The total integral of a function equals to the integral of a function of its Lebesgue measure? The descirption on title might be confusing, but I am doing Bass's Real Analysis for Graduate Student, excercise 11.3. The problem is ask us to prove $$\int_{-\infty}^\infty |f(x)| \, dx = \int_0^\infty m( \{x:|f(x)| \geq t\}) \, dt$$
I don't know how to understand the integral on the right-hand-side. Anyone can give a hint on how to approach? Thanks!
 A: Hint: write $$\int_{-\infty}^\infty |f(x)|\,dx = \int_{-\infty}^\infty \int_{-\infty}^\infty 1_{0 \leq t \leq |f(x)|}\,dt\,dx.$$
What happens to the right hand side if you apply the Fubini-Tonelli theorem?
A: Suppose $g$ is an everywhere non-negative function. A simple function is a function $s$ whose set of values $\{s(x) : -\infty<x<\infty\}$ is finite. Then the Lebesgue integral is by definition
$$
\int_{-\infty}^\infty g(x) \,dx = \sup\left\{ \int_{-\infty}^\infty s(x)\,dx : s \text{ is a non-negative simple function }\le g \right\}.
$$
Each of these integrals of a simple function is a sum of finitely many terms. Suppose the positive values of a simple function $s$ are $0<s_1<\cdots<s_n$ and for $i=1,\ldots,n,$ $m_i = m(\{ x : s(x) = s_i \}).$ Then
\begin{align}
\int_{-\infty}^\infty s(x)\,dx = {} & s_1 m_1 + \cdots + s_n m_n \\ = {} & s_1(m_1+ m_2+m_3+\cdots+m_n) \\
& {} + (s_2-s_1)(m_2+m_3 + \cdots + m_n) \\
& {} + (s_3-s_2)(m_3+\cdots+ m_n) \\
& {} \,\,\,\vdots \\[10pt]
= {} & s_1 m(\{ x : s(x)\ge s_1 \}) \\
& {} + (s_2-s_1) m(\{x: s\ge s_2\}) \\
& {} + (s_3-s_2) m(\{x: s\ge s_3\}) \\
& {} \,\,\,\vdots
\end{align}
Now the problem is to show that the integral on the right side of the proposed equality is the supremum over all simple functions, of this last sum.
Note that $t \mapsto m(\{x: s(x) \ge t\})$ is a simple function, and the measure of the set of points where it is equal to $m(\{x:s(x)\ge s_k\})$ is $s_k-s_{k-1}.$
