The Hilbert stress-energy tensor is defined as $$T_{\mu\nu}=-2 \frac{1}{\sqrt{g}}\frac{\delta S_M}{\delta g^{\mu\nu}}.$$
Given the name one expects that it transform as a tensor, but how to prove this directly from this definition?
In the above $S_M$ is an action, and $g$ is the determinant of the metric.
I think it best to write the functional derivative unambiguously as the explicit expression $$ \delta S = -\frac 12 \int_M \delta g^{\mu\nu} T_{\mu\nu}\sqrt{g}d^dx, $$ where $S$ is the (scalar) action functional. Then, as $\delta g^{\mu\nu}$ is a tensor, and $\sqrt{g}d^dx$ an invariant, it is clear that $T_{\mu\nu}$ is a tensor.
The potential ambiguity is whether to include the $\sqrt{g}$ in the measure in the integral definition of $\delta S/\delta g^{\mu\nu}$ or not.
I just noticed that I had answered this a couple years ago:
Is $T^{\mu\nu}=-\frac{2}{\sqrt{-g}}\frac{\delta S_m}{\delta g_{\mu\nu}}$ a true tensor or a density?
