Suppose \(f\in L^1(\mathbb R^n,m)\) satisfies \[ \left|\int_E f\,dm\right|\le m(E) \] for every Lebesgue-measurable set \(E\). Prove that \[ |f|\le1 \] almost everywhere.
Solution
Define the finite signed or complex measure \[ \nu(E)=\int_E f\,dm. \] The hypothesis says \[ |\nu(E)|\le m(E) \] for every measurable set \(E\).
Let \(A\) be measurable and let \[ A=\bigsqcup_{j=1}^N A_j \] be any finite measurable partition. Then \[ \sum_{j=1}^N|\nu(A_j)| \le\sum_{j=1}^N m(A_j) =m(A). \] Taking the supremum over all finite measurable partitions of \(A\) gives \[ |\nu|(A)\le m(A), \] where \(|\nu|\) is the total variation measure of \(\nu\).
Since \(\nu\) has density \(f\) with respect to Lebesgue measure, its total variation has density \(|f|\): \[ |\nu|(A)=\int_A|f|\,dm. \] Therefore \[ \int_A|f|\,dm\le m(A) \] for every measurable \(A\).
Fix \(\varepsilon>0\) and let \[ A_\varepsilon=\{x:|f(x)|>1+\varepsilon\}. \] Then \[ (1+\varepsilon)m(A_\varepsilon) <\int_{A_\varepsilon}|f|\,dm \le m(A_\varepsilon). \] Hence \[ m(A_\varepsilon)=0. \] Taking the union over positive rational \(\varepsilon\) gives \[ m\{x:|f(x)|>1\}=0. \] Thus \(|f|\le1\) almost everywhere.