Q:

Is energy conserved in general relativity? Does $\nabla_aT^{ab}_{\rm matter}=0$ represent the conservation of energy and momentum?

For example, the radiation dominated cosmology, the energy density of radiation is proportional to $a^{-4}$ and the volume is proportional to $a^3$, where $a$ is the scale factor. So the total energy of radiation is propotional to $a^{-1}$. So where is the loss of energy of radiation? Is it because the gravitational field has the energy?
Does $\nabla_aT^{ab}_{\rm matter}=0$ represent the conservation of energy and momentum of matter field in GR?

A:

After taking a look at the answer by Jim, I am not sure in my knowledge at the moment. However, lets try to figure out the details. I claim, that energy-momentum tensor of matter in GR is not conserved by itself since matter always interacts with gravitational field and the total energy should be taken into account instead. 
Vanishing of the covariant divergence $\nabla_a T^{ab}=0$ exactly reflects this feature. Consider this equation integrated over a 4-dimensional volume
\begin{equation}
\begin{aligned}
0=&\int_V d^4x \sqrt{-g}\;\nabla_a T^{ab}=\int_V d^4x \nabla_a(\sqrt{-g}\; T^{ab})=\int_V d^4x \partial_a(\sqrt{-g}\; T^{ab})+\ldots\\
=&\int_{\partial V} d\Sigma_a T^{ab}+\ldots,
\end{aligned}
\end{equation}
where dots represent whatever Christoffel symbols appear there. Hence, we see, that if we choose the surface $\partial V$ in the usual way when only its $x^0=const$ parts contribute to the integral we will get the usual conservation law deformed by the connection terms
\begin{equation}
0=P^b(x^0=t_2)-P^b(x^0=t_1)+\dots.
\end{equation}
So, since the difference of 4-momentum at different time does not vanish, the energy momentum tensor of matter is not conserved by itself.
However, if we take into account the contribution from gravity that is calculated in the usual way
\begin{equation}
T^{grav}_{ab}=\frac{1}{\sqrt{-g}}\frac{\delta S_{EH}}{\delta g^{ab}}=R_{ab}-\frac12g_{ab}R,
\end{equation}
we see the well known feature of GR that the total energy-momentum tensor vanishes due to the Einstein equations
\begin{equation}
T_{ab}+T_{ab}^{grav}=0.
\end{equation}
That is kind of obvious, because the total energy momentum tensor is obtained by variation of the total action wrt $g_{ab}$ and therefore gives exactly EOM for gravitational field with a source.
It seems, that the loss of energy of radiation goes into energy of the gravitational field. In addition, you may read the 2 volume of Landau-Lofshitz to find out how people define energy-momentum pseudo-tensor of gravitational field that does not cancel the matter energy-momentum tensor and therefore is more useful for some applications.

A:

The energy of radiation falls off like $a^{-1}$ because space is expanding. As space expands, the peaks of an electromagnetic wave expand with it, which makes them get farther apart. This means the wavelength of radiation increases as space expands, thus the frequency decreases. Since energy is $E=h\nu$, if the frequency decreases proportional to $a^{-1}$, then the energy also falls off like $a^{-1}$. This is called cosmological redshift.
Also, as mentioned in the comments, $\nabla_{\mu}T^{\mu}_{~\nu}=0$ means that the energy-momentum tensor is conserved. It is the GR equivalent of the conservation of energy and the conservation of momentum laws.

