Let \(f:\mathbb R\to[0,\infty)\) be measurable, and let \[ \varphi:[0,\infty)\to[0,\infty) \] be nondecreasing and absolutely continuous on \([0,T]\) for every finite \(T\). Assume \(\varphi(0)=0\). Prove that \[ \int_{\mathbb R}\varphi(f(x))\,dx = \int_0^\infty m(\{x:f(x)>t\})\varphi'(t)\,dt. \]
Solution
Since \(\varphi\) is nondecreasing and absolutely continuous on every compact interval, one has \[ \varphi'(t)\ge0 \qquad\text{for almost every }t\ge0, \] and for every \(s\ge0\), \[ \varphi(s)-\varphi(0)=\int_0^s\varphi'(t)\,dt. \] Because \(\varphi(0)=0\), \[ \varphi(s)=\int_0^\infty \mathbf1_{\{t<s\}}\varphi'(t)\,dt. \] Substituting \(s=f(x)\) gives \[ \varphi(f(x)) = \int_0^\infty \mathbf1_{\{t<f(x)\}}\varphi'(t)\,dt. \] The integrand is nonnegative and measurable on \(\mathbb R\times[0,\infty)\), so Tonelli’s theorem yields \[ \begin{aligned} \int_{\mathbb R}\varphi(f(x))\,dx &=\int_{\mathbb R}\int_0^\infty \mathbf1_{\{t<f(x)\}}\varphi'(t)\,dt\,dx\\ &=\int_0^\infty\int_{\mathbb R} \mathbf1_{\{f(x)>t\}}\,dx\,\varphi'(t)\,dt\\ &=\int_0^\infty m(\{x:f(x)>t\})\varphi'(t)\,dt. \end{aligned} \] Both sides may be \(+\infty\); Tonelli’s theorem gives the equality in \([0,\infty]\).