Suppose that \(f\in L^1(\mathbb R)\). Prove that for every \(\epsilon>0\) there exists \(\delta>0\) such that \[ \int_A |f|<\epsilon \] for every measurable set \(A\) with \(m(A)<\delta\).
Solution
Fix \(\epsilon>0\). Since \(|f|\in L^1(\mathbb R)\), the truncations \[ |f|\mathbf 1_{\{|f|\le M\}} \] increase pointwise to \(|f|\) as \(M\to\infty\). Monotone convergence gives \(\int|f|\mathbf 1_{\{|f|\le M\}}\to\int|f|<\infty\), hence \[ \int_{\{|f|>M\}}|f|\longrightarrow0. \] Choose \(M>0\) so large that \[ \int_{\{|f|>M\}}|f|<\frac\epsilon2. \] Set \[ \delta=\frac{\epsilon}{2M}. \] If \(A\) is measurable and \(m(A)<\delta\), then \[ \begin{aligned} \int_A|f| &=\int_{A\cap\{|f|\le M\}}|f|+\int_{A\cap\{|f|>M\}}|f|\\ &\le M m(A)+\int_{\{|f|>M\}}|f|\\ &<M\delta+\frac\epsilon2 =\epsilon. \end{aligned} \] Thus the Lebesgue integral of an \(L^1\) function is absolutely continuous with respect to Lebesgue measure in the stated sense.