Controlling Many Increments at Once
Absolute continuity is a condition on sums, not just on individual increments. A function may change little across each short interval, but a family can contain many such intervals. The relevant question is whether the sum of all endpoint changes must be small whenever the sum of the interval lengths is small. An estimate for that entire sum is often the most direct way to verify absolute continuity.
Let \(a<b\) be real numbers. The functions in this section are real-valued functions \(f:[a,b]\to\mathbb{R}\). For a finite family of intervals \([u_j,v_j]\subseteq[a,b]\), we allow \(u_j=v_j\), and require the interiors of distinct intervals to be disjoint. Endpoints may be shared. The total length is \(\sum_j(v_j-u_j)\), and the total increment is \(\sum_j|f(v_j)-f(u_j)|\).
The absolute values matter: increases and decreases cannot cancel each other. Also, the definition requires control uniformly over every eligible finite family, not merely a bound for one interval chosen in advance.
A General Interval-Sum Criterion
Proof. Fix \(\varepsilon>0\). The limit condition gives a \(\delta>0\) such that \(0<t<\delta\) implies \(\Phi(t)<\varepsilon\); decrease \(\delta\), if necessary, so that \(\delta\leq b-a\). Take any eligible finite family whose total length \(s=\sum_j(v_j-u_j)\) is less than \(\delta\). If \(s>0\), the assumed estimate gives \[ \sum_j|f(v_j)-f(u_j)|\leq\Phi(s)<\varepsilon. \] If \(s=0\), every interval has \(u_j=v_j\), so every increment is zero and the same strict conclusion holds. This is the definition of absolute continuity. \(\square\)
In applications, one often has a simpler bound \(\sum_j|f(v_j)-f(u_j)|\leq C\sum_j(v_j-u_j)\) for a fixed finite \(C\geq0\). The proposition applies with \(\Phi(t)=Ct\), which tends to zero as \(t\downarrow0\). If \(C>0\), choosing \(\delta=\varepsilon/C\) makes total length less than \(\delta\) imply total increment less than \(\varepsilon\). If \(C=0\), all the increments are zero, so any positive \(\delta\) works.
Worked Interval-Sum Estimates
Worked Example: A Direct Estimate for the Square Function
Let \(f(x)=x^2\) on \([0,1]\). For any \(0\leq u\leq v\leq1\), factoring the difference gives \[ |f(v)-f(u)|=v^2-u^2=(v-u)(v+u)\leq2(v-u), \] because \(v+u\leq2\). Therefore, for any finite family of intervals with pairwise disjoint interiors, \[ \sum_j|f(v_j)-f(u_j)|\leq2\sum_j(v_j-u_j). \] For any \(\varepsilon>0\), take \(\delta=\varepsilon/2\). If the total length is less than \(\delta\), the total increment is less than \(2\delta=\varepsilon\). Thus \(f\) is absolutely continuous by the interval-sum criterion.
For a concrete check, take \([0,1/4]\) and \([1/2,3/4]\). Their total length is \(1/4+1/4=1/2\). Their increments are \[ (1/4)^2-0^2=1/16,\qquad (3/4)^2-(1/2)^2=9/16-4/16=5/16. \] The total increment is \(1/16+5/16=6/16=3/8\), and the bound gives \(2(1/2)=1\), so \(3/8\leq1\). Direct substitution into the endpoint values gives the same two increments.
Worked Example: A Common Bound on the Derivative
Suppose \(f:[a,b]\to\mathbb{R}\) is continuous on \([a,b]\), differentiable on \((a,b)\), and satisfies \(|f'(x)|\leq M\) for every \(x\in(a,b)\), where \(M\) is a finite nonnegative constant. For any \(u<v\) in \([a,b]\), the mean value theorem applies on \([u,v]\): continuity holds on that closed interval and differentiability holds in its interior. Thus there is a point \(c\in(u,v)\) such that \(f(v)-f(u)=f'(c)(v-u)\). Consequently, \[ |f(v)-f(u)|\leq M(v-u). \] For \(u=v\), both sides are zero, so this estimate holds for every subinterval. Summing it over any finite family with pairwise disjoint interiors gives \[ \sum_j|f(v_j)-f(u_j)|\leq M\sum_j(v_j-u_j). \] If \(M>0\), choose \(\delta=\varepsilon/M\); total length less than \(\delta\) makes the total increment less than \(\varepsilon\). If \(M=0\), the estimate says every endpoint increment is zero, and any \(\delta>0\) suffices. In either case \(f\) is absolutely continuous.
