Let \(f:\mathbb R^n\to[0,\infty]\) and define \[ \mathcal A:=\{(x,t)\in\mathbb R^n\times\mathbb R:0\le t\le f(x)\}. \]
Prove that \(f\) is Lebesgue measurable if and only if \(\mathcal A\) is Lebesgue measurable in \(\mathbb R^{n+1}\).
If \(f\) is measurable, prove \[ m_{n+1}(\mathcal A) =\int_{\mathbb R^n}f(x)\,dx =\int_0^\infty m_n(\{x:f(x)\ge t\})\,dt. \]
Solution
1. If \(f\) is measurable, then \(\mathcal A\) is measurable.
Define \[ H(x,t):=f(x)-t. \] The map \((x,t)\mapsto f(x)\) is measurable, as is \((x,t)\mapsto t\), so \(H\) is measurable. Therefore \[ \mathcal A =(\mathbb R^n\times[0,\infty))\cap H^{-1}([0,\infty]) \] is Lebesgue measurable.
2. If \(\mathcal A\) is measurable, then \(f\) is measurable.
For every fixed \(x\in\mathbb R^n\), the vertical section is \[ \mathcal A_x :=\{t:(x,t)\in\mathcal A\} =[0,f(x)]. \] Hence \[ f(x)=m_1(\mathcal A_x) =\int_{\mathbb R}\mathbf1_{\mathcal A}(x,t)\,dt. \] Since \(\mathbf1_{\mathcal A}\) is a nonnegative measurable function on \(\mathbb R^{n+1}\), Tonelli’s theorem implies that the function \[ x\longmapsto \int_{\mathbb R}\mathbf1_{\mathcal A}(x,t)\,dt \] is measurable. Thus \(f\) is measurable.
3. Compute the measure of the subgraph.
Assume \(f\) is measurable. By Step 1, \(\mathcal A\) is measurable, so Tonelli gives \[ \begin{aligned} m_{n+1}(\mathcal A) &=\int_{\mathbb R^{n+1}}\mathbf1_{\mathcal A}(x,t)\,d(x,t)\\ &=\int_{\mathbb R^n}\left(\int_{\mathbb R}\mathbf1_{\mathcal A}(x,t)\,dt\right)dx\\ &=\int_{\mathbb R^n}f(x)\,dx. \end{aligned} \]
4. Derive the layer-cake formula.
For \(t\ge0\), the horizontal section is \[ \mathcal A^t :=\{x:(x,t)\in\mathcal A\} =\{x:f(x)\ge t\}. \] A second application of Tonelli gives \[ \begin{aligned} m_{n+1}(\mathcal A) &=\int_0^\infty\left(\int_{\mathbb R^n}\mathbf1_{\mathcal A}(x,t)\,dx\right)dt\\ &=\int_0^\infty m_n(\{x:f(x)\ge t\})\,dt. \end{aligned} \] Combining with Step 3 yields \[ \boxed{ m_{n+1}(\mathcal A) =\int_{\mathbb R^n}f(x)\,dx =\int_0^\infty m_n(\{f\ge t\})\,dt.} \]