I am having some trouble in seeing exactly how the Riemannian density form gives rise to a measure on $\text{Borel(M)}$.
Let $(M,g)$ be a Riemannian manifold. We have the Riemannian density $\mu_g$. Now we can define $\int f \mu_g $ for any compactly supported continuous function.
I guess the way to introuduce a measure on $\text{Borel(M)}$ is as follows:
For an open subset $U \subseteq M$ we would like to define $\mu(U) = \int _U \mu_g$, where the integral is taken on the submanifold $U$ with the induced metric from $(M,g)$.
The problem is that the conatant function $1_U$ is not compactly supporteed on $U$ so I am not sure how to define it.*
Even if we did that, how do we extend the measure to the whole Borel sigma-algebra? (via some machinery of measure theory?) How do we go on proving sigma-additivity of the measure?
Any explanation or a detailed reference would be welcomed.
*Maybe using somehow the positivity of the integral, if $U$ is contained in a compact set $K$ we can try using a sequence of approximating bump functions which are $1$ on $U$ and decrease faster and faster outside $U$ and are all equal to $0$ outside of $K$. Maybe if the sequence of integrals converge the limit is $\mu(U)$?
But this does not work if $U$ is not contained in a compact set.
Given that you have defined your integral for functions with compact support, you can extend the set of functions it acts on by using a Partition of Unity:
This is a set of functions $\phi_i$ which can be made to have compact support (see e.g. Chapter 4 - Partitions of Unity and Smooth Functions) so that integral of a general function $f$ is defined via: $$\int_X f d\mu_g = \int_X f \sum_i \phi_i d\mu_g = \sum_i \int_X f \phi_i d\mu_g$$