Absolute Continuity

F(x)=x2F(x) = x^2 is absolutely continuous — when the intervals' total length goes to 0, so does ∑∣ΔF∣\sum |\Delta F|. The Cantor function is uniformly continuous but not absolutely continuous: total length (23)n→0\left(\frac23\right)^n \to 0 while ∑∣Δc∣=1\sum |\Delta c| = 1 stays frozen. Absolutely continuous functions are exactly the indefinite integrals of Lebesgue-integrable functions.

F absolutely continuous   ⟺  ∀ε>0 ∃δ>0:∑(bi−ai)<δ⇒∑∣F(bi)−F(ai)∣<εF \text{ absolutely continuous } \iff \forall \varepsilon > 0\, \exists \delta > 0: \sum (b_i - a_i) < \delta \Rightarrow \sum |F(b_i) - F(a_i)| < \varepsilon
F(x)=F(a)+∫axf dx(f∈L1)  ⟺  F absolutely continuousF(x) = F(a) + \int_a^x f\,dx\quad (f \in L^1) \iff F \text{ absolutely continuous}

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Examples & Exercises

Let f∈L1[a,b]f \in L^1[a, b] and F(x)=F(a)+∫axfF(x) = F(a) + \int_a^x f. Prove FF is absolutely continuous. First prove the lemma: for every ε>0\varepsilon > 0 there is δ>0\delta > 0 such that m(E)<δm(E) < \delta implies ∫E∣f∣<ε\int_E |f| < \varepsilon.

∫E∣f∣≤ε2+M⋅m(E)<ε,∑i∣F(bi)−F(ai)∣≤∫⋃i[ai,bi]∣f∣<ε\int_E |f| \le \frac{\varepsilon}{2} + M \cdot m(E) < \varepsilon, \qquad \sum_{i} |F(b_i) - F(a_i)| \le \int_{\bigcup_i [a_i,b_i]} |f| < \varepsilon
F(x)=x2F(x) = x^2, δ=0.2\delta = 0.2: ∑∣ΔF∣=1−(1−0.2)2=0.36\sum |\Delta F| = 1 - (1 - 0.2)^2 = 0.36 — as the intervals' total length 0.20.2 shrinks, so does ∑∣ΔF∣\sum |\Delta F| (AC ✓).
Show proof
  1. Lemma (absolutely continuous integral): use that the tails of ∣f∣|f| vanish. By dominated convergence (from the Lebesgue scene), choose MM with ∫∣f∣>M∣f∣<ε2\int_{|f| > M} |f| < \frac{\varepsilon}{2}.
  2. For any measurable EE, splitting ∣f∣=∣f∣⋅1{∣f∣>M}+∣f∣⋅1{∣f∣≤M}|f| = |f| \cdot \mathbf{1}_{\{|f| > M\}} + |f| \cdot \mathbf{1}_{\{|f| \le M\}} gives ∫E∣f∣≤∫∣f∣>M∣f∣+M⋅m(E)<ε2+M⋅m(E)\int_E |f| \le \int_{|f| > M} |f| + M \cdot m(E) < \frac{\varepsilon}{2} + M \cdot m(E). So ∫E∣f∣<ε\int_E |f| < \varepsilon whenever m(E)<δ:=ε2Mm(E) < \delta := \frac{\varepsilon}{2M}.
  3. Now let the disjoint [ai,bi][a_i, b_i] have total length below δ\delta. On each interval ∣F(bi)−F(ai)∣=∣∫aibif∣≤∫aibi∣f∣|F(b_i) - F(a_i)| = \left|\int_{a_i}^{b_i} f\right| \le \int_{a_i}^{b_i} |f|, so summing gives ∑i∣F(bi)−F(ai)∣≤∑i∫aibi∣f∣=∫⋃i[ai,bi]∣f∣<ε\sum_i |F(b_i) - F(a_i)| \le \sum_i \int_{a_i}^{b_i} |f| = \int_{\bigcup_i [a_i,b_i]} |f| < \varepsilon (applying the lemma to E=⋃i[ai,bi]E = \bigcup_i [a_i,b_i]) — i.e. FF is absolutely continuous.
  4. Numerical check (F(x)=x2F(x) = x^2, δ=0.2\delta = 0.2): packing the intervals at [1−δ,1][1 - \delta, 1] gives ∑∣ΔF∣=1−(1−0.2)2=0.36\sum |\Delta F| = 1 - (1 - 0.2)^2 = 0.36 — matching the caption, and 2δ−δ2→02\delta - \delta^2 \to 0 as δ→0\delta \to 0 verifies the "absolute" condition.

