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I want to prove that if $B^3_R$ is a 3-ball of radius $R$ around $0$, then

$$ \int_{B^3_R}||x||^p dx = 4\pi\frac{R^{p+3}}{p+3}$$

I of want to use Fubini's theorem here to switch to spherical coordinates, but the part that is confusing me is the following: Fubini's theorem talks about the product measure of two spaces, but if I understand correctly the Lebesgue measure on a product is not quite the product measure. How relevant is this, and what does a formal proof look like?

Thanks a lot!

Thomas Bakx
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    You use the change of variables theorem to switch to spherical coordinates. The domain of the spherical coordinate parametrization is something like $(0,R)\times (0,\pi)\times (0,2\pi)$. Then, you use Fubini (because the domain is now a product). Also, there is version of Fubini for completion of measures (see Folland's text for the precise statement). – peek-a-boo Dec 20 '21 at 22:07
  • Okay, but then I should regard the set you quoted as having the product measure, and not the Lebesgue measure on the corresponding subset of $\mathbb{R}^3$? I guess this is what confuses me, because then when I apply the substitution theorem I suddenly go from Lebesgue measure to something else. Does this make sense? – Thomas Bakx Dec 20 '21 at 22:10
  • See the end of my first comment for the general situation. Anyway, the sets here are very nice so on $\Bbb{R}^3$ once you restrict Lebesgue measure to the Borel sigma-algebra (which is in fact the 3-fold product of Borel sigma-algebras of $\Bbb{R}$, since the reals are second-countable) this is equal to the product of one-dimensional Lebesgue measures restricted to Borel sigma-algebra of $\Bbb{R}$. – peek-a-boo Dec 20 '21 at 22:15
  • So, I can just say that the integral of my function with respect to the Lebesgue measure equals the integral with respect to the Borel measure? Under what conditions does this hold exactly, and does it have anything to do with my function? – Thomas Bakx Dec 20 '21 at 22:22
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    If you have Borel functions and integrating over Borel sets, you can do this. Here, you have continuous functions and intervals (open sets), so everything is nice. – peek-a-boo Dec 20 '21 at 22:26
  • Thanks! If you can summarize this in an answer and maybe provide a reference for these things I can accept it :) – Thomas Bakx Dec 20 '21 at 22:32
  • The approach in Folland is somewhat reversed: it proves that there exists a measure $\sigma$ on $S^{n-1}$ such that Lebesgue measure on $\mathbb{R}^n$ agrees with the product measure of $\rho^{n-1} d\rho$ and $\sigma$. (And in the proof, the construction of $\sigma$ is that the measure of a subset $A$ of $S^{n-1}$ is defined to be $n$ times the Lebesgue measure of the sector spanned by that subset, i.e. the measure of ${ \lambda x \mid \lambda \in [0,1], x \in A }$.) – Daniel Schepler Dec 20 '21 at 22:37

2 Answers2

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In general suppose you have a set $X$ with two $\sigma$-algebras, $\mathfrak{M}\subset \mathfrak{N}$ and a measure $\nu$ on the larger one, and $\mu$ is the resitriction to the smaller one. Suppose $f:X\to\Bbb{C}$ is $\mathfrak{M}$-$\mathcal{B}(\Bbb{C})$ measurable (i.e with respect to the smaller $\sigma$-algebra), then it is almost immediate from definitions that for any set $E\in\mathfrak{M}$ in the smaller $\sigma$-algebra, $f$ is integrable over $E$ with respect to $\mu$ if and only if it is with respect to $\nu$, in which case \begin{align} \int_Ef\,d\mu&=\int_Ef\,d\nu. \end{align} Unwind the definitions: clearly true for indicators of sets in $\mathfrak{M}$, hence for non-negative linear combinations, hence all non-negative $\mathfrak{M}$-measurable functions, hence all integrable ones.

For each integer $n\geq 1$, let $\mathcal{L}_n,\mathcal{B}_n, \lambda_n,\beta_n$ be the Lebesgue, Borel $\sigma$-algebras and measures respectively, i.e $\beta_n:=\lambda_n|_{\mathcal{B}_n}$. Then, we have $\beta_n= (\beta_1)^{\otimes n}$ is the $n$-fold product measure (and $\mathcal{B}_n=(\mathcal{B}_1)^{\otimes n}$ because $\Bbb{R}$ is second-countable. See Folland's Chapter 1 for these details).

Because of these facts, after changing variables, you can use Fubini naively on the Borel sets. Anyway, you should note that Folland also states (and outlines as an exercise) a version of Fubini valid for complete measures (i.e $\mu,\nu$ $\sigma$-finite and complete, and $\lambda$ the completion of $\mu\otimes \nu$), so take a look at his book for that. Also, the general integral calculation becomes almost obvious now: for $n$-dimensional balls of radius $R$, and $p>-n$, \begin{align} \int_{B_R^n(0)}\|x\|^p_2\,dx&=A_{n-1}\frac{R^{p+n}}{p+n}=\frac{2\pi^{n/2}}{\Gamma(n/2)}\frac{R^{p+n}}{p+n}. \end{align}

One thing to note is that, as mentioned in the comments, Folland introduces the spherical measure in a rather ad-hoc/reversed way. The way I prefer to think of the surface measure is as the positive measure induced by the Riemannian metric on $S^{n-1}$ induced from $\Bbb{R}^{n}$. Also, Folland's text has a statement and proof of the change of variables theorem for Lebesgue integrals, so you can look at that as well (if you have access to extra measure-theoretic mumbo jumbo, i.e Chapter 3 of Folland/chapter 6 of Rudin's RCA, you can also follow along this answer I wrote)

peek-a-boo
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Spherical coordinates in $\mathbb R^3$: $$ dx_1\,dx_2\,dx_3=r^2\,dr\,d\omega $$ where $\int_{\|x\|=1}d\omega=4\pi$, the area of a unit sphere.

Hence $$ \int_{\|x\|\le R}\|x\|^p\,dx_1\,dx_2\,dx_3=\int_0^R\left(\int_{\|x\|=1}r^pr^2\,d\omega\right)\,dr=\int_0^R 4\pi r^{p+2}dr=\frac{4\pi R^{p+3}}{p+3} $$

Yiorgos S. Smyrlis
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