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.

  • $\begingroup$ Well, I've never seen the definitions before, but if I had to define those terms on a manifold, that's exactly how I would define them. For the definition to make sense, you need to check that the transition maps are measurable, and take nullsets to nullsets. $\endgroup$ – Mario Carneiro Jun 17 '15 at 19:39
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    $\begingroup$ Your statements are correct (and can be used as definitions), and they make sense for smooth manifolds even without a Riemannian metric. (In order to define the measure $\mu$, you need the metric, but measure zero is independent of the metric.) $\endgroup$ – Lukas Geyer Jun 17 '15 at 19:46

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.


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