Absolute continuity of the Lebesgue integral

Remarks

Proof idea: let $f_N = \min(\abs{f}, N)$. By the monotone convergence theorem, $\int (\abs{f} - f_N) \, d\mu \to 0$, so one can fix $N$ with $\int (\abs{f} - f_N) < \varepsilon / 2$. Then $\int_A \abs{f} \le \varepsilon / 2 + N \mu(A)$, and $\delta = \varepsilon / (2N)$ works.

Example: on $[a, b]$ with Lebesgue measure, the function $F(x) = \int_a^x f(t) \, dt$ is absolutely continuous. Indeed, a finite union of disjoint intervals of total length less than $\delta$ has measure less than $\delta$.

Integrability is needed. For $f(x) = 1/x$ on $(0, 1]$ with Lebesgue measure, the set $(0, \delta / 2)$ has measure less than $\delta$ but $\int_0^{\delta/2} dx / x = \infty$.

Going deeper

For a student

In the language of measures, $\nu(A) = \int_A \abs{f} \, d\mu$ defines a finite measure that is absolutely continuous with respect to $\mu$, in the $\varepsilon$-$\delta$ sense. The Radon–Nikodym theorem gives a converse: every finite measure absolutely continuous with respect to a σ-finite measure $\mu$ has this form. The uniform version of this property, for a whole family of functions, is part of uniform integrability.