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Pertti Mattila's Fourier Analysis and Hausdorff Dimension makes the following claim:

One checks easily that $\sigma^{n-1}$ is the weak limit of the measures $\delta^{-1} \mathcal L^n\vert_{B(0, 1+\delta)\setminus B(0,1)}$ as $\delta\to 0$.

In other words, we want to show that for every $\phi\in C_0(\Bbb R^n)$, $$\lim_{\delta\to 0} \delta^{-1}\int \phi\, d\mathcal L^n\vert_{B(0, 1+\delta)\setminus B(0,1)} = \int \phi \, d\sigma^{n-1}$$

How does one do this? Thanks for any help!

Note $1$: The surface measure $\sigma^{n-1}$ on the sphere $S^{n-1}$ is defined in Folland text, in the following way:

Consider the homeomorphism $\Phi:\Bbb R^n\setminus\{0\} \to (0,\infty)\times S^{n-1}$ given by $\Phi(x) = (r,x')$ where $r = |x|$ and $x' = \frac{x}{|x|}$. Let $m_\ast$ be the Borel measure induced on $(0,\infty)\times S^{n-1}$ by the Lebesgue measure on $\Bbb R^n$, via $\Phi$, i.e., $$m_\ast(E) := m(\Phi^{-1}(E))$$ We define the measure $\rho = \rho_n$ on $(0,\infty)$ by $$\rho(E) := \int_E r^{n-1}\, dr$$ There exists a unique Borel measure $\sigma = \sigma_{n-1}$ on $S^{n-1}$ such that $$m_\ast = \rho \times\sigma$$ If $f$ is Borel measurable on $\Bbb R^n$ and $f\ge 0$, or $f\in L^1(m)$, then $$\int_{\Bbb R^n} f(x)\, dx = \int_0^\infty \int_{S^{n-1}} f(rx') r^{n-1}\, d\sigma(x') \, dr$$

Note $2$: $\mathcal L^n$ and $m$ both denote the $n$-dimensional Lebesgue measure. I did not bother using same notation everywhere, since the first half of the post is borrowed from Folland's book, while the second half is from Mattila's book.

stoic-santiago
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1 Answers1

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Suppose $\phi\in\mathcal{C}_0(\mathbb{R}^n)$. Then $$\begin{align} \frac{1}{\delta}\int_{B(0;1+\delta)\setminus B(0;1)} \phi(x)\,dx&=\int_{\mathbb{S}_{n-1}}\frac{1}{\delta}\Big(\int^{1+\delta}_1 r^{n-1}\phi(r\mathbf{u})\,dr\Big)\sigma_{n-1}(d\mathbf{u})\\ &=\int_{\mathbb{S}_{n-1}}\frac{1}{\delta}\Big(\int^{1+\delta}_1 r^{n-1}\big(\phi(r\mathbf{u})-\phi(\mathbf{u})\big)\,dr\Big)\sigma_{n-1}(d\mathbf{u})+\\ &\qquad \qquad +\frac{(1+\delta)^n-1}{n\delta}\int_{\mathbb{S}_{n-1}}\phi(\mathbf{u})\,\sigma_{n-1}(d\mathbf{u}) \end{align}$$

Since $\phi\in\mathcal{C}_0$, $\phi$ is uniformly continuous. Hence, the conclusion follows by passsing to the limit $\delta\rightarrow0$ by standard $\varepsilon-\delta$ arguments. For example, given $\varepsilon>0$ there is $\delta_\varepsilon>0$ such that $|x-y|<\delta_\varepsilon$ implies $|\phi(x)-\phi(y)|<\varepsilon$. Thus for $\delta<\delta_\varepsilon$, $|\phi(r\mathbf{u})-\phi(\mathbf{u})|<\varepsilon$ for all $1\leq r<1+\delta$ and $u\in\mathbb{S}_{n-1}$; hence $$ \left|\int_{\mathbb{S}_{n-1}}\frac{1}{\delta}\Big(\int^{1+\delta}_1 r^{n-1}\big(\phi(r\mathbf{u})-\phi(\mathbf{u})\big)\,dr\Big)\sigma_{n-1}(d\mathbf{u})\right|\leq \frac{(1+\delta)^n-1}{n\delta}\varepsilon\sigma_{n-1}(\mathbb{S}_{n-1})$$ All this shows that $$\lim_{\delta\rightarrow0}\int_{\mathbb{S}_{n-1}}\frac{1}{\delta}\Big(\int^{1+\delta}_1 r^{n-1}\big(\phi(r\mathbf{u})-\phi(\mathbf{u})\big)\,dr\Big)\sigma_{n-1}(d\mathbf{u})=0$$

Mittens
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  • Could you please explain the first equality? Can't see where it's coming from. Thanks! – stoic-santiago May 26 '22 at 01:39
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    That is polar coordinates: $\int_{\mathbb{R}^n}f(x),dx=\int^\infty_0\int_{S_{n-1}}f(r u)r^{n-1},du,\sigma_{n-1}(du)$. In the case of your problem, set $f(x)=\phi(x)\mathbb{1}{B(0;1+\delta)\setminus B(0;1)}(x)$. Notice that $\mathbb{1}{B(0;b)\setminus B(0;a)}(x)=\mathbb{1}_{[(a,b)}(|x|)$. The formula for polar coordinates is also an exercise in Folland's book where polar coordinates are introduced (Rudin's RCA also has that formula as an exercise). – Mittens May 26 '22 at 01:49
  • Did you also apply Fubini's theorem at some point, to pull out $\int_{S_{n-1}}$? As you can see, in $$\int_{\mathbb{R}^n}f(x),dx=\int^\infty_0\int_{S_{n-1}}f(r u)r^{n-1},du,\sigma_{n-1}(du)$$ the integral over $S_{n-1}$ is inside. If yes, how did you apply it? I believe the requirement is that we work with $\sigma$-finite measure spaces - then we can swap the integrals. – stoic-santiago May 26 '22 at 02:24
  • @connected-subgroup Yes it is Fubinis theorem. $S^{n - 1}$ has finite surface area, so Fubinis theorem works here. – Mason May 26 '22 at 02:41
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    Thank you! Please allow me to confirm the last step. We have $$\left|\int_{\mathbb{S}{n-1}}\frac{1}{\delta}\Big(\int^{1+\delta}_1 r^{n-1}\big(\phi(r\mathbf{u})-\phi(\mathbf{u})\big),dr\Big)\sigma{n-1}(d\mathbf{u})\right|\leq \frac{(1+\delta)^n-1}{n\delta}\varepsilon\sigma_{n-1}(\mathbf{S}{n-1})$$ where we push $\delta\to 0$, to get $$\limsup{\delta\to 0} \left|\int_{\mathbb{S}{n-1}}\frac{1}{\delta}\Big(\int^{1+\delta}_1 r^{n-1}\big(\phi(r\mathbf{u})-\phi(\mathbf{u})\big),dr\Big)\sigma{n-1}(d\mathbf{u})\right| \leq \varepsilon\sigma_{n-1}(\mathbf{S}_{n-1})$$ (1/2) – stoic-santiago May 26 '22 at 03:34
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    Since this holds for all $\varepsilon > 0$, we have $$\lim_{\delta\to 0} \left|\int_{\mathbb{S}{n-1}}\frac{1}{\delta}\Big(\int^{1+\delta}_1 r^{n-1}\big(\phi(r\mathbf{u})-\phi(\mathbf{u})\big),dr\Big)\sigma{n-1}(d\mathbf{u})\right| = 0$$ (2/2) – stoic-santiago May 26 '22 at 03:34