Volume of a paracompact manifold It is stated, without proof, in Wald (1984) (General Relativity) that given any connected manifold $M$ (which is by definition paracompact), one may define a volume measure $\mu$ such that $\mu[M]$ is finite. Wald stresses the fact that manifolds are paracompact and thus this is possible. This makes me think it has something to do with a partition of unity.
This is my naive guess at how to approach this. Select on $M$ a countable atlas $(U_\alpha,\phi_\alpha)$. (Note that $\phi_\alpha:U_\alpha\longrightarrow O_\alpha\subset \mathbb{R}^4$. Perhaps the manifold being 4-dimensional plays a role.) I can measure the hypervolume of each $O_\alpha$ by defining 
$$\mu_E(O_\alpha)=\int_{O_\alpha}\operatorname{vol}(\mathbb{R}^4)$$ 
where $\operatorname{vol}(\mathbb{R}^4)$ is the volume form on 4-dimensional Euclidean space and $E$ stands of course for Euclidean. Then perhaps I use the partition of unity to sum these up. I'm not sure how to sum them really or how to make the sum finite. Also how do I define $\mu$ such that I can also act upon a subset $D\subset M$ and find the "volume" of $D$?
Any help would be greatly appreciated.
 A: Let us assume, as you said in the comment, that $M$ is also connected (it suffices to work with manifolds which have at most countably many components). Then one proves:
Lemma. A paracompact connected smooth $n$-dimensional manifold admits a countable, locally finite open cover $\{U_j: j\in J\}$ by open subsets whose closures are diffeomorphic to the closed unit ball $B^n$ in $R^n$. (Countable here refers to the fact that $J$ is contained in the set of natural numbers.) 
The proof is not hard, but does take some effort. I will assume the lemma since the solution is already quite long. 
I will also assume that $M$ is oriented, otherwise, I have to digress into densities. Choice of orientation will give me a collection of maps $f_j$ for which transition maps are orientation-preserving. Given a cover as above, one proceeds as follows. Let $\{\eta_j: j\in J\}$ be a partition of unity corresponding to this cover and let $f_j: cl(U_j)\to cl(B^n)$ be diffeomorphisms. Then for $B^n$ we take a bump-function $b: B^n\to (0,\infty)$, such that the extension
$$
\tilde{b}= \begin{cases} b(x), & x\in B^n\\
0, & x\in R^n - B^n
\end{cases}
$$
is smooth. By rescaling we can assume that 
$$
\int_{B^n} b =1. 
$$
 Now, let $\omega$ be the standard volume form on $R^n$ and let $\omega_j:= f_j^*(b_j \omega)$. We extend this form by $0$ to the rest of $M$, and retain the notation $\omega_j$ for the extension. For each $j$ we define its $j$-volume:
$$
a_j=\int_{B_j} \omega_j. 
$$
One is now tempted to define the volume density on $M$ as the infinite sum:
$$
\sum_{j\in J} \eta_j \omega_j. 
$$
Such a form would indeed work as a smooth volume form (positivity follows from the fact that compositions $f_i\circ f_j^{-1}$ are orientation-preserving), but its integral over $M$ is, in general, infinite. However, 
$$
\int_{B_j} \eta_j \omega_j \le Vol(B^n). 
$$
Therefore, I will use the density 
$$
\omega= \sum_{j\in J} \frac{1}{j^2} \eta_j \omega_j
$$
and obtain
$$
\int_M \omega \le \sum_{j\in J}  \frac{1}{j^2} <\infty. 
$$
