Absolute Continuity

Lemma (absolutely continuous integral): apply dominated convergence to the truncations $|f| \cdot \mathbf{1}_{\{|f| \le M\}}$. They converge pointwise to $|f|$ as $M \to \infty$, with dominating function $|f| \in L^1$ (the hypothesis).

Hence $\int |f| \cdot \mathbf{1}_{\{|f| \le M\}} \to \int |f|$, i.e. the tail $\int_{|f| > M} |f| \to 0$ as $M \to \infty$.

So some $M$ has $\int_{|f| > M} |f| < \frac{\varepsilon}{2}$, and enlarging $M$ only shrinks the tail, so take $M \ge 1$.

For any measurable $E$, splitting $|f| = |f| \cdot \mathbf{1}_{\{|f| > M\}} + |f| \cdot \mathbf{1}_{\{|f| \le M\}}$ gives $\int_E |f| \le \int_{|f| > M} |f| + M \cdot m(E) < \frac{\varepsilon}{2} + M \cdot m(E)$. Since $M \ge 1 > 0$ above, $\delta := \frac{\varepsilon}{2M}$ is well defined, and $\int_E |f| < \varepsilon$ whenever $m(E) < \delta$.

Now let the disjoint $[a_i, b_i]$ have total length below $\delta$. On each interval $|F(b_i) - F(a_i)| = \left|\int_{a_i}^{b_i} f\right| \le \int_{a_i}^{b_i} |f|$, so summing gives $\sum_i |F(b_i) - F(a_i)| \le \sum_i \int_{a_i}^{b_i} |f| = \int_{\bigcup_i [a_i,b_i]} |f| < \varepsilon$ (applying the lemma to $E = \bigcup_i [a_i,b_i]$) — i.e. $F$ is absolutely continuous.

Numerical check ($F(x) = x^2$, $\delta = 0.2$): packing the intervals at $[1 - \delta, 1]$ gives $\sum |\Delta F| = 1 - (1 - 0.2)^2 = 0.36$ — matching the caption, and $2\delta - \delta^2 \to 0$ as $\delta \to 0$ verifies the "absolute" condition.

Negate absolute continuity: we show that for $\varepsilon = \frac12$, no $\delta > 0$ can satisfy the condition. The stage-$n$ Cantor set consists of $2^n$ closed intervals of length $3^{-n}$ (each interval is trisected and its middle removed).

Their total length is $2^n \cdot 3^{-n} = \left(\frac23\right)^n \to 0$, so for large $n$ the total length is below $\delta$. But the Cantor function $c$ rises by exactly $2^{-n}$ on each stage-$n$ interval (fixing the next binary digit of its base-2 expansion).

Hence $\sum |\Delta c| = 2^n \cdot 2^{-n} = 1 \not< \frac12 = \varepsilon$ — the sum stays frozen at 1 no matter how small the total length gets.

This directly violates the definition of absolute continuity, so $c$ is not absolutely continuous.

Since $c$ is uniformly continuous, uniform continuity does not imply absolute continuity.

Numerical check ($n = 4$): total length $\left(\frac23\right)^4 = \frac{16}{81} \approx 0.1975$, $\sum |\Delta c| = 1$ — matching the caption. Even at $n = 10$ the total length shrinks to $\approx 0.0173$ while $\sum |\Delta c|$ stays 1.

Given $\varepsilon > 0$, set $\delta = \frac{\varepsilon}{L}$ (if $L = 0$ then $F$ is constant and there is nothing to prove). Take disjoint $[a_i, b_i]$ with total length below $\delta$.

On each interval $|F(b_i) - F(a_i)| \le L(b_i - a_i)$, so summing gives $\sum_i |F(b_i) - F(a_i)| \le L \sum_i (b_i - a_i) < L\delta = \varepsilon$ — exactly the definition of absolute continuity.

Numerical check ($F(x) = x^2$, $L = 2$ on $[0, 1]$): at $\delta = 0.2$, $\sum |\Delta F| = 1 - (1 - 0.2)^2 = 0.36 < L\delta = 0.4$ — matching the caption; the Lipschitz constant directly supplies the $\delta$ of absolute continuity.