Folland’s real analysis, theorem 2.44

measure-theoryreal-analysis

I'm confused about the last part of the proof of theorem 2.44 in Folland's real analysis, which states

$GL(n, \mathbb{R})$ (the "general linear" group) is the group of invertible
linear transformations of $\mathbb{R}^n$. Every $T\in GL(n, \mathbb{R})$ can be written as the product of finitely many transformations of three "elementary" types:
$$T_1(x_1, …, x_j, …, x_n) = (x_1, …, c x_j, …, x_n) (c\neq 0)$$
$$T_2(x_1, …, x_j, …, x_n) = (x_1, …, x_j + cx_k, …, x_n) (k\neq j)$$
$$T_3(x_1, …, x_j, …, x_k, …, x_n) = (x_1, …, x_k, …, x_j, …, x_n)$$

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I understand the proof of b when $E$ is a Borel set, while I get stuck when $E$ is a Lebesgue measurable set. The book mentions this tersely, as underlined in red, but I do not understand. In my attempt:

$E$ is Lebesgue measurable set, so is of format $E=B\cup N$ where $B, F \in \mathcal{B}_{\mathbb{R}^n}, N\subset F, m(F)=0$. If $B\cap N \neq \emptyset$, $B:= B\setminus N$. We have
$$m(E) = m(B) + m(N) = m(B)$$
$$m(T(B))=|det T|m(B)$$

My goal is to show $m(T(E))=|det T|m(E)$. Any help would be appreciated!

Best Answer

You are very close to the solution.

$E$ is Lebesgue measurable set, so is of format $E=B\cup N$ where $B, F \in \mathcal{B}_{\mathbb{R}^n}, N\subset F, m(F)=0$.

Let $N_1:= N\setminus B$. Then $E=B\cup N_1$, and $N_1 \subseteq N \subseteq F$ and $m(F)=0$. We have

  1. $m(E) = m(B) + m(N_1) = m(B)$;
  2. $m(T(B))=|det T|m(B)$
  3. $m(T(F)) = m(F) =0$

Note that $ T(B) \subseteq T(E) = T(B \cup N_1) = T(B) \cup T(N_1) \subseteq T(B) \cup T(F)$.

So, using items $2$ and $3$ above, we have

\begin{align} |det T|m(B) & = m(T(B)) \leqslant m(T(E)) \leqslant m(T(B) \cup T(F)) \leqslant \\ &\leqslant m(T(B)) + m(T(F)) =|det T|m(B) + 0 = |det T|m(B) \end{align}

So, combining with item $1$ above, we have $ m(T(E)) = |det T|m(B) = |det T|m(E)$.

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