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Let $M$ be a Riemannian manifold (smooth) with a volume form $dV.$

Question 1. Let $f$ be a smooth function on $M,$ $f:M\rightarrow\mathbb{R}$. If $f$ has a compact support then we know that $\intop_{M}fdV$ is well-defined as in the book by J.Lee. In this case, by the compactness of $supp\left(f\right)$, we can cover it by a finite number of charts and use partition of unity to make the def. through. My first question is, if we don't require that $f$ has a compact support, can we still use the def. in the Lee's book to $\intop_{M}fdV$? That is, $\intop_{M}fdV$ is still well-defined in this way, even it may not exist.

Question 2. How can we define the space $L^{p}\left(M\right)$ or the Sobolev space $W^{k,p}\left(M\right)?$ I look at in some textbooks and see that, in order to define these spaces, firstly we consider smooth functions in appropriate norms and then take the completion of it under these norms. However, this approach seems artifical to me. My second question is, do we have a clear definition of a measure on $M$ (may be related to the metric) so that $L^{p}\left(M\right)$ is defined in the same manner as in the Lebesgue theory?

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    $\begingroup$ 12 questions and no accepted answer? Hm! $\endgroup$ – Alex M. Mar 30 '17 at 19:27
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For (1): No, that approach no longer works if $f$ no longer has compact support. Remember that the convenience of compactness resides in the fact that you may cover a compact set by finitely many domains of charts, which (together with a partition of unity) allows you to define the integral of $f$ as a finite sum of integrals on these chart domains. If the support of $f$ is not compact, you will no longer have a finite sum (and possible not even countably many terms), and then you're in trouble, because how do you sum that nightmare?

For (2): Similarly to how a measure is produced in the Riesz-Markov theorem, one may naturally construct a (unique) measure associated to a volume form: if $E \subseteq M$ is Borel, just define $\mu (E) = \sup \{ \int _M f \ \Bbb d V \mid f \in \mathcal C _0 ^\infty (M), \ \text{supp } f \subseteq E, \ 0 \le f \le 1 \}$ - exactly like in the theorem, but using smooth functions instead of just continuous ones. Notice, though, that you don't need to do this if you don't want to, because the Lebesgue (or Sobolev) spaces that you get are identical to those obtained by completing $\mathcal C _0 ^\infty (M)$ in those respective norms. Even if this looks artificial to you, it is in fact a very powerful procedure, because it allows you to prove theorems by verifying them on smooth compactly-supported functions (which are very well behaved), and then extend these properties to the whole function space by an argument based upon density - a very powerful and convenient procedure.

Alternatively, you may consider the Hausdorff measure associated to the natural distance on $M$.

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  • $\begingroup$ Thanks, your answer is quite satisfactory. Can we have a reference for (2) in your answer? $\endgroup$ – Binjiu Mar 31 '17 at 0:04
  • $\begingroup$ @Binjiu: For the Hausdorff measure approach see mathoverflow.net/questions/87930/…, math.stackexchange.com/questions/107099/… and math.stackexchange.com/questions/1232726/…. The Riesz-Markov approach is sketched in chapter 9 of "Generalized Curvatures" by Jean-Marie Morvan and remark 10.4 in "From Calculus to Cohomology" by Madsen & Tornehave. $\endgroup$ – Alex M. Apr 1 '17 at 11:52
  • $\begingroup$ You mean the Hausdorff measure and the measure mentioned by the Riesz-Markov theorem yield to the same measure to define $L^{p}(M)$? $\endgroup$ – Binjiu Apr 3 '17 at 12:11
  • $\begingroup$ @Binjiu: Yes, save possibly for a multiplicative constant that is easy to fix, though, by requiring the two measures to agree on some predetermined set. (In general, the two measures might be a multiple of each other.) $\endgroup$ – Alex M. Apr 4 '17 at 11:33

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