Let $(M,g)$ be a Riemannian manifold (which doesn't have to be orientable). As far as I know, the metric $g$ induces a "canonical" measure $\mu$ and so one can talk about sets $U\subset M$ of measure zero and measurable functions $f\colon M\rightarrow \mathbb{R}$.
My knowledge of $\mu$ and its properties is rather sketchy (I plan on understanding it better in the future) but for now I'm looking for "easy" criteria that tell me if a set has measure zero or a function is measurable.
That is, is the following true?
A subset $U\subset M$ has measure zero (i.e. is a null set) iff for every chart $(V,y)$ of $M$, the set $y(V\cap U)$ has measure zero in $\mathbb{R}^n$ (w.r.t. lebesgue measure on $\mathbb{R}^n$).
A function $f\colon M\rightarrow \mathbb{R}$ is measurable iff for every chart $(V,y)$ of $M$, the function $f\circ y^{-1}\colon y(V)\rightarrow \mathbb{R}$ is measurable.
Edit: I also appreciate any references on that matter.
Any topological space can be equipped with its Borel sigma algebra, which is the sigma algebra generated by the open sets. I've seen this denoted by $\mathcal{B}(M)$. Then a function $f\colon M\to \mathbb R$ is measurable if $f^{-1}(E)\in \mathcal{B}(M)$ for any $E\in \mathcal{B}(\mathbb R)$.
For measure 0 sets, there is some work to be done to verify that your definition is well-defined. Specifically, it is a priori possible for a set to have Lebesgue measure 0 in one chart but not in another. But since the change of coordinates maps between charts are $C^1$ diffeomorphisms, you can show that null sets remain null under coordinate change.
By the way, you can directly write down the measure by integrating a density.