Power Functions and Piecewise Examples of Absolute Continuity

Testing Absolute Continuity with Increments

Absolute continuity controls the total change of a function across many short, disjoint intervals at once. This is stronger than continuity: continuity controls the change on each short interval separately, while absolute continuity requires a uniform bound on the sum of all those changes. For power functions, this condition remains valid even when the derivative becomes unbounded at an endpoint. For piecewise functions, it provides a practical way to check whether the pieces fit together without a jump.

Throughout, $m$ denotes Lebesgue measure on $[0,1]$, and functions are real-valued. All absolute-continuity statements below are pointwise statements on the whole interval $[0,1]$, not merely almost-everywhere statements.

Definition: A function $f:[0,1]\to\mathbb{R}$ is absolutely continuous if, for every $\varepsilon>0$, there is a $\delta>0$ such that every finite collection of intervals $[a_k,b_k]\subseteq[0,1]$ with pairwise disjoint interiors and $\sum_k(b_k-a_k)<\delta$ satisfies $\sum_k|f(b_k)-f(a_k)|<\varepsilon$. Intervals of zero length contribute zero.

We will use a sufficient test that translates this definition into an integral bound. It is deliberately a sufficient condition: this tutorial does not assert that every absolutely continuous function must satisfy the particular pointwise increment bound in the test.

Proposition: Suppose $f:[0,1]\to\mathbb{R}$, and $g:[0,1]\to[0,\infty]$ is measurable with $\int_0^1g\,dm<\infty$. If, for every $0\le u\le v\le1$, $|f(v)-f(u)|\le\int_u^v g\,dm$, then $f$ is absolutely continuous on $[0,1]$.

Proof. Fix $\varepsilon>0$. For each positive integer $j$, let $g_j=\min(g,j)$. Then $g_j$ is measurable, $0\le g_j\uparrow g$, and $\int g<\infty$. The Monotone Convergence Theorem therefore gives $\int g_j\to\int g$. By linearity of the integral for integrable functions, $\int(g-g_j)=\int g-\int g_j\to0$. Choose $j$ large enough that $\int(g-g_j)\,dm<\varepsilon/2$, and set $K=j$, so $0\le g_j\le K$. Choose $\delta=\varepsilon/(2K)$ if $K>0$; if $K=0$, then $g=0$ almost everywhere by the tail bound and the claim follows directly from the increment hypothesis. For any finite collection of intervals as in the definition, let $E$ be their union. Their interiors are disjoint, so $m(E)=\sum_k(b_k-a_k)<\delta$; shared endpoints have measure zero. The assumed increment bound and additivity of the integral give $$ \sum_k|f(b_k)-f(a_k)| \le \int_E g\,dm =\int_E g_j\,dm+\int_E(g-g_j)\,dm \le K m(E)+\int(g-g_j)\,dm <\frac{\varepsilon}{2}+\frac{\varepsilon}{2} =\varepsilon. $$ This is the definition of absolute continuity. $\square$

Power Functions on the Unit Interval

Theorem: For every $\alpha>0$, the function $f(x)=x^\alpha$ is absolutely continuous on $[0,1]$. Its derivative on $(0,1]$ is $g(x)=\alpha x^{\alpha-1}$, and this nonnegative function is integrable on $[0,1]$, with $\int_0^1g\,dm=1$.

Proof. Define $g(0)=0$; this choice at one point does not affect measurability or the integral. For $0<u<v\le1$, elementary calculus gives $$ \int_u^v \alpha x^{\alpha-1}\,dx=v^\alpha-u^\alpha. $$ If $u=0$, take the limit as $u\downarrow0$: since $u^\alpha\to0$, the improper integral is $\int_0^v g\,dm=v^\alpha$. In particular, $\int_0^1g\,dm=1$, so $g$ is integrable. The displayed identity gives $|f(v)-f(u)|=\int_u^v g\,dm$ for all $0\le u\le v\le1$, because $x^\alpha$ is increasing. The proposition now implies absolute continuity. $\square$

The derivative behaves differently across the parameter ranges, but the conclusion does not change. When $0<\alpha<1$, $g$ is unbounded near zero; when $\alpha=1$, it is constant; and when $\alpha>1$, it extends continuously to zero. Integrability, rather than boundedness of the derivative, is what makes the increment test work.

Worked Example: The Square-Root Function

For $f(x)=\sqrt{x}$, take $\alpha=1/2$. On $(0,1]$, $g(x)=1/(2\sqrt{x})$, and $\int_0^1g\,dm=1$. To see directly how the bound handles the singularity, let $E\subseteq[0,1]$ be a finite union of intervals with $m(E)<\delta$, where $0<\delta\le1$. On $[0,\delta]$, the integral of $g$ is $\sqrt{\delta}$. On $[\delta,1]$, $g(x)\le1/(2\sqrt{\delta})$. Thus

\int_Eg\,dm \le \int_0^\delta g\,dm+\frac{m(E)}{2\sqrt{\delta}} <\sqrt{\delta}+\frac{\delta}{2\sqrt{\delta}} =\frac{3}{2}\sqrt{\delta}.

For $\varepsilon=1/10$, choose $\delta=1/400$. The bound is $\frac32\sqrt{1/400}=\frac32\cdot\frac1{20}=\frac3{40}<\frac1{10}$. The proposition therefore gives absolute continuity. Check: on the interval $[1/4,1]$, the endpoint increment is $1-1/2=1/2$, and the derivative integral is $\int_{1/4}^1(2\sqrt{x})^{-1}\,dx=1-\frac12=\frac12$, as required.

