Carefully state Tonelli’s theorem for a nonnegative function \(F(x, t)\) on \(\RR^n\cross \RR\).
Let \(f:\RR^n\to [0, \infty]\) and define \[ \mca \definedas \theset{(x, t) \in \RR^n\cross \RR \suchthat 0\leq t \leq f(x)} .\]
Prove the validity of the following two statements:
- \(f\) is Lebesgue measurable on \(\RR^{n} \iff \mca\) is a Lebesgue measurable subset of \(\RR^{n+1}\).
- If \(f\) is Lebesgue measurable on \(\RR^n\) then \[ m(\mathcal{A})=\int_{\mathbb{R}^{n}} f(x) d x=\int_{0}^{\infty} m\left(\left\{x \in \mathbb{R}^{n}\suchthat f(x) \geq t\right\}\right) d t .\]
Solution
Let \(F\colon \RR^n \times \RR \to [0,\infty]\) be Lebesgue measurable on \(\RR^{n+1}\). Then for a.e. \(x \in \RR^n\) the slice \(t \mapsto F(x,t)\) is measurable on \(\RR\), the function \(x \mapsto \int_\RR F(x, t)\,dt\), defined a.e., is measurable on \(\RR^n\), and \[ \int_{\RR^{n+1}} F = \int_{\RR^n}\left(\int_\RR F(x, t)\,dt\right)dx, \] all integrals possibly infinite. The same holds with the roles of \(x\) and \(t\) exchanged.
1. If \(f\) is measurable, then \(\mca\) is measurable.
\(\mca = \theset{(x,t) : t \ge 0} \cap \theset{(x,t) : f(x) - t \ge 0}\), where \(f(x) - t\) is taken in \([-\infty, \infty]\). The function \((x,t) \mapsto f(x)\) is measurable on \(\RR^{n+1}\) because \(f\) is measurable (cylinder functions of measurable functions are measurable; see E-JJ746), and \((x,t) \mapsto t\) is continuous, so their difference is measurable.
2. If \(\mca\) is measurable, then \(f\) is measurable.
For every \(x\), the slice \(\theset{t : (x,t) \in \mca}\) is \([0, f(x)]\), which has length \(f(x)\). By part (a) applied to \(F = \chi_\mca\), the function \(x \mapsto m_1([0, f(x)]) = f(x)\) is measurable.
3. If \(f\) is measurable, then \(m(\mca) = \int_{\RR^n} f(x)\,dx = \int_0^\infty m\theset{x : f(x) \ge t}\,dt\).
By step 1, \(\chi_\mca\) is measurable. Part (a) with the \(t\)-integral inside gives \(m(\mca) = \int_{\RR^n}\int_\RR \chi_\mca(x,t)\,dt\,dx = \int_{\RR^n} f(x)\,dx\), since the slice at \(x\) has length \(f(x)\). Part (a) with the \(x\)-integral inside gives \(m(\mca) = \int_\RR\int_{\RR^n}\chi_\mca(x,t)\,dx\,dt\). For \(t < 0\) the slice \(\theset{x : (x,t) \in \mca}\) is empty, and for \(t \ge 0\) it is \(\theset{x : f(x) \ge t}\), so this equals \(\int_0^\infty m\theset{f \ge t}\,dt\).
4. Q.E.D.
Steps 1 and 2 are (b)1, and step 3 is (b)2.