1
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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
  • 1
    $\begingroup$ 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
1
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.