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It seems well-known that the volume form natually induces a canonical measure on a Riemannian manifold. For example, this page mentions this fact. I am wondering the construction of such a measure. For my background, I learned some real analysis and read Tu's An Introduction to Manifolds.

In Tu's book, the integral over a smooth manifold is defined for smooth forms (actually, it is no harm to extend the definition for continuous forms) via the usual Riemann integral. For an $n$-form $\omega$ on a coordinate chart $(U,\phi)$, the integral is defined by $$ \int_U \omega = \int_{\phi(U)} (\phi^{-1})^* \omega $$ where the right hand side is the Riemann integral. For an $n$-form on a smooth manifold, we use an open cover of coordinate open sets and a partition of unity subordinate to the cover to define the integral of $\omega$.

For an oriented $n$-dimensional Riemannian manifold, the volume form can be written as $$vol = \sqrt{g} dx^1 \wedge \dots \wedge dx^n$$ Then, we can use a partition of unity to define the integral of a smooth (or continuous) function via Riemann integral. In particular, the integral of a function $f$ is defined by $$ \int_M f = \int_M fvol$$ where the RHS is the integral over an $n$-form.

Here is my idea about the construction of the canonical measure $\mu$ on the Borel $\sigma$-algebra of a Riemannian manifold $M$. For a measurable set $V$ contained in a chart $(U,\phi)$, since a hemeomorphism maps a Borel set to a Borel set, we can define $\mu(V)$ by setting $$ \mu(V) = \int_V vol = \int_{\phi(V)} (\phi^{-1})^* vol $$ where the RHS is now intepreted as a Lebesgue integral. Then for any Borel set $V$, we choose an open cover consisting of coordinate open sets and a partition of unity to define $\mu(V)$. After checking $\mu(V)$ is independent of the cover and a partition of unity, a positive measure $\mu$ is defined so that we can talk about sets of measure zero and $L^2$ functions.

Is this the right way to construct the canonical measure on the Borel $\sigma$-algebta of an oriented Riemannian manifold? If so, is this the most commonly used measure for geometric analysis (e.g. Jost's Riemannian Geometry and Geometric Analysis)?

Justin Lien
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    Yes, this is all correct, expect you are ignoring the orientation issue. For non-orientable manifolds, you will be defining a volume density, not a volume form. – Moishe Kohan Feb 09 '22 at 17:17
  • Yes, this is fine for oriented manifolds (otherwise there is no volume form as mentioned above). But note that orientation is unnecessary. For each chart $(U,\phi)$ we can consider the expression $\sqrt{|g|}:U\to\Bbb{R}$ which is the square root of the (absolute value) determinant of the matrix whose entries are the components of $g$ with respect to the chart $(U,\phi)$. All we need is this function as the "weight" in the definition of the measure. This gives you a measure on $U$, which you can then patch together using a partition of unity. – peek-a-boo Feb 10 '22 at 09:18
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    Alternatively, you can avoid partitions of unity, and just do it directly, as I explain here. Modulo smoothness issues, yes, if nothing else is said, this is the measure we're referring to when talking about Riemannian manifolds.

    Note btw that this construction is not specific to the Riemannian metric. If you have a symplectic manifold $(M,\omega)$, you can repeat the same thing. More generally, any $(0,2)$-tensor field gives rise to a measure in this fashion. Even more generally, any scalar density gives rise to a measure.

     – peek-a-boo Feb 10 '22 at 09:20

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