Integration in Polar Coordinates (Fubini/Tonelli)

[joypav]

Let $S^{n-1} = \left\{ x \in R^2 : \left| x \right| = 1 \right\}$ and for any Borel set $E \in S^{n-1}$ set $E* = \left\{ r \theta : 0 < r < 1, \theta \in E \right\}$. Define the measure $\sigma$ on $S^{n-1}$ by $\sigma(E) = n \left| E* \right|$.

With this definition the surface area $\omega_{n-1}$ of the sphere in $R^n$ satisfies $\omega_{n-1} = n \gamma_n = \frac{2 \pi^{n/2}}{\Gamma (n/2)}$, where $\gamma_n$ is the volume of the unit ball in $R^n$. Prove that for all non-negative Borel functions $f$ on $R^n$,

$\int_{R^n} f(x) dx = \int_0^{\infty} r^{n-1} (\int_{S^{n-1}} f(r \theta ) d \sigma (\theta) ) dr $.

In particular, if $f$ is a radial function, then ($f$ radial $\implies f(x) = f(\left| x \right|)$)

$\int_{R^n} f(x) dx = \frac{2 \pi^{n/2}}{\Gamma (n/2)} \int_0^{\infty} r^{n-1} f(r) dr = n \gamma_n \int_0^{\infty} r^{n-1} f(r) dr$.

I admit my head is spinning over this problem. Can someone give me an idea of the proof or explain the idea of it more clearly...
Also, this problem is in the section for Fubini/Tonelli Theorems.

 

[caffeinemachine]

Let $S^{n-1} = \left\{ x \in R^2 : \left| x \right| = 1 \right\}$ and for any Borel set $E \in S^{n-1}$ set $E* = \left\{ r \theta : 0 < r < 1, \theta \in E \right\}$. Define the measure $\sigma$ on $S^{n-1}$ by $\sigma(E) = n \left| E* \right|$.

With this definition the surface area $\omega_{n-1}$ of the sphere in $R^n$ satisfies $\omega_{n-1} = n \gamma_n = \frac{2 \pi^{n/2}}{\Gamma (n/2)}$, where $\gamma_n$ is the volume of the unit ball in $R^n$. Prove that for all non-negative Borel functions $f$ on $R^n$,

$\int_{R^n} f(x) dx = \int_0^{\infty} r^{n-1} (\int_{S^{n-1}} f(r \theta ) d \sigma (\theta) ) dr $.

In particular, if $f$ is a radial function, then ($f$ radial $\implies f(x) = f(\left| x \right|)$)

$\int_{R^n} f(x) dx = \frac{2 \pi^{n/2}}{\Gamma (n/2)} \int_0^{\infty} r^{n-1} f(r) dr = n \gamma_n \int_0^{\infty} r^{n-1} f(r) dr$.

I admit my head is spinning over this problem. Can someone give me an idea of the proof or explain the idea of it more clearly...
Also, this problem is in the section for Fubini/Tonelli Theorems.

$\newcommand{\set}[1]{\{#1\}}$
$\newcommand{\vp}{\varphi}$
$\newcommand{\mc}{\mathcal}$
$\newcommand{\R}{\mathbf R}$Let $(X, \mc X, \mu)$ be a measure space and $(Y, \mc Y)$ be a measurable space and let $\vp:X\to Y$ be a measurable function.
We define the push-forward of $\mu$ via $\vp$ as a measure on $(Y, \mc Y)$, which we denote by $\vp_*\mu$, by declaring $(\vp_*\mu)(E)=\mu(\vp^{-1}(E))$ for all $E\in \mc Y$.

A key theorem, which is easy to prove, is that if $g:Y\to \R$ is a measurable function such that $g\circ f$ is an integrable function on $X$, the we have
\begin{equation}
\int_X g\circ \vp\ d\mu = \int_Y g\ d(\vp_*\mu)
\end{equation}

We will use this theorem to solve the problem at hand.
We have a natural map $\vp: (0, \infty)\times S^{n-1}\to \R^n\setminus\set{0}$ which sends $(r, \theta)$ to $r\theta$.
Write $\nu$ to denote the Lebesgue measure on $\R^n\setminus \set{0}$.
Define a measure $\mu$ on $(0, \infty)\times S^{n-1}$ as $\mu=\vp^{-1}_*\nu$.
The main observation is that if $\lambda$ is the Lebesgue measure on $\R$, then the measure $\mu$ is the product $\alpha\times \sigma$, where $\alpha$ is given by $d\alpha= x^{n-1} d\lambda$.

