change of variable in Lebesgue integration

Question

I was encountering such an equation in Lebesgue integration

$$ \int_{B(x_{0}, r)} \left ( \frac{\|x_{0} - x\|_2}{r} \right)^{2} \lambda(dx) = n \cdot \lambda(B(0,1)) \int_{0}^{r} (\frac{\rho}{r})^{2} \rho^{n-1} d\rho $$

where $x$ is a n-dimensional vector, $\lambda$ is Lebesgue measure on $\mathbb{R}^{n}$ and $B(c,r) = \{x \in \mathbb{R}^{n} : ||x - c||_2 \leq r \}$.

So it is obvious we use change of variable by replacing $||x_0 - x||_{2}$ to $\rho$. However, I cannot strictly construct the path from $\lambda(dx)$ to $\lambda (d B(0, \rho))$.

Any thoughts on this would be helpful.

Answer 1

Any Lebesgue integral in $\mathbb{R}^n$ can be rewritten in polar form. More precisely we can identify $(0 , \infty) \times S^{n-1} $ with $\mathbb{R}^n - \{0\}$ under the homeomorphism $\phi : (r, x) \mapsto rx$. Then the Lebesgue measure restricted to the latter space is the product of a radial measure on $(0, \infty)$ given by $E \mapsto \int_E r^{n-1} dr$ and a surface measure $\sigma$ on $S^{n-1}$. The surface measure has a very natural definition. If $E$ is a measureable subset of $S^{n-1}$ then we can consider the sector $E_1$ obtained by taking the set of all lines from the origin to $E$, and $\sigma(E)$ is defined as $n\lambda(E_1)$ where $\lambda$ is the Lebesgue measure.

For your integral we have $$\begin{align*} \int_{B(x_0, r)} \left(\frac{\|x - x_0\|}{r}\right)^2 dx &= \int_{B(0, r)} \left(\frac{\|x\|}{r}\right)^2 dx \quad \text{using a change of variable}\\ &= \int_{S^{n-1}} \int_0^r \left(\frac{\rho}{r}\right)^2 \rho^{n-1}d\rho d\sigma \quad \text{writing as a polar integral}\\ &= \left(\int_0^r \left(\frac{\rho}{r}\right)^2 \rho^{n-1}d\rho \right) \left(\int_{S^{n-1}} d\sigma \right) \\ &= \left(\int_0^r \left(\frac{\rho}{r}\right)^2 \rho^{n-1}d\rho \right) n \lambda(B(0, 1)) \end{align*}$$

Answer 2

Here is a direct approach using simple functions that sidesteps the change of variables theorem.

Write $B(x_0,r) = \cup_{k=1}^N A_{k,N} $, where $A_{1,N} = B(x_0, {1 \over N}r) $, $A_{k,N}=B(x_0, {k \over N}r) \setminus B(x_0, {k-1 \over N}r) $ for $k=2,...,N$.

Define the simple function $f_N(x) = \sum_{k=1}^N ({k-1 \over N })^2 1_{A_{k,N}}(x) $ and note that $\int f_N d \lambda = \sum_{k=2}^N ({k-1 \over N })^2 (\lambda B(x_0, {k \over N}r) - \lambda B(x_0, {k-1 \over N}r) ) = r^n\lambda B(0,1) \sum_{k=2}^N ({k-1 \over N })^2 (({k\over N})^n - ({k-1\over N})^n )$.

It follows from the dominated convergence theorem (DCT) that $\int f_N d \lambda \to \int_{B(x_{0}, r)} \left ( \frac{\|x_{0} - x\|_2}{r} \right)^{2} d\lambda(x) $.

Note that for $x,x-h \in [0,1]$ we have $x^n-(x-h)^n = nx^{n-1}h -{1 \over 2!} n(n-1) \xi^{n-1}h^2$ for some $\xi \in [x,x+h]$. In particular, there is some $M$ such that $|x^n-(x-h)^n - nx^{n-1}h | \le M h^2$.

For the second integral, take the simple function $g_N(\rho) = \sum_{k=1}^N ({k-1 \over N })^2 r^{n-1} ({k-1 \over N })^{n-1} 1_{[{k-1\over N}r, {k \over N}r)}(\rho)$ and note that $\int g_N d \lambda = \sum_{k=1}^N ({k-1 \over N })^2 r^{n-1} ({k-1 \over N })^{n-1} r {1 \over N} = r^n\sum_{k=2}^N ({k-1 \over N })^2 ({k-1 \over N })^{n-1} {1 \over N} $.

As above, the DCT gives $n \lambda B(0,1) \int g_N d \lambda \to n \lambda B(0,1) \int_{0}^{r} (\frac{\rho}{r})^{2} \rho^{n-1} d\rho $.

Finally, we have $|\int f_N d \lambda -n \lambda B(0,1) \int g_N d \lambda | \le r^n\lambda B(0,1) M \sum_{k=2}^N ({k-1 \over N })^2 {1 \over N^2}$.

It is straightforward to show that the right hand side $\to 0$.