## Lebesgue Measure and Linear Transformations

Lemma 1. Let $U\subset\mathbb{R}^{n}$ be open. If $F: U\rightarrow\mathbb{R}^{n}$ is locally Lipschitz on $U$, then the set $F(A)$ is measurable.

Proof. Let $A\subset U$ be measurable. Replacing $A$ by $A\cap[-N,N], N\in\mathbb{N}$, we may assume that $A$ is bounded. Observe that by the inner regularity of the Lebsgue measure, we may write

$\displaystyle A=B\cup\bigcup_{j=1}^{\infty}K_{j}$

where the sets $K_{j}$ are compact and $B$ is a set of measure zero. Since the set $F(\bigcup_{j=1}^{\infty}K_{j})=\bigcup_{j=1}^{\infty}F(K_{j})$ is a Borel set, being the union of compact sets, it suffices to show that $F(B)$ is measurable. Fix $\varepsilon>0$. Since $B$ has measure zero, we can cover $B$ by a sequence of cubes $Q_{j}$ with edge lengths $r_{j}$ and with sum of measures less than $\varepsilon$. Since $F$ is locally Lipschitz, it has Lipschitz constant $L$ on the set $U\cap[-N,N]$. Hence, $F(Q_{j})$ is contained in a ball of radius $L\sqrt{n}r_{j}, \forall j$. Hence, $F(Q_{j})$ is contained a cube with edge length $2L\sqrt{n}r_{j}$. Thus,

$\displaystyle \lambda_{n}(F(B))\leq\sum_{j=1}^{\infty}\lambda_{n}(F(Q_{j}))\leq\sum_{j=1}^{\infty}\left(2L\sqrt{n}r_{j}\right)^{n}=\left(2L\sqrt{n}\right)^{n}\varepsilon$

Since $\varepsilon>0$ was arbitrary, we conclude that $F(B)$ has measure zero, and therefore necessarily measurable. $\Box$

Recall that we say that a $n\times n$ real matrix $A$ is orthogonal if $A$ is invertible and $A^{-1}=A^{T}$. There are a number of properties which equivalently characterize an orthogonal matrix. We summarize them with the following proposition.

Proposition 2. The following are equivalent:

1. The matrix $A:\mathbb{R}^{n}\rightarrow\mathbb{R}^{n}$ is orthogonal.
2. The columns of $A$ form an orthonormal basis.
3. $\langle{Ax,Ay}\rangle=\langle{x,y}\rangle$, for all vectors $x,y\in\mathbb{R}^{n}$.
4. The linear transformation defined by $A$ is an isometry.

Proof. In what follows, $\left\{e_{1},\ldots,e_{n}\right\}$ denotes the standard basis of $\mathbb{R}^{n}$, and $\left\{Ae_{1},\ldots,Ae_{n}\right\}$ denotes the column vectors of the matrix $A$. Suppose (1) holds. Then, for $1\leq i,j\leq n$,

$\displaystyle\langle{Ae_{i},Ae_{j}}\rangle=\langle{A^{T}Ae_{i},e_{j}}\rangle=\langle{A^{-1}Ae_{i},e_{j}}\rangle=\langle{e_{i},e_{j}}\rangle=\begin{cases}{1}&{i=j}\\ {0}&{i\neq j}\end{cases}$

Now suppose that (2) holds. By hypothesis, for any vectors $x,y\in\mathbb{R}^{n}$, we can write

$\displaystyle x=\alpha_{1}e_{1}+\cdots+\alpha_{n}e_{n},\indent y=\beta_{1}e_{1}+\cdots+\beta_{n}e_{n}$

Using the bilinearity of the inner product, we obtain that

$\begin{array}{lcl}\displaystyle\langle{Ax,Ay}\rangle=\sum_{i=1}^{n}\sum_{j=1}^{n}\alpha_{i}\beta_{j}\langle{Ae_{i},Ae_{j}}\rangle&=&\displaystyle\sum_{k=1}^{n}\alpha_{k}\beta_{k}\\&=&\displaystyle\sum_{i=1}^{n}\sum_{j=1}^{n}\alpha_{i}\beta_{j}\langle{e_{i},e_{j}}\rangle\\&=&\displaystyle\langle{x,y}\rangle\end{array}$

