Recursion formula for volume of an unit ball in $\mathbb{R^n}$

Question

Let $B^n\subset \mathbb{R^n}$ be an unit ball in $\mathbb{R^n}$ and $V_n(A)$ be the volume of $A\subset \mathbb{R^n}$.Then, prove

\begin{equation} V_n(B^n)=2 V_{n-1} (B^{n-1}) \displaystyle\int_0^1 (1-u^2)^{\frac{n-1}{2}} du. \end{equation}

using Fubini's theorem.

My idea is here: Let $m_n$ be $n$- dimentional Lebesgue measure. Then, $$V_n(B^n)=m_n (B^n)=\displaystyle \int_{\mathbb{R^n}} \chi_{B^n} d\mu .$$

Since the claim says "using Fubini", I think I have to separate $\int_{R^n}$ as $\int_{R^n}=\int_{R^{n-1}}\int_{R} $. So I have $$\int_{\mathbb{R^n}} \chi_{B^n} d\mu=\int_{R^{n-1}} \left(\int_{R} \chi_{B^n} \ dy \right)dx.$$

I don't know how I have to proceed.

Answer 1

Its exactly the right thing to do, you just need to push through. A key lemma is

if $T:\mathbb R^d\to\mathbb R^d$ is a linear map and $\mu$ is $\Bbb R^d$ Lebesgue measure then for all measurable subsets $U$, $$ \mu(TU) =|\det T|\mu (U).$$

You should try to use it a little before reading on. $$\int_{\Bbb R^n} \chi_{B^n} dx = \int_{[-1,1]^n} \chi_{B^n} dx = \int_{-1}^1 \int_{[-1,1]^{n-1}}\chi_{B^n}(x',x_n) dx' dx_n =: \int_{-1}^1 v(x_n) dx_n $$ and then the above lemma on the scaling property of Lebesgue measure gives \begin{align} v(u) &= \mu_{\mathbb R^{n-1}}(\{x'\in \mathbb R^{n-1}: |x'| \le (1-u^2)^{1/2}\}) \\ &= (1-u^2)^{(n-1)/2} \mu_{\mathbb R^{n-1}}( \{|x'|\le 1\}) \\&= (1-u^2)^{(n-1)/2} V_{n-1}(B^{n-1}), \end{align} hence the result.