Finding integral of exact form on punctured 3-space

Question

I'm trying to solve the following quals problem, and I think I have the general idea, but there's one step I'm not sure about, and I highlight it below.

The problem:

Let $M=\mathbb{R}^3\setminus0$. Let $\Sigma_1\subset M$ be the unit sphere, and $\sigma_1$ its volume form such that \begin{equation} \int_{\Sigma_1}\sigma_1 = 1 \end{equation} Suppose $\rho:M\rightarrow\Sigma_1$ is the radial retraction, and $\omega=\rho^\ast(\sigma_1)$. If $f:\mathbb{R}^3\rightarrow\mathbb{R}$ is a smooth compactly-supported function, then \begin{equation} \int_M df\wedge\omega \end{equation} depends only on $f(0)$.

My attempt is as follows: since $f$ is compactly supported, there's an $R$ such that $f\equiv0$ on $\Sigma_R$. Define \begin{align} A_r &= \{x\in M: r\le|x|\le R\}\\ B_r &= \{x\in M: |x|\le r\} \end{align} By Stokes' theorem then \begin{equation} \int_{A_r}df\wedge\omega = \int_{\Sigma_r}f\omega \end{equation} Because $\rho:\Sigma_r\rightarrow\Sigma_1$ is a diffeomorphism, it must be that $\rho^\ast(\sigma_1)=\sigma_r$ on $\Sigma_r$. So the integral on the right (and hence the left) satisfies \begin{equation} m_r \le \int_{\Sigma_r}f\omega \le M_r \end{equation} with $m_r=\min_{B_r}\{f(x)\}$ and $M_r=\max_{B_r}\{f(x)\}$.

Here's the part I'm unsure of. I want to claim \begin{equation} \int_M df\wedge\omega = \lim_{r\rightarrow0^+}\int_{A_r} df\wedge\omega = f(0) \end{equation} I believe I can prove the second equality via some squeeze-type argument, but how can I rigorously shows the first equality holds? It seems intuitive to me, but I'm not even sure how to prove the limit exists. I'm also concerned because this seems a much stronger conclusion than the problem asked for: the original integral was supposed to depend on $f(0)$, and I've shown it is $f(0)$!

Answer 1

I think this might work.

Write everything in spherical coordinates, then $\omega=\frac{1}{4\pi^2}\sin\theta d\theta d\phi$, and $df=\partial_rfdr+\partial_\theta fd\theta+\partial_\phi fd\phi$. It follows that: \begin{align} \int_Mdf \wedge \omega=&\frac{1}{4\pi}\int_0^{2\pi}\int_0^{\pi}\int_0^\infty (\partial_rf(r,\theta,\phi))\sin\theta drd\theta d\phi \end{align} Now we can integrate with respect to $r$ holding all other variables constant, so regardless of whether or not $f$ depends on $\theta$ and $\phi$, we know that $\int_0^\infty\partial_rf dr=f(0)$. However, $f(0)$ doesn't depend on $\theta$ or $\phi$ (as for all values of $\theta$ and $\phi$, $f$ must be the same at the orign, so this is a constant and we have that: $$\int_Mdf \wedge \omega=f(0)$$ Moreover, regarding my previous comment, for $f$ to have compact support on $M$, we must have that $f(0)=0$, as other wise the support of $f$ in $M$ can't be compact as it will be a compact set minus a point. It follows that this is not a counter example to stokes theorem which is nice.

Edit: A priori I guess there’s some finicky Lebesque integration stuff one “should” know and be careful about when integrating non compactly supported forms, but since $f$ has compact support on $\mathbb R^3$ everything is “probably” fine.