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Assume a compact connected manifold $M$ is given. Assume we have a local parametrization $f:U \to M$ such that $M \setminus f(U)$ has zero measure in $M$ and call this property **. Then it is enough to have this single parametrization to calculate the volume of $M$ (example: remove a point from a sphere). While there is the construction of a partition of unity to calculate the volume in case one needs more charts, I think I never saw an example of a calculation where $M$ did not have the property ** and one really had to construct a partition of unity. (Simple textbook examples like spheres, torus etc. all seem to have property **).

So can one classify all such manifolds having property **?

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    $\begingroup$ Every compact manifold has a chart whose complement is measure zero. I do not know an easy proof of this. $\endgroup$ Commented Jan 9, 2016 at 16:13
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    $\begingroup$ @Mare: On a Riemannian manifold the exponential map from an arbitrary point $p$ is surjective. Fix $p$, and let $U$ be the largest open set in $T_{p}M$ that contains $p$ and whose image does not hit the cut locus. The exponential map furnishes a parametrization $\exp:U \to M$ whose image is dense and has full measure. (There are details to check, but that's one sketch.) $\endgroup$ Commented Jan 9, 2016 at 18:33
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    $\begingroup$ sounds interesting, hope you could make an answer out of it filling some details ;) $\endgroup$ Commented Jan 9, 2016 at 19:16
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    $\begingroup$ @MoisheKohan: You have closed this question just 2 hours after the OP placed a bounty on it. This means that (s)he will lose the 50 reputation points offered, but not of his/her fault. Do you think we can fix this unintended waste of reputation points? $\endgroup$ Commented Mar 7 at 18:00
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    $\begingroup$ I see that the question is open again, but I don't think the reason for closure was very good in the first place: the question seemed different from the referenced one, there is an answer from which this also follows, but I don't think this question really was answered there. In particular, I think you will have to be very lucky to find that answer on the site if you have this question. I do think it would be a good answer to this question to point out that it can be concluded from Andrew's argument using the referenced result of Ito and Tanaka. $\endgroup$ Commented Mar 10 at 19:16

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