Prove the Cantor function cc is not absolutely continuous: the 2n2^n stage-nn Cantor intervals have total length (23)n\left(\frac23\right)^n but ∑∣Δc∣=1\sum |\Delta c| = 1.

total length=2n⋅3−n=(23)n→0,∑∣Δc∣=2n⋅2−n=1≮12\text{total length} = 2^n \cdot 3^{-n} = \left(\tfrac23\right)^n \to 0, \qquad \sum |\Delta c| = 2^n \cdot 2^{-n} = 1 \not< \tfrac12
n=4n = 4: 242^4 intervals, total length (23)4=1681≈0.1975→0\left(\frac23\right)^4 = \frac{16}{81} \approx 0.1975 \to 0, yet ∑∣Δc∣=1\sum |\Delta c| = 1 — always 1, so not AC ✗.
Show proof
  1. Negate absolute continuity: we show that for ε=12\varepsilon = \frac12, no δ>0\delta > 0 can satisfy the condition. The stage-nn Cantor set consists of 2n2^n closed intervals of length 3−n3^{-n} (each interval is trisected and its middle removed).
  2. Their total length is 2n⋅3−n=(23)n→02^n \cdot 3^{-n} = \left(\frac23\right)^n \to 0, so for large nn the total length is below δ\delta. But the Cantor function cc rises by exactly 2−n2^{-n} on each stage-nn interval (fixing the next binary digit of its base-2 expansion).
  3. Hence ∑∣Δc∣=2n⋅2−n=1≮12=ε\sum |\Delta c| = 2^n \cdot 2^{-n} = 1 \not< \frac12 = \varepsilon — the sum stays frozen at 1 no matter how small the total length gets. This directly violates the definition of absolute continuity, so cc is not absolutely continuous. Since cc is uniformly continuous, uniform continuity does not imply absolute continuity.
  4. Numerical check (n=4n = 4): total length (23)4=1681≈0.1975\left(\frac23\right)^4 = \frac{16}{81} \approx 0.1975, ∑∣Δc∣=1\sum |\Delta c| = 1 — matching the caption. Even at n=10n = 10 the total length shrinks to ≈0.0173\approx 0.0173 while ∑∣Δc∣\sum |\Delta c| stays 1.

Prove that F:[a,b]→RF:[a, b] \to \mathbb{R} with Lipschitz constant LL is absolutely continuous (use δ=εL\delta = \frac{\varepsilon}{L}).

∑i∣F(bi)−F(ai)∣≤L∑i(bi−ai)<L⋅εL=ε\sum_i |F(b_i) - F(a_i)| \le L \sum_i (b_i - a_i) < L \cdot \frac{\varepsilon}{L} = \varepsilon
F(x)=x2F(x) = x^2 is L=2L = 2-Lipschitz on [0,1][0, 1]: at δ=0.2\delta = 0.2, ∑∣ΔF∣=0.36<Lδ=0.4\sum |\Delta F| = 0.36 < L\delta = 0.4 — the Lipschitz bound ∣F(b)−F(a)∣≤L(b−a)|F(b) - F(a)| \le L(b - a) directly yields absolute continuity.
Show proof
  1. Given ε>0\varepsilon > 0, set δ=εL\delta = \frac{\varepsilon}{L} (if L=0L = 0 then FF is constant and there is nothing to prove). Take disjoint [ai,bi][a_i, b_i] with total length below δ\delta.
  2. On each interval ∣F(bi)−F(ai)∣≤L(bi−ai)|F(b_i) - F(a_i)| \le L(b_i - a_i), so summing gives ∑i∣F(bi)−F(ai)∣≤L∑i(bi−ai)<Lδ=ε\sum_i |F(b_i) - F(a_i)| \le L \sum_i (b_i - a_i) < L\delta = \varepsilon — exactly the definition of absolute continuity.
  3. Numerical check (F(x)=x2F(x) = x^2, L=2L = 2 on [0,1][0, 1]): at δ=0.2\delta = 0.2, ∑∣ΔF∣=1−(1−0.2)2=0.36<Lδ=0.4\sum |\Delta F| = 1 - (1 - 0.2)^2 = 0.36 < L\delta = 0.4 — matching the caption; the Lipschitz constant directly supplies the δ\delta of absolute continuity.

10. Integration