Let \(f : [0,1] \to [0,1]\) be a continuous function and \(\phi : \mathbb{R} \to \mathbb{R}\) be a \(C^1\) function with \(\phi(0) = 0\). Prove
\[ \int_0^1 \phi(f(x))\,dx =\int_0^\infty m(\{x\in[0,1]:f(x)>t\})\phi'(t)\,dt. \]
Solution
1. Write the weighted value pointwise as an integral over its level parameter.
For every \(x\in[0,1]\), the Fundamental Theorem of Calculus and \(\phi(0)=0\) give \[ \phi(f(x)) =\int_0^{f(x)}\phi'(t)\,dt =\int_0^\infty \mathbf1_{\{t<f(x)\}}\phi'(t)\,dt. \] Because \(0\le f(x)\le1\), the integrand vanishes for \(t>1\).
2. Justify interchanging the integrals.
Since \(\phi'\) is continuous, it is bounded on \([0,1]\). Thus \[ \int_0^1\int_0^\infty \left|\mathbf1_{\{t<f(x)\}}\phi'(t)\right|\,dt\,dx \le \|\phi'\|_{L^\infty([0,1])}<\infty. \] Therefore Fubini’s theorem applies. Hence \[ \begin{aligned} \int_0^1\phi(f(x))\,dx &=\int_0^1\int_0^\infty \mathbf1_{\{t<f(x)\}}\phi'(t)\,dt\,dx\\ &=\int_0^\infty \left(\int_0^1\mathbf1_{\{f(x)>t\}}\,dx\right) \phi'(t)\,dt\\ &=\int_0^\infty m(\{x\in[0,1]:f(x)>t\})\phi'(t)\,dt. \end{aligned} \] This proves the stated identity.