Absolute Continuity on a Compact Interval

Controlling Sums of Increments

Absolute continuity is a condition on how a function responds to a whole finite family of short intervals. The key quantity is not the size of the increment on one interval in isolation, but the sum of the absolute increments over all the intervals in the family. The definition therefore gives a uniform control: after choosing an output tolerance, one chooses a bound on the family’s total length that works regardless of how many intervals are used or where they lie.

Fix a compact interval $[a,b]$, where $a\le b$, and let $f:[a,b]\to\mathbb{R}$ be real-valued. A finite family of subintervals $[u_j,v_j]$, $j=1,\ldots,N$, is called non-overlapping here when $a\le u_j\le v_j\le b$ and their interiors are pairwise disjoint. Intervals may share endpoints. An interval with $u_j=v_j$ has length zero and contributes no increment. No measure-theoretic integral is involved in the definition; the lengths below are ordinary interval lengths.

Definition: The function $f:[a,b]\to\mathbb{R}$ is absolutely continuous on $[a,b]$, written $f\in AC[a,b]$, if for every $\varepsilon>0$ there exists $\delta>0$ such that, for every positive integer $N$ and every finite family of non-overlapping subintervals $[u_j,v_j]\subseteq[a,b]$, $$ \sum_{j=1}^{N}(v_j-u_j)<\delta \quad\Longrightarrow\quad \sum_{j=1}^{N}|f(v_j)-f(u_j)|<\varepsilon. $$

The order of the quantifiers matters. First, an arbitrary $\varepsilon>0$ is given. The function must then allow a choice of $\delta>0$. That same $\delta$ must work for every permitted number $N$ of intervals and every permitted placement of them. In particular, $\delta$ may depend on $f$, $[a,b]$, and $\varepsilon$, but it cannot depend on the family selected afterward. The hypothesis concerns the sum of the interval lengths, not a bound on each length separately.

The conclusion also controls a sum: it requires the sum of the absolute increments to be less than $\varepsilon$. The absolute values prevent increments of opposite signs from cancelling. Thus the condition is stronger than requiring a small net change from the first endpoint to the last. The strict inequalities are part of this version of the definition: the total length must be $<\delta$, and the resulting sum must be $<\varepsilon$.

For comparison, uniform continuity asks for a bound on the change in $f$ across each individual pair of sufficiently close points. Absolute continuity asks for control over the accumulated changes across any finite non-overlapping family, provided its total length is sufficiently small. The two notions are distinct; their precise relationship is a separate question.

Constant Functions

Proposition: Every constant real-valued function on a compact interval $[a,b]$ belongs to $AC[a,b]$, including the case $a=b$.

Proof. Let $f(x)=c$ for every $x\in[a,b]$, where $c\in\mathbb{R}$. For every subinterval $[u_j,v_j]$, we have $f(v_j)=c=f(u_j)$, so $|f(v_j)-f(u_j)|=0$. Therefore every finite family satisfies $$ \sum_{j=1}^{N}|f(v_j)-f(u_j)|=0. $$ Given any $\varepsilon>0$, choose $\delta=1$. Whenever the total interval length is less than $\delta$, the sum of increments is $0<\varepsilon$. This also holds when $a=b$, because every subinterval then has length zero and its increment is zero. Hence $f\in AC[a,b]$. $\square$

Worked Example: A Constant on a Nontrivial Interval

Let $f:[-2,3]\to\mathbb{R}$ be $f(x)=7$. Test the definition using $\varepsilon=\tfrac{1}{10}$, and take the family $[-2,-1]$, $[0,\tfrac12]$, and $[2,3]$.

The intervals lie in $[-2,3]$ and have disjoint interiors. Their lengths are $$ -1-(-2)=1,\qquad \frac12-0=\frac12,\qquad 3-2=1, $$ so the total length is $1+\tfrac12+1=\tfrac52$. This particular family would satisfy the definition’s length hypothesis only for a $\delta$ greater than $\tfrac52$; it is not needed to use such a large $\delta$, because the increment conclusion holds for every family, with no length restriction.

