Let \(f \in L^1(\mathbb{R})\) and define \(E_\alpha = \{x : |f(x)| > \alpha\}\). Prove that
\[\int_{\mathbb{R}} |f(x)| \, dx = \int_0^\infty m(E_\alpha) \, d\alpha.\]
Solution
For every \(x\in\mathbb R\), \[ |f(x)|=\int_0^\infty \mathbf1_{\{\alpha<|f(x)|\}}\,d\alpha. \] The integrand \[ (x,\alpha)\longmapsto \mathbf1_{\{\alpha<|f(x)|\}} \] is nonnegative and measurable on \(\mathbb R\times(0,\infty)\). Tonelli’s theorem therefore gives \[ \begin{aligned} \int_{\mathbb R}|f(x)|\,dx &=\int_{\mathbb R}\int_0^\infty \mathbf1_{\{\alpha<|f(x)|\}}\,d\alpha\,dx\\ &=\int_0^\infty\int_{\mathbb R} \mathbf1_{\{|f(x)|>\alpha\}}\,dx\,d\alpha\\ &=\int_0^\infty m(E_\alpha)\,d\alpha. \end{aligned} \] Thus \[ \boxed{\int_{\mathbb R}|f|=\int_0^\infty m(E_\alpha)\,d\alpha.} \]