Enumerable unions and Lebesgue measure; Result from Macbeath & Rogers

Question

Let $f:(0,1] \rightarrow \mathbb{R}$ be measurable function and $T \subset SL(n, \mathbb{R})$ such that the set $T(1) = \{ \lambda A: A \in T, \lambda \in (0,1] \}$ is measurable. Then: $\int_{T(1)} f(det(A)) d \mu(A) = \mu(T(1)) \int_0^1 f(t^{1/n})dt$. This is a lemma by Macbeath & Rogers. Regarding the proof, they show first that the equation holds if $f$ is the indicator function for a open interval by straightforward computation. However, subsequently it just says "Taking enumerable unions of intervals, the result follows for the characteristic function of any Lebesgue measurable set [...]". This is where I need help. The equation clearly holds for enumerable unions of intervals by monotone convergence theorem, but how does the statement follows for arbitrary measurable sets?

Answer 1

For any $\epsilon>0$, any measurable set $E$ can be covered by a countable family of intervals $I_n$ such that $\sum_{n}\mu(I_n) \le \mu(E)+\epsilon$. Let $U=\bigcup_n I_n$. We know the statement holds for $U$, and we have $E\subset U$ with $\mu(U\setminus E)<\epsilon$. In the language of characteristic functions, this implies $\|\chi_E-\chi_U\|<\epsilon$ with the norm being $L^1(\mu)$ norm. Since $\epsilon>0$ is arbitrary, the conclusion follows.