Let \(f\in L^1(\mathbb R^n)\). Prove that for every \(\varepsilon>0\) there exists \(\delta>0\) such that if \(A\) is measurable and \(|A|<\delta\), then \[ \left|\int_A f\right|<\varepsilon. \]
Solution
Fix \(\varepsilon>0\). Since \(f\in L^1\), choose \(M>0\) so that \[ \int_{\{|f|>M\}}|f|<\frac{\varepsilon}{2}. \] Set \[ \delta=\frac{\varepsilon}{2M}. \] If \(|A|<\delta\), then \[ \begin{aligned} \left|\int_A f\right| &\le\int_A|f|\\ &\le\int_{A\cap\{|f|\le M\}}|f|+\int_{\{|f|>M\}}|f|\\ &\le M|A|+\frac{\varepsilon}{2}\\ &<\varepsilon. \end{aligned} \] Thus the Lebesgue integral is absolutely continuous with respect to Lebesgue measure.