On the three intervals the absolute increments are $$ |7-7|=0,\qquad |7-7|=0,\qquad |7-7|=0. $$ Their sum is $0<\tfrac{1}{10}$. More generally, choosing $\delta=1$ works for every finite family whose total length is less than $1$, because every term in its increment sum is zero. Check: the computed increment sum $0$ is below the specified tolerance $\tfrac{1}{10}$.

Affine Functions

Proposition: Every affine function $f(x)=cx+d$, with $c,d\in\mathbb{R}$, is in $AC[a,b]$ on every compact interval $[a,b]$.

Proof. Fix $\varepsilon>0$ and consider an arbitrary finite family of non-overlapping subintervals $[u_j,v_j]\subseteq[a,b]$. For every $j$, $$ |f(v_j)-f(u_j)| =|cv_j+d-(cu_j+d)| =|c|(v_j-u_j). $$ Adding these equalities gives $$ \sum_{j=1}^{N}|f(v_j)-f(u_j)| =|c|\sum_{j=1}^{N}(v_j-u_j). $$ If $c=0$, this sum is zero for every family, so $\delta=1$ works. If $c\ne0$, choose $\delta=\varepsilon/|c|>0$. Whenever $\sum_j(v_j-u_j)<\delta$, multiplication by $|c|$ gives $$ \sum_{j=1}^{N}|f(v_j)-f(u_j)| =|c|\sum_{j=1}^{N}(v_j-u_j) <|c|\delta =\varepsilon. $$ In both cases, the required implication holds for every finite family, proving $f\in AC[a,b]$. $\square$

Worked Example: Choosing Delta for an Affine Function

Let $f:[0,3]\to\mathbb{R}$ be $f(x)=-4x+2$. For $\varepsilon=\tfrac{3}{10}$, find a $\delta$ that verifies the definition and check it on the family $[0,\tfrac{1}{100}]$ and $[1,\tfrac{3}{200}]$.

The slope is $c=-4$, so $|c|=4$. The proof gives the choice $$ \delta=\frac{\varepsilon}{|c|} =\frac{3/10}{4} =\frac{3}{40}. $$ The total length of the specified family is $$ \frac{1}{100}+\left(\frac{3}{200}-1\right) $$ which is not a valid expression for a subinterval length: the second interval as written would have its left endpoint larger than its right endpoint. We therefore use the valid family $[0,\tfrac{1}{100}]$ and $[1,\tfrac{101}{100}]$, both contained in $[0,3]$. Its total length is $$ \frac{1}{100}+\left(\frac{101}{100}-1\right) =\frac{1}{100}+\frac{1}{100} =\frac{1}{50}. $$ Since $\tfrac{1}{50}<\tfrac{3}{40}$, the hypothesis is satisfied. The endpoint increments are $$ \left|f\left(\frac{1}{100}\right)-f(0)\right| =|-4/100|=\frac{1}{25}, \qquad \left|f\left(\frac{101}{100}\right)-f(1)\right| =|-4/100|=\frac{1}{25}. $$ Their sum is $\tfrac{2}{25}$, and $\tfrac{2}{25}<\tfrac{3}{10}$. Check: the general identity gives $|c|$ times the total length, namely $4\cdot\tfrac{1}{50}=\tfrac{2}{25}$, agreeing with the endpoint calculation.

A Fixed Jump Cannot Be Made Small

Proposition: The function $s:[0,1]\to\mathbb{R}$ defined by $s(x)=0$ for $0\le x<\tfrac12$ and $s(x)=1$ for $\tfrac12\le x\le1$ is not in $AC[0,1]$.

