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I asked this question yesterday Lebesgue measure of a subspace of lower dimension is 0, Matt S answer didn't really convince me and after trying to find a solution without using determinants and the decomposition of a linear map without getting anything simple I gave up and decided to just do it the other way.

This is the notation and the theorems we assume:

$m$ is the (complete) Lebesgue measure over $R^k$, a a $k$-cell is a set $W$ of the form ($a$ and $b$ are points of $R^k$) $W=\{x\in R^k:a_i<x_i<b_i\,i=1,2,...,k.\}$ or any set obtained by replacing any or all of the signs $<$ by $\le$, $Vol(W)=(b_1-a_1)...(b_n-a_n)$ and if $W$ is any type of cell then $m(W)=vol(W)$. Any linear transformation $T:R^k\to R^k$ can be written as a finite product of one of the following linear transformations:

  1. A Linear transformation that just permutes the standard basis $\{e_1,...,e_k\}$
  2. $T(e_1)=t e_1$ and $T(e_i)=e_i$ for $i=2,...,k$ for some real $t$.
  3. $T(e_1)=e_1+e_2$ and $T(e_i)=e_i$ for $i=2,...,k$.

If T is of any type as above except for 2. and $t=0$ (i.e. $T$ is not a homeomorphism, the other ones are homeomorphisms), $E$ is a lebesgue measurable set then $T(E)$ is also a lebesgue measurable set and \begin{equation} \tag{1} m(T(E))=|\det T|m(E) \end{equation}

Everything above is assumed.

We now want to prove (1) for all linear transformations $T:R^k\to R^k$. Let's start by proving it for the type 2. with $t=0$. We see that $\det T=0$ and $T(R^k)=\{0\}\times R^{k-1}$ is a lebesgue measurable (in fact closed) set. We prove that $m(0\times R^{k-1})$. Let $a_n=(0,-n,-n,...,-n)$ and $b_n=(0,n,n,...,n)$ and define $W_n=\{x\in R^k:{a_n}_i\le x_i\le {b_n}_i, i=1,2,..,k\}$. It's clear that each $W_n$ is a compact $k$-cell, $m(W)=Vol(W)=0$ and $\cup_{n=1}^\infty W_n=\{0\}\times R^{k-1}$ and that $W_n$ is an increasing sequence of sets. We thus have $m(\{0\}\times R^{k-1})=\lim_{n\to \infty}m(W_n)=\lim_{n\to\infty}0=0$. So we now have proved (1) for $E=R^k$, if $E$ is any subset of $R^k$ then $T\subset T(R^k)$ and then $T(E)$ is a Lebesgue measurable because $m$ is complete and $m(T(R^k))=0$ and then (1) still holds.

Let $T:R^k\to R^k$ be any linear transformation and write $T=T_1T_2...T_n$ where each $T_i$ is of some of the three types above and $E$ be a Lebesgue measurable set, we inductively get that $T(E)$ is another Lebesgue measurable set. We also inductively get: \begin{align} m(T(E))&=m(T_1(T_2(...T_n(E)...))\\ &=|\det T_1|m(T_2(...(T_n(E)...))\\ &...\\ &=|\det T_1||\det T_2|...|\det T_n|m(E)\\ &=|det T|m(E) \end{align}

We now have (*) proved for all linear maps $T:R^k\to R^k$.

I wanna check if this is right and to find an answer to my other question that is as simple as possible.

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1 Answers1

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Another way to think of this is as follows. Suppose $T$ is not onto, and let $\{b_1,b_2,\dotsc,b_p\}$ be a basis for the range of $T$. Extend to a basis $\{b_1,b_2,\dotsc,b_k\}$ for $\mathbf{R}^k$, and define $U$ as the following transformation: $U(e_i)=b_i$, where $\{e_1,e_2,\dotsc,e_k\}$ is the standard basis for $\mathbf{R}^k$. Note that $U$ is invertible, and since $m(U(E))=\Delta(U)m(E)$, it suffices to show the hyperplane spanned by $\{e_1,e_2,\dotsc,e_p\}$ has measure $0$. The following $k$-cell: $[0,1]^p\times\{0\}^{k-p}$ has measure $0$ (its volume is $0$), and the hyperplane spanned by $\{e_1,e_2,\dotsc,e_p\}$ is a countable union of translations of this $k$-cell.

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