Let \(f:\mathbb{R}\to\mathbb{R}\) be a nonnegative integrable function.
Show that \(\sin\circ f\) is integrable.
Use Fubini’s theorem to show that \[ \int_{[0,\infty)} m(\{x:f(x)\ge y\})\cos y\,dy=\int_{\mathbb{R}}\sin\bigl(f(x)\bigr)\,dx. \]
Solution
1. Prove that \(\sin\circ f\) is integrable.
Since \(f\ge0\) and \[ |\sin t|\le t \qquad(t\ge0), \] we have \[ |\sin(f(x))|\le f(x) \] for almost every \(x\). Because \(f\in L^1(\mathbb R)\), \[ \int_{\mathbb R}|\sin(f(x))|\,dx \le \int_{\mathbb R}f(x)\,dx<\infty. \] Hence \(\sin\circ f\in L^1(\mathbb R)\).
2. Set up an absolutely integrable function on the product space.
Define \[ H(x,y):=\mathbf1_{\{0\le y\le f(x)\}}\cos y, \qquad (x,y)\in\mathbb R\times[0,\infty). \] This function is measurable. Moreover, \[ |H(x,y)|\le \mathbf1_{\{0\le y\le f(x)\}}. \] By Tonelli’s theorem, \[ \begin{aligned} \int_{\mathbb R}\int_0^\infty |H(x,y)|\,dy\,dx &\le \int_{\mathbb R}\int_0^{f(x)}1\,dy\,dx\\ &= \int_{\mathbb R}f(x)\,dx<\infty. \end{aligned} \] Thus \(H\in L^1(\mathbb R\times[0,\infty))\), so Fubini’s theorem applies.
3. Integrate in the two orders.
Integrating first in \(y\) gives \[ \int_0^\infty H(x,y)\,dy = \int_0^{f(x)}\cos y\,dy = \sin(f(x)). \] Hence \[ \iint H(x,y)\,dy\,dx = \int_{\mathbb R}\sin(f(x))\,dx. \]
On the other hand, for fixed \(y\ge0\), \[ \int_{\mathbb R}H(x,y)\,dx = \cos y\,m(\{x:f(x)\ge y\}). \] Therefore Fubini yields \[ \boxed{ \int_0^\infty m(\{x:f(x)\ge y\})\cos y\,dy = \int_{\mathbb R}\sin(f(x))\,dx.} \]