Proof. To negate the definition, it is enough to exhibit one $\varepsilon_0>0$ such that for every $\delta>0$ there is a permitted family with total length less than $\delta$ but increment sum at least $\varepsilon_0$. Set $\varepsilon_0=\tfrac12$. Given any $\delta>0$, choose $$ h=\min\left\{\frac{\delta}{2},\frac14\right\}. $$ Then $h>0$, $h<\delta$, and $h\le\tfrac14$, so $[\tfrac12-h,\tfrac12]\subseteq[0,1]$. This one-interval family has total length $h<\delta$, while $$ \left|s\left(\frac12\right)-s\left(\frac12-h\right)\right| =|1-0|=1\ge\frac12. $$ Thus no $\delta>0$ works for $\varepsilon_0=\tfrac12$. Therefore $s\notin AC[0,1]$. $\square$

Worked Example: Testing a Step Function at Its Jump

Use the preceding step function and test $\delta=\tfrac{1}{20}$ with the interval $[\tfrac{49}{100},\tfrac12]$. Take $\varepsilon=\tfrac12$.

The interval is contained in $[0,1]$, and its length is $$ \frac12-\frac{49}{100}=\frac{50}{100}-\frac{49}{100}=\frac{1}{100}<\frac{1}{20}. $$ The left endpoint is strictly below $\tfrac12$, so $s(\tfrac{49}{100})=0$; the right endpoint equals $\tfrac12$, so $s(\tfrac12)=1$. Hence the increment sum for this one interval is $|1-0|=1$, which is not less than $\varepsilon=\tfrac12$. This disproves the implication for this particular $\delta$. The proposition’s proof strengthens the test by producing such an interval for every positive $\delta$. Check: the interval length is $\tfrac{1}{100}$, while its increment is $1$; shrinking the interval does not shrink the jump.

Quantifier Pitfall: The Total Length Matters

A frequent error is to check that each individual interval is short and then conclude that the sum of the increments is small. The definition does not use that hypothesis. It requires the sum of all the interval lengths to be less than $\delta$. The distinction is essential because many individually short intervals can have a large combined length and a large combined increment.

For a concrete calculation, take $f(x)=x$ on $[0,1]$, let $\varepsilon=\tfrac{1}{10}$, and use the valid choice $\delta=\tfrac{1}{10}$. Consider the ten intervals $$ \left[\frac{9k}{100},\frac{9k+9}{100}\right], \qquad k=0,1,\ldots,9. $$ They lie in $[0,1]$ and have pairwise disjoint interiors. Each has length $\tfrac{9}{100}<\delta$, but the total length is $$ 10\cdot\frac{9}{100}=\frac{9}{10}, $$ which is not less than $\delta=\tfrac{1}{10}$. Each increment of $f(x)=x$ equals the interval’s length, so the sum of the increments is also $\tfrac{9}{10}$. This does not contradict absolute continuity: the definition’s total-length hypothesis was not met. In contrast, for any finite family with total length less than $\tfrac{1}{10}$, the sum of increments for $f(x)=x$ equals that total length and is therefore less than $\varepsilon$.

Absolute continuity of a function is also distinct from absolute continuity of a measure. Here the term names a property of a real-valued function on an interval, defined by sums of endpoint increments. For measures, absolute continuity is the relation $\nu\ll\mu$, meaning that every $\mu$-null measurable set is also $\nu$-null. The definitions concern different objects and should not be conflated.

Check Your Understanding

For each question, keep track of the full finite-family condition, including whether the total length is strictly less than $\delta$.

Write the quantifier order in the definition of $AC[a,b]$. Which quantities may the chosen $\delta$ depend on, and which may it not depend on?
For $f(x)=5$ on $[-1,2]$, give a valid $\delta$ for an arbitrary $\varepsilon>0$ and verify the increment sum for an arbitrary finite family.
For $f(x)=3x-2$ on $[0,4]$, determine a valid $\delta$ in terms of $\varepsilon$ and verify the definition directly.
Negate the definition and use that form to show that a step function with a jump of size $2$ at an interior point is not absolutely continuous.
Explain why ten intervals each shorter than $\delta$ do not necessarily satisfy the hypothesis in the definition.