Scale Invariance of the Einstein-Hilbert Functional Define the normalised Einstein Hilbert functional for an n-dimensional Psuedo-Riemannian manifold ($M, g$) as
$\mathcal{E}(g) = \frac{{\int_{\mathcal{M}}S_g \Omega_g}}{\left({\int_{\mathcal{M}} \Omega_g}\right)^{\frac{n-2}{n}}}$.
where $S_g$ is the scalar curvature and $\Omega_g$ is the Riemannian volume form for $g$. I'd like to show that this is scale invariant, ie. $\mathcal{E}(g) = \mathcal{E}(cg)$ for $c \in \mathbb{R}$.
Defining $\Omega_{cg}$ and $S_{cg}$ in the obvious way and using the rule on scalar curvature $S_{cg}=\frac{1}{c}S_g$ I get
$\frac{\mathcal{E}(g)}{{\mathcal{E}(cg)}} = \frac{{\int_{\mathcal{M}}S_g \Omega_g}}{{\int_{\mathcal{M}}S_{cg} \Omega_{cg}}}$ $\frac{{\left({\int_{\mathcal{M}} \Omega_{cg}}\right)^{\frac{n-2}{n}}}}{{\left({\int_{\mathcal{M}} \Omega_g}\right)^{\frac{n-2}{n}}}}$ = $ \frac{S_g(vol)_g}{\frac{S_g}{c}(vol)_{cg}} \left(\frac{(vol)_c}{(vol)_{cg}}\right)^{\frac{n-2}{n}}$.
But $(vol)_{cg} = c^{n}(vol)_g$ leaving us with
$\frac{\mathcal{E}(g)}{{\mathcal{E}(cg)}} = \frac{c}{c^n} \frac{1}{c^{n-2}} = c^{-1} \neq 1$.
I'm wondering what I have done wrong here?
 A: You've made two mistakes.
The first mistake is that $\int_M S_g\Omega_g \neq S_g\operatorname{vol}_g$; the left hand side is a number, while the right hand side is a function unless $S_g$ is constant, in which case you do get an equality.
Secondly, $\Omega_{cg} = c^{\frac{n}{2}}\Omega_g$ not $c^n$ (think about what happens when you replace $g$ by $cg$ in this local expression).
With these things in mind, note that
$$\int_M S_{cg}\Omega_{cg} = \int_M c^{-1}S_g c^{\frac{n}{2}}\Omega_g = c^{\frac{n}{2}-1}\int_MS_g\Omega_g = c^{\frac{n-2}{n}}\int_M S_g\Omega_g$$
and
$$\left(\int_M\Omega_{cg}\right)^{\frac{n-2}{n}} = \left(\int_Mc^{\frac{n}{2}}\Omega_g\right) = \left(c^{\frac{n}{2}}\int_M\Omega_g\right)^{\frac{n-2}{n}} = c^{\frac{n-2}{2}}\left(\int_M\Omega_g\right)^{\frac{n-2}{n}}.$$
Therefore
$$\mathcal{E}(cg) = \dfrac{\displaystyle\int_M S_{cg}\Omega_{cg}}{\left(\displaystyle\int_M\Omega_{cg}\right)^{\frac{n-2}{n}}} = \dfrac{c^{\frac{n-2}{n}}\displaystyle\int_M S_g\Omega_g}{c^{\frac{n-2}{2}}\left(\displaystyle\int_M\Omega_g\right)^{\frac{n-2}{n}}} = \dfrac{\displaystyle\int_M S_g\Omega_g}{\left(\displaystyle\int_M\Omega_g\right)^{\frac{n-2}{n}}} = \mathcal{E}(g).$$