(3)$\Longrightarrow$(4) is obvious. To see that (4)$\Longrightarrow$(1), we will prove the more general result that any isometry of $\mathbb{R}^{n}$ is an affine transformation. Recall that an isometry of Euclidean space is a function $\phi:\mathbb{R}^{n}\rightarrow\mathbb{R}^{n}$ which preserves Euclidean distance: $\left\|\phi(x)-\phi(y)\right\|=\left\|x-y\right\|$, where $\left\|\cdot\right\|$ denotes the Euclidean norm. It turns out that every isometry of $\mathbb{R}^{n}$ is an affine map such that

$\displaystyle\phi(x)=Ax+b,\indent\forall x\in\mathbb{R}^{n}$

and $A$ is orthogonal. To see this, note that by the polarization identity, the map $A:=\phi-\phi(0)$ peserves the Euclidean inner product. Hence, for any $\alpha,\beta\in\mathbb{R}$ and $x,y\in\mathbb{R}^{n}$, we see that

$\begin{array}{lcl}\displaystyle\left\|A(\alpha x+\beta y)-\alpha A(x)-\beta A(y)\right\|^{2}&=&\displaystyle\left\|A(\alpha x+\beta y)\right\|^{2}\\&-&\displaystyle2\langle{A(\alpha x+\beta y),\alpha A(x)+\beta A(y)}\rangle+\left\|\alpha A(x)+\beta A(y)\right\|^{2}\\&=&\displaystyle2\left\|\alpha x+\beta y\right\|^{2}-2\langle{\alpha x+\beta y,\alpha x}\rangle-2\langle{\alpha x+\beta y,\beta y}\rangle\\&=&\displaystyle0\end{array}$

So, $A$ is linear, equivalently $\phi$ is affine. That $A$ is orthogonal is an immediate consequence of the preservation of the inner product under $A$. $\Box$

Although every orthogonal matrix necessarily has determinant satisfying $(\det A)^{2}=1$, this condition is not sufficient for a matrix to be orthogonal. Indeed, consider the $2\times 2$ matrix

$\displaystyle A=\begin{bmatrix}{2}&{0}\\{0}&{\frac{1}{2}}\end{bmatrix}$,

which has both determinant $1$ and orthogonal columns.

If $T:\mathbb{R}^{n}\rightarrow\mathbb{R}^{n}$ is a linear transformation given by an orthogonal matrix, then $T$ is tautologically invertible and moreover, its inverse is given by an orthogonal matrix. Consider the pushforward of the Lebesgue measure, which we denote by $\mu:=T_{*}\lambda_{n}$. If we can show that $\mu$ is a translation-invariant measure on the measure space $(\mathbb{R}^{n},\mathcal{B}(\mathbb{R}^{n}))$, and $\mu([0,1]^{n})=1$, then the uniqueness of the Lebesgue measure (see Proposition 5 here) proves that $\mu=\lambda_{n}$. Replacing $T$ by $S=T^{-1}$, we obtain the desired invariance property of Lebesgue measure.

For any Borel set $B\subset\mathbb{R}^{n}$ and $x\in\mathbb{R}^{n}$, we have that

$\displaystyle\mu(x+B)=\lambda_{n}\left(T^{-1}(x+B)\right)=\lambda_{n}\left(T^{-1}x+T^{-1}B\right)=\lambda_{n}\left(T^{-1}B\right)=\mu(B)$,

where we use the translation-invariance of the Lebesgue measure in the penultimate equality. By Theorem 6 here, we conclude that $\mu=\kappa\lambda_{n}$, for some nonzero real constant $\kappa$. To see that $\kappa=1$, observe that since $T$ is orthogonal,

$\displaystyle\left\|Tx\right\|^{2}=\langle{Tx,Tx}\rangle=\langle{T^{-1}Tx,x}\rangle=\left\|x\right\|^{2}<1,\indent\forall\left\|x\right\|<1$

If $B_{1}(0)$ is the open unit ball in $\mathbb{R}^{n}$, then $T^{-1}(B_{1}(0))=B_{1}(0)$. We conclude that

$\displaystyle\kappa\cdot\lambda_{n}\left(B_{1}(0)\right)=\mu\left(B_{1}(0)\right)=\lambda_{n}\left(B_{1}(0)\right)$

and therefore $\kappa=1$.