for example, with Maxwell’s theory where $delta S_{EMF} + delta S_{matter} = 0$ each therm are non zero.

So, why so for gravity?
I really thought that only the sum of two $delta S $ is zero, but not by themselves.

Source http://www.roma1.infn.it/teongrav/VALERIA/TEACHING/ONDE_GRAV_STELLE_BUCHINERI/AA2006_2007/DISPENSE/variational_principle.pdf