Integrals over manifolds minus sets of measure zero

Question

I am reading about calculating the integral of scalar curvature of manifolds which have conical singularities - that is, are locally products $\mathcal{C}_\alpha\times \Sigma$, where $\mathcal{C}_\alpha$ are cones with conical singularities of angle deficit $\alpha$. For example, Fursaev and Soloduhkin (1994) or Troyanov (1991) and I have something which is probably a rather trivial question.

Putting the singularity question aside for a moment, I want to make sure I understand how to calculate something like

$$\int_{M\setminus \Sigma}R,$$

where $\Sigma$ is a 2-dimensional manifold embedded in $M$ (which is higher dimensional, for my purposes we can take this to be a 4-manifold). Certainly, I agree with the statement

$$\int_\Sigma R=0$$

Because $\Sigma$ is a set of measure zero in $M$. And certainly as sets $M$ and $\Sigma$ we could write

$$\int_M R=\int_{M\setminus \Sigma}R + \int_{\Sigma}R$$

So I want to make sure that the following is correct:

$$\int_{M\setminus \Sigma}R=\int_{M} R.$$

Of course, in the context of the singularity papers above, you want to remove the singular set $\Sigma$ and integrate $R$ over only the domain in which it is smoothly defined, $M\setminus \Sigma$. But, if I am doing this on something like a sphere with constant curvature I want to make sure I can write something like

$$\int_{\mathbb{S}^4\setminus \Sigma}R=\int_{\mathbb{S}^4}R(\mathbb{S}^4)=V(\mathbb{S}^4)R(\mathbb{S}^4)$$

Rather than

$$\int_{\mathbb{S}^4\setminus \Sigma}R=R(\mathbb{S}^4)V(\mathbb{S}^4\setminus \Sigma).$$

Again, it seems to me that $V(\mathbb{S}^4)=V(\mathbb{S}^4\setminus \Sigma)$.

Answer 1

Let me offer a way to extend the definition of the integral and answer your question. Let $M$ be a smooth manifold and assume for simplicity it is orientable (otherwise, replace volume forms by densities in what follows). Choose a volume form $\omega \in \Omega^{\text{top}}(M)$ on $M$ (for example, if $M$ comes with a metric this can be the Riemannian volume form). Using $\omega$ and abusing notation a little, we can define the integral of a smooth compactly supported function on $M$ by

$$ \int_{M} f := \int_M f\omega. $$

The map $f \mapsto \int_M f = \int_M f \omega$ gives us a positive linear function from $C^{\infty}_c(M)$ to $\mathbb{R}$ and so, after extending this map to $C_c(M)$, the Riesz representation theorem gives us a unique regular Borel measure $\mu_{\omega}$ such that $$\int_{M} f\omega = \int_M f d\mu_{\omega} $$ for all smooth (or continuous) compactly supported functions. Working with this measure, we can write stuff like

$$ \int_{M} R = \int_{M} R(\chi_{\Sigma} + (1 - \chi_{\Sigma})) d\mu_{\omega} = \int_{\Sigma} R d\mu_{\omega} + \int_{M \setminus \Sigma} R d\mu_{\omega} = \int_{M \setminus \Sigma} R d\mu_{\omega} $$

when $\Sigma$ has zero measure. Note that the notion of a zero measure set can be defined using coordinate charts without constructing any particular measure. If $\Sigma$ is Borel, all the measures $\mu_{\omega}$ that we constructed will satisfy $\mu_{\omega}(\Sigma) = 0$.