Absolute Continuity of Indefinite Integrals

Small Sets and Indefinite Integrals

Let \(f\) be integrable on a compact interval \([a,b]\). Its indefinite Lebesgue integral is the function \(F(x)=\int_a^x f(t)\,dt\). The key point is that the integral of \(|f|\) over a measurable set becomes small when the set has sufficiently small measure. This remains true even if \(f\) is unbounded; it is the estimate that turns small total interval length into small total change in \(F\).

Throughout, \([a,b]\) is equipped with the Lebesgue sigma-algebra and the restriction of Lebesgue measure \(m\). This is a complete finite measure space, with \(m([a,b])=b-a\). Unless \(a=b\), assume \(a<b\). The function \(f:[a,b]\to\mathbb{R}\) is measurable and satisfies \(\int_{[a,b]}|f|\,dm<\infty\), so \(f\in L^1([a,b])\). The integral and the resulting \(F\) are real-valued. Changing \(f\) on a null set does not change any of these integrals or \(F\).

The definition controls a sum of changes across any finite family of intervals, not just the change across one interval. The following estimate supplies precisely that stronger control.

Absolute Continuity of the Lebesgue Integral on Small Sets

Proof. Set \(g_n=\min(|f|,n)\) for each positive integer \(n\). These are nonnegative measurable functions, \(g_n\uparrow |f|\) everywhere, and \(\int |f|\,dm<\infty\). By the Monotone Convergence Theorem, \(\int g_n\,dm\uparrow\int|f|\,dm\). Since both limiting integrals are finite, it follows that \(\int(|f|-g_n)\,dm=\int|f|\,dm-\int g_n\,dm\to0\).

For every real \(y\ge0\), \(y\chi_{\{y>2n\}}\le2(y-\min(y,n))\): if \(y\le2n\), the left side is zero; if \(y>2n\), then \(2(y-n)\ge y\). Substituting \(y=|f(t)|\) and integrating gives \(\int_{\{|f|>2n\}}|f|\,dm\le2\int(|f|-g_n)\,dm\). Choose \(n\ge1\) large enough that this last bound is less than \(\varepsilon/2\), and set \(\delta=\varepsilon/(4n)\). For any measurable \(E\subseteq[a,b]\), split its integral into the parts where \(|f|\le2n\) and \(|f|>2n\). Then $$ \int_E|f|\,dm \le 2n\,m(E)+\int_{\{|f|>2n\}}|f|\,dm. $$ If \(m(E)<\delta\), the first term is less than \(2n\delta=\varepsilon/2\), and the second term is less than \(\varepsilon/2\). Thus \(\int_E|f|\,dm<\varepsilon\), as required. \(\square\)

The truncation argument is important: it first controls the part of \(f\) above a large threshold by its integrable tail, then controls the bounded part using the measure of \(E\). In particular, the lemma does not assume that \(f\) is bounded.

The Indefinite Integral Is Absolutely Continuous

Proof. If \(a=b\), the domain has one point and the defining interval-sum condition is satisfied for every \(\varepsilon>0\). Suppose \(a<b\), and fix \(\varepsilon>0\). Apply the lemma to \(|f|\), or equivalently to \(f\), to obtain \(\delta>0\) such that \(m(E)<\delta\) implies \(\int_E|f|\,dm<\varepsilon\) for every Lebesgue measurable \(E\subseteq[a,b]\).

Take any finite family of pairwise disjoint subintervals \([u_i,v_i]\subseteq[a,b]\) with \(\sum_i(v_i-u_i)<\delta\). Additivity of the Lebesgue integral on adjacent intervals gives \(F(v_i)-F(u_i)=\int_{[u_i,v_i]}f\,dm\); including or excluding endpoints does not change the integral because finite sets have measure zero. The integral triangle inequality therefore gives $$ \sum_i|F(v_i)-F(u_i)| \le \sum_i\int_{[u_i,v_i]}|f|\,dm =\int_E|f|\,dm, $$ where \(E\) is the union of these intervals, with any shared endpoints counted just once. Since the intervals have disjoint interiors, \(m(E)=\sum_i(v_i-u_i)<\delta\). The lemma now yields \(\int_E|f|\,dm<\varepsilon\). This is exactly the definition of absolute continuity of \(F\). \(\square\)

The proof uses no bound on the values of \(f\). Its essential input is integrability of \(|f|\), which ensures that sets of small measure carry little integral. As an immediate consequence of the definition, \(F\) is continuous: for a single interval \([u,v]\), choosing its length less than \(\delta\) gives \(|F(v)-F(u)|<\varepsilon\).

A Common Pitfall: Using a Supremum That Does Not Exist as a Finite Bound

For bounded \(f\), the estimate \(|F(v)-F(u)|\le\|f\|_\infty(v-u)\) is useful. But it cannot be applied to the unbounded integrand in the preceding example: there is no finite \(\|f\|_\infty\). A concrete incorrect calculation would treat the supremum as \(1\), conclude \(\int_0^{1/16}t^{-1/2}\,dt\le1/16\), and thereby claim that the integral over that interval is at most its length. In fact, $$ \int_0^{1/16}t^{-1/2}\,dt=2\sqrt{1/16}=\frac12, \qquad m([0,1/16])=\frac1{16}. $$ The proposed bound fails. The correction is not to force a supremum estimate, but to use the small-set lemma: for this integrable \(f\), sufficiently small measure guarantees a small integral, with a threshold chosen using the tail of \(f\). Thus the theorem covers unbounded \(L^1\) integrands, not merely bounded ones.