Alternative expression of norm L1

Question

Let $(X, \Sigma , \mu)$ a space with measure and $f \in L^1(X,\mu)$. Prove that $$ ||f||_{L^1(X,\mu)} = \int_{0}^{+\infty} \mu(A_r) dr $$

Where:

$$A_r = \{x \in X: |f(x)|\ge r\}$$

I tried to solve this exercise, but I couldn't. I was unable to relate the integral in X to the real integral. Can someone help me?

Answer 1

This is an exercise in Fubini.

\begin{align} \|f\|_{L^1(X,\mu)} &=\int_X |f(x)|\textrm{d}\mu(x)=\int_X \int_0^{|f(x)|} 1\textrm{d}r\textrm{d}\mu(x)=\int_X \int_0^{\infty} 1_{r\leq|f(x)|}\textrm{d}r\textrm{d}\mu(x)\\ &=\int_0^{\infty} \int_X 1_{r\leq |f(x)|} \textrm{d}\mu(x) \textrm{d}r=\int_0^{\infty} \mu(A_r)\textrm{d}r \end{align}