Let \(f:\\RR^n\\to\\RR\) be measurable. Show that \[ \int_{\RR^n} \abs{ f} = \int_0^{\infty } m(A_t)\dt && A_t \da \ts{x\in \RR^n \st \abs{f(x)} > t} .\]
Solution
For every \(x\), the nonnegative function \(t\mapsto \mathbf{1}_{A_t}(x)\) is the indicator of \(0\leq t<\abs{f(x)}\). Therefore \[ \abs{f(x)} = \int_0^\infty \mathbf{1}_{A_t}(x)\,\dt. \] The integrand is measurable and nonnegative on \(\RR^n\times[0,\infty)\). Tonelli’s theorem then permits the order of integration to be exchanged: \[ \int_{\RR^n}\abs{f(x)}\,\dx = \int_{\RR^n}\int_0^\infty \mathbf{1}_{A_t}(x)\,\dt\,\dx = \int_0^\infty \int_{\RR^n}\mathbf{1}_{A_t}(x)\,\dx\,\dt = \int_0^\infty m(A_t)\,\dt. \] Both sides may be \(+\infty\); no integrability assumption is needed.