For \(f(x)=e^x\) on \([0,1]\), the derivative is \(e^x\), and \(e^x\leq e\) on this interval. Thus the estimate holds with \(M=e\). For the intervals \([0,1/4]\) and \([1/2,1]\), the total length is \(1/4+1/2=3/4\), while the total increment is \((e^{1/4}-1)+(e-e^{1/2})\). Applying the derivative estimate separately yields \(e^{1/4}-1\leq e/4\) and \(e-e^{1/2}\leq e/2\), so the total increment is at most \(3e/4\), exactly the bound \(M\) times the total length.
Check: each displayed exponential increment is the endpoint difference \(f(v)-f(u)\), and the two interval lengths sum to \(3/4\). The bound \(3e/4\) is therefore obtained from \(M=e\) and this family’s actual total length.
Worked Example: An Interval Crossing a Change of Slope
Define \(f:[0,1]\to\mathbb{R}\) by \[ f(x)= \begin{cases} 2x,&0\leq x\leq1/2,\\ 3/2-x,&1/2\leq x\leq1. \end{cases} \] The two formulas agree at \(x=1/2\), where both give \(1\). On an interval \([u,v]\) entirely to the left of \(1/2\), the endpoint increment is \(2(v-u)\). On an interval entirely to the right, its absolute value is \(v-u\), which is at most \(2(v-u)\). If \(u<1/2<v\), split the increment at \(1/2\) and use the triangle inequality: \[ |f(v)-f(u)|\leq|f(v)-f(1/2)|+|f(1/2)-f(u)| =(v-1/2)+2(1/2-u)\leq2(v-u). \] The last inequality follows since \(v-1/2\leq2(v-1/2)\). The same estimate holds when an endpoint equals \(1/2\). Summing over a finite family gives the bound \(2\sum_j(v_j-u_j)\), so the function is absolutely continuous, with \(\delta=\varepsilon/2\).
For the intervals \([0,1/4]\) and \([1/4,3/4]\), the interiors are disjoint and the total length is \(1/4+1/2=3/4\). Their increments are \(f(1/4)-f(0)=1/2-0=1/2\) and \(f(3/4)-f(1/4)=3/4-1/2=1/4\). Their sum is \(3/4\), which is no greater than the estimate \(2(3/4)=3/2\). Evaluating the defining formulas at all four endpoints confirms the calculation.
What a Failed Estimate Looks Like
A common mistake is to check that increments become small on short intervals and then assume the function is absolutely continuous. That only addresses each interval separately. Absolute continuity asks for a bound on the sum over an arbitrary finite family, and a jump can persist even on intervals of arbitrarily small length.
Worked Example: A Jump Defeats the Interval-Sum Condition
On \([0,1]\), define \(H(x)=0\) for \(0\leq x<1/2\), and \(H(x)=1\) for \(1/2\leq x\leq1\). For each integer \(n\geq2\), the interval \([1/2-1/n,1/2]\) lies in \([0,1]\) and has length \(1/n\). Its left endpoint is strictly less than \(1/2\), so \(H(1/2-1/n)=0\); its right endpoint equals \(1/2\), so \(H(1/2)=1\). Thus its endpoint increment is exactly \(1\).
Given any \(\delta>0\), choose an integer \(n\geq2\) large enough that \(1/n<\delta\). This one-interval family then has total length less than \(\delta\), but its total increment is \(1\). Taking \(\varepsilon=1/2\) shows that no \(\delta\) can satisfy the definition of absolute continuity. The endpoint convention in the definition of \(H\) is essential here: the interval ends at the jump, where the function has value \(1\).
The practical method is to seek one inequality that applies to every interval in the family, then sum it before choosing \(\delta\). A direct algebraic factorization, a derivative bound with the mean value theorem, or a careful split at a change of formula can each supply that inequality. The final choice of \(\delta\) must then match the summed estimate and the strict inequality in the definition.
Check Your Understanding
For each question, track both the individual increments and their sum over the full family.
- If \(|f(v)-f(u)|\leq 5(v-u)\) for every \(a\leq u\leq v\leq b\), what choice of \(\delta\) verifies absolute continuity for a given \(\varepsilon>0\)?
- Why does the derivative argument require continuity on each closed subinterval as well as differentiability in its interior?
- For \(f(x)=x^2\) on \([0,1]\), verify the bound \(|f(v)-f(u)|\leq2(v-u)\) by factoring \(v^2-u^2\).
- In the jump-function example, why must the interval end at \(1/2\), rather than at a point strictly below \(1/2\), to obtain increment \(1\)?
- Explain why a bound on each individual short-interval increment is not by itself an estimate on the sum over a finite family.