So we have
\begin{equation}
\int_{\R^n}f\ d\nu =\int_{\R^n\setminus \set{0}} f\ d\nu = \int_{(0, \infty)\times S^{n-1}} f\circ \vp\ d(\alpha\times \sigma)
\end{equation}

Now by Fubini
\begin{equation}
\int_{(0, \infty)\times S^{n-1}} f\circ \vp\ d(\alpha\times \sigma) = \int_0^\infty \left(\int_{S^{n-1}} f\circ \vp(r, \theta) \ d\sigma(\theta)\right) d\alpha(r)
\end{equation}
which is, by the fact that $d\alpha=x^{n-1} \ d\lambda$, is

\begin{equation}
\int_0^\infty \left(\int_{S^{n-1}} f\circ \vp(r, \theta) \ d\sigma(\theta)\right) d\alpha(r)=\int_0^\infty r^{n-1} \left(\int_{S^{n-1}} f(r\theta) \ d\sigma(\theta)\right) d\lambda(r)
\end{equation}

 

[joypav]

Thank you!
Makes sense. I wonder why we haven't seen the "push forward"? (Wondering)

In the very first integral, is that meant to be $d\lambda$?

 

[caffeinemachine]

Thank you!
Makes sense. I wonder why we haven't seen the "push forward"? (Wondering)

In the very first integral, is that meant to be $d\lambda$?

I am using $\nu$ to denote the Lebesgue measure on $\mathbb R^n$ and $\lambda$ to denote the Lebesgue measure on $\mathbb R$.

 

[laurabon]

$\newcommand{\set}[1]{\{#1\}}$
$\newcommand{\vp}{\varphi}$
$\newcommand{\mc}{\mathcal}$
$\newcommand{\R}{\mathbf R}$Let $(X, \mc X, \mu)$ be a measure space and $(Y, \mc Y)$ be a measurable space and let $\vp:X\to Y$ be a measurable function.
We define the push-forward of $\mu$ via $\vp$ as a measure on $(Y, \mc Y)$, which we denote by $\vp_*\mu$, by declaring $(\vp_*\mu)(E)=\mu(\vp^{-1}(E))$ for all $E\in \mc Y$.

A key theorem, which is easy to prove, is that if $g:Y\to \R$ is a measurable function such that $g\circ f$ is an integrable function on $X$, the we have
\begin{equation}
\int_X g\circ \vp\ d\mu = \int_Y g\ d(\vp_*\mu)
\end{equation}

We will use this theorem to solve the problem at hand.
We have a natural map $\vp: (0, \infty)\times S^{n-1}\to \R^n\setminus\set{0}$ which sends $(r, \theta)$ to $r\theta$.
Write $\nu$ to denote the Lebesgue measure on $\R^n\setminus \set{0}$.
Define a measure $\mu$ on $(0, \infty)\times S^{n-1}$ as $\mu=\vp^{-1}_*\nu$.
The main observation is that if $\lambda$ is the Lebesgue measure on $\R$, then the measure $\mu$ is the product $\alpha\times \sigma$, where $\alpha$ is given by $d\alpha= x^{n-1} d\lambda$.

So we have
\begin{equation}
\int_{\R^n}f\ d\nu =\int_{\R^n\setminus \set{0}} f\ d\nu = \int_{(0, \infty)\times S^{n-1}} f\circ \vp\ d(\alpha\times \sigma)
\end{equation}

Now by Fubini
\begin{equation}
\int_{(0, \infty)\times S^{n-1}} f\circ \vp\ d(\alpha\times \sigma) = \int_0^\infty \left(\int_{S^{n-1}} f\circ \vp(r, \theta) \ d\sigma(\theta)\right) d\alpha(r)
\end{equation}
which is, by the fact that $d\alpha=x^{n-1} \ d\lambda$, is

\begin{equation}
\int_0^\infty \left(\int_{S^{n-1}} f\circ \vp(r, \theta) \ d\sigma(\theta)\right) d\alpha(r)=\int_0^\infty r^{n-1} \left(\int_{S^{n-1}} f(r\theta) \ d\sigma(\theta)\right) d\lambda(r)
\end{equation}

sorry if I am re opening this , I have the same problem , can you explain why $\mu$ is the product $\alpha\times \sigma$, where $\alpha$ is given by $d\alpha= x^{n-1} d\lambda$.? Really thanks