Prove the absolute continuity of the Lebesgue integral; in other words, prove that if \(f\) is integrable on \(\mathbb{R}^d\), then for every \(\epsilon > 0\) there exists \(\delta > 0\) such that
\[\int_E |f| < \epsilon \quad \text{whenever} \quad m(E) < \delta.\]
Solution
1. Truncate the integrable function.
Fix \(\varepsilon>0\). Since \(f\in L^1(\mathbb R^d)\), \[ \int_{\{|f|>M\}}|f|\,dx\longrightarrow0 \qquad(M\to\infty). \] Choose \(M>0\) so large that \[ \int_{\{|f|>M\}}|f|<\frac\varepsilon2. \]
2. Control the integral on a small measurable set.
Let \[ \delta:=\frac{\varepsilon}{2M}. \] If \(m(E)<\delta\), split \[ E=E\cap\{|f|\le M\}\;\cup\;E\cap\{|f|>M\}. \] Then \[ \begin{aligned} \int_E|f| &\le \int_{E\cap\{|f|\le M\}}|f| +\int_{\{|f|>M\}}|f|\\ &\le M\,m(E)+\frac\varepsilon2\\ &<M\delta+\frac\varepsilon2 =\varepsilon. \end{aligned} \] Thus \[ \boxed{ m(E)<\delta\Longrightarrow\int_E|f|<\varepsilon.} \]