Absolute Continuity of Lebesgue Integral

The following is a wonderful property of the Lebesgue Integral, also known as absolute continuity of Lebesgue Integral. Basically, it means that whenever the domain of integration has small enough measure, then the integral will be arbitrarily small.

Suppose $f$ is integrable.
Given $\epsilon>0$ , there exists $\delta>0$ such that for all measurable sets $B\subseteq E$ with $|B|<\delta$ , $|\int_B f\,dx|<\epsilon$ .

Proof:
Define $A_k=\{x\in E: \frac 1k\leq|f(x)|<k\}$ for $k\in\mathbb{N}$ . Each $A_k$ is measurable and $A_k\nearrow A:=\bigcup_{k=1}^\infty A_k$ . Note that $\displaystyle \int_E |f|=\int_{\{f=0\}}|f|+\int_A |f|+\int_{\{f=\infty\}}|f|=\int_A |f|.$

Let $f_k=|f|\chi_{A_k}$ . Then $\{f_k\}$ is a sequence of non-negative functions such that $f_k\nearrow |f|\chi_A$ . By Monotone Convergence Theorem, $\lim_{k\to\infty}\int_E f_k=\int_E |f|\chi_A$ , that is, $\displaystyle \lim_{k\to\infty}\int_{A_k}|f|\,dx=\int_A |f|\,dx=\int_E |f|\,dx.$

Let $N>0$ be sufficiently large such that $\int_{E\setminus A_N}|f|\,dx<\epsilon/2$ .

Let $\delta=\frac{\epsilon}{2N}$ , and suppose $|B|<\delta$ . Then
$\begin{aligned} |\int_B f\,dx|&\leq\int_B |f|\,dx\\ &=\int_{(E\setminus A_N)\cap B}|f|\,dx+\int_{A_N\cap B}|f|\,dx\\ &\leq\int_{E\setminus A_N}|f|\,dx+\int_{A_N\cap B}N\,dx\\ &<\epsilon/2+N\cdot|A_N\cap B|\\ &\leq\epsilon/2+N\cdot|B|\\ &<\epsilon/2+N\cdot\frac{\epsilon}{2N}\\ &=\epsilon. \end{aligned}$