Worked Example: A Power with Bounded Derivative

Take $f(x)=x^{3/2}$. Its derivative $g(x)=\frac32\sqrt{x}$ on $[0,1]$ satisfies $0\le g(x)\le3/2$. For any finite family of intervals with total length less than $\delta$, the increment identity gives

\sum_k|f(b_k)-f(a_k)| =\sum_k\int_{a_k}^{b_k}\frac32\sqrt{x}\,dx \le\frac32\sum_k(b_k-a_k) <\frac32\delta.

For $\varepsilon=1/10$, choose $\delta=1/20$; then the total increment is less than $\frac32\cdot\frac1{20}=\frac3{40}<\frac1{10}$. This is also consistent with the direct endpoint check: $\int_0^1\frac32\sqrt{x}\,dx=[x^{3/2}]_0^1=1$, equal to $f(1)-f(0)=1$. Here a bounded derivative makes the estimate especially simple, but the square-root example shows that boundedness is not required.

A Continuous Piecewise Example

For a piecewise-defined function, first check that the formulas agree at every join. Then seek a bound on increments on each piece. A derivative that is integrable on each piece can provide such a bound, provided the function is continuous at the joins and the increment identities hold up to the piece endpoints.

Worked Example: A Square Root Followed by a Line

Define $f:[0,1]\to\mathbb{R}$ by

f(x)= \begin{cases} \sqrt{x},&0\le x\le \frac14,\\[2pt] \frac12+\frac23\left(x-\frac14\right),&\frac14<x\le1. \end{cases}

At the join, the first formula gives $\sqrt{1/4}=1/2$, and the second formula also gives $1/2+\frac23(1/4-1/4)=1/2$. Thus there is no jump. The function is increasing. On the first piece, its increments are integrals of $g(x)=1/(2\sqrt{x})$; on the second, they are integrals of $g(x)=2/3$. Set $g(1/4)=0$, which does not affect any integral. The integral over the whole interval is finite:

\int_0^1g\,dm =\int_0^{1/4}\frac{1}{2\sqrt{x}}\,dx+\int_{1/4}^1\frac23\,dx =\frac12+\frac23\cdot\frac34 =\frac12+\frac12=1.

For $u\le v$ lying on one piece, the relevant derivative integral equals $f(v)-f(u)$. If $u<1/4<v$, split at $1/4$; continuity there gives $f(v)-f(u)=[f(1/4)-f(u)]+[f(v)-f(1/4)]=\int_u^{1/4}g\,dm+\int_{1/4}^vg\,dm=\int_u^vg\,dm$. Thus the increment hypothesis of the proposition holds for every interval.

For a direct small-set check, let $E$ be a finite union of intervals with $m(E)<\delta$, and choose $0<t\le1/4$. On $[0,t]$, the integral of the first-piece derivative is $\sqrt{t}$; on $[t,1/4]$, it is bounded by $1/(2\sqrt{t})$; on the linear piece it is $2/3$. Therefore $\int_Eg\,dm\le\sqrt{t}+\delta/(2\sqrt{t})+2\delta/3$. Given $\varepsilon>0$, choose $t\le1/4$ with $\sqrt{t}<\varepsilon/3$, then choose $\delta>0$ with $\delta/(2\sqrt{t})<\varepsilon/3$ and $2\delta/3<\varepsilon/3$. The total increment is then less than $\varepsilon$, so $f$ is absolutely continuous. Check: $f(1)=1/2+(2/3)(3/4)=1$, and $f(1)-f(0)=1$, matching the computed integral of $g$.

Why an Integrable Derivative Is Not Enough by Itself

Worked Example: A Jump at the Join

Define $h(x)=x$ for $0\le x<1/2$, and $h(x)=x+1$ for $1/2\le x\le1$. Away from $x=1/2$, the derivative exists and equals $1$, so the derivative exists almost everywhere and is integrable. Nevertheless, $h$ is not absolutely continuous.

Indeed, for $0<s<1/2$, the single interval $[1/2-s,1/2+s]$ has length $2s$, while its endpoint increment is $$ |h(1/2+s)-h(1/2-s)| =|(3/2+s)-(1/2-s)| =1+2s. $$ Given any proposed $\delta>0$, choose $s>0$ with $s<\min(1/2,\delta/2)$. The interval length is then less than $\delta$, but its increment is greater than $1$. Taking $\varepsilon=1$ disproves the definition of absolute continuity. The error in using the almost-everywhere derivative alone is that it misses the jump: the derivative integral across this interval is $2s$, not the endpoint increment $1+2s$.

These examples give a reliable procedure for this class of problems:

Check that the piecewise formulas agree at each join; a jump immediately rules out absolute continuity.
Find an integrable nonnegative bound whose integral controls each interval increment.
For finitely many disjoint intervals, add the bounds over their union, rather than trying to control each interval separately.
When a derivative is unbounded, estimate its integral near the singular endpoint instead of requiring a uniform derivative bound.

Check Your Understanding

Use the interval-sum definition and the increment estimates from this tutorial.

For $f(x)=x^{1/3}$, find an integrable derivative on $(0,1]$ and verify that its integral over $[0,1]$ is $1$.
For $f(x)=x^2$, use a uniform derivative bound to give a $\delta$ that works for $\varepsilon=1/10$.
In the piecewise square-root example, verify the increment identity for an interval whose left endpoint is $1/8$ and right endpoint is $1/2$, splitting at $1/4$.
Explain why the jump example is not absolutely continuous even though its derivative equals $1$ almost everywhere.
State the increment-bound proposition and identify where integrability of $g$ enters its proof.