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I am exploring whether it is possible to define a Borel or Lebesgue measure on smooth (finite-dimensional) manifolds, similar to how it is done on Euclidean spaces $\mathbb R^n$ . My intuition suggests that since manifolds are locally homeomorphic to $\mathbb R^n$ via charts, one might "glue" measures defined on local charts to construct a global measure.

Specific Questions:

(A) Do we need any additional conditions on a smooth manifold so as to equip it with a Lebesgue measure? Does this require additional structure on the manifold such as, Riemannian metric, volume form, or partition of unity?

(B) If such a measure exists, how is it rigorously defined? Is it achieved via gluing the Lebesgue measures on the charts, integration of volume forms, or another method?

(C) How does orientability affect the construction? For example, can measures be defined on non-orientable manifolds like the Möbius strip or $\mathbb {RP}^n$ as well?

(D) Could such a measure allow us to generalize theorems like Sard’s theorem or enable measure-theoretic proofs in differential geometry?

I am familiar with the Lebesgue measure on $\mathbb {R}^n$ and the use of partitions of unity in manifold theory to patch local quantities. I’ve seen the Riemannian volume form. I wonder if this form gives rise to a measure. I also wonder if a Lebesgue measure can exist without a metric. Any kind of reference or insight is welcome.

Thanks in advance.

Kishalay Sarkar
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    What do you mean by Lebesgue-like measure? If you ask for Borel regular measures, there are many such by the Riesz representation Theorem. – Mememaster696 May 22 '25 at 08:25
  • @Mememaster696 I am looking for a Lebesgue measure. – Kishalay Sarkar May 22 '25 at 09:32
  • @Mememaster696 For example, if it is a Riemannian manifold, I would like the lengths, areas, volumes to satisfy the properties of length as in Lebesgue measure and require countable sets to have measure zero. – Kishalay Sarkar May 22 '25 at 09:34
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    I have answered almost all these questions before in various levels of generality in various answers. Start here and go through the various links/related questions. – peek-a-boo May 22 '25 at 10:58
  • @peek-a-boo From the link you've provided, I found that one can define a collection of measurable sets of a smooth manifold $M$ by setting $A\subset M$; lying in the chart $(U,\varphi)$ is measurable if and only if $\varphi (A\cap U)$ is measurable.

    However, unless there is a metric on the manifold, one cannot think about any way to define a natural Lebesgue measure on it, right? (Because of the patching problem.)

     – Kishalay Sarkar May 22 '25 at 11:14
  • @peek-a-boo However, in case of a Riemannian Manifold, we can use the Riemannian Volume Form to define a measure on $M$. – Kishalay Sarkar May 22 '25 at 11:14
  • @peek-a-boo In particular we can define, $\lambda_g(A)= \int_{\varphi(A)}\sqrt{|det(\varphi^*g)|}d\lambda_{\varphi(U)}$, where $A\subset U$, with $(U,\varphi)$ a chart. – Kishalay Sarkar May 22 '25 at 11:21
  • @peek-a-boo Where $\lambda_{\varphi(U)}$ is the restriction of the Lebesgue measure $\lambda=du_1\wedge du_2\dots \wedge du_n$ of $\mathbb R^n={(u_1,\dots, u_n): u_i\in \mathbb R}$ to the open set $\varphi(U)$. – Kishalay Sarkar May 22 '25 at 11:23
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    yes, in general a manifold has no “natural” measure (but the notion of “measure-zero” is well-defined). Next, you seem to be confusing a measure with a volume form. Note that these are separate concepts. Sure, every top-degree differential form $\omega$ induces in a natural manner, a measure on the manifold (via $A\mapsto \int_A|\omega|$… so actually the general idea is that a scalar density gives rise to a measure, and that by taking absolute values, we get from top-forms to scalar densities, so really it’s like top-form —> non-negative scalar density —> non-negative measure). – peek-a-boo May 22 '25 at 12:16
  • @peek-a-boo I am not confusing scalar density, measure and volume form. I am saying that if we have a volume form, we can cook up a natural measure. – Kishalay Sarkar May 22 '25 at 12:19
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    the reason I’m insisting on keeping a clean separation between volume forms vs measures is that the former (by definition) only exist on orientable manifolds, while the latter doesn’t need any orientability assumptions (see similar remarks in this answer of mine). In particular, if you equip with $\Bbb{RP}^n$ with a Riemannian metric it always has a naturally-induced measure, but it is orientable iff $n$ is odd. – peek-a-boo May 22 '25 at 12:19
  • Let us continue this discussion in chat. – peek-a-boo May 22 '25 at 12:19

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