Why an Absolutely Continuous Function With Zero Derivative Is Constant

From a Derivative Condition to a Global Conclusion

A derivative describes behavior near one point, while being constant is a statement about every pair of points in an interval. The bridge between these scales is absolute continuity: it controls the sum of changes in the function across many short, disjoint intervals. When the derivative vanishes almost everywhere, a covering argument lets us use the derivative estimate on most of an interval and use absolute continuity to control what remains.

Let \([a,b]\) be a compact interval with \(a<b\), equipped with Lebesgue measure \(m\) restricted to \([a,b]\). All functions in this tutorial are real-valued. A function \(f:[a,b]\to\mathbb{R}\) is absolutely continuous if, for every \(\varepsilon>0\), there is a \(\delta>0\) such that for every finite family of pairwise disjoint subintervals \([u_j,v_j]\subseteq[a,b]\),

“The derivative is zero almost everywhere” means that there is a Lebesgue-null set \(N\subseteq[a,b]\) such that, at every \(x\in(a,b)\setminus N\), the finite derivative \(f'(x)\) exists and equals zero. The theorem below makes a pointwise conclusion, not merely an almost-everywhere one.

The Interval-Covering Argument

The proof uses a tagged partition adapted to a point-dependent radius. A tagged partition of \([a,b]\) consists of points \(a=x_0\le x_1\le\cdots\le x_n=b\) and tags \(\xi_i\in[x_{i-1},x_i]\). Given a positive function \(\gamma:[a,b]\to(0,\infty)\), the partition is called \(\gamma\)-fine if each interval \([x_{i-1},x_i]\) is contained in \((\xi_i-\gamma(\xi_i),\xi_i+\gamma(\xi_i))\). The radius can be chosen separately at every point: small near exceptional points and suited to the derivative at the other points.

Proof. Let \(S\) be the set of \(c\in[a,b]\) for which \([a,c]\) has a finite \(\gamma\)-fine tagged partition; the empty partition shows \(a\in S\). Put \(c_*=\sup S\). If \(c_*<b\), choose \(d>0\) with \(d<\gamma(c_*)/2\) and \(d<b-c_*\). By the definition of supremum, some \(c\in S\) satisfies \(c>c_*-d\). Append to a fine partition of \([a,c]\) the interval \([c,c_*+d]\), tagged by \(c_*\). Its endpoints are within distance \(2d<\gamma(c_*)\) of the tag, so the appended interval is contained in the required open neighborhood. This gives a fine partition reaching beyond \(c_*\), a contradiction. Therefore \(c_*=b\). If \(b\in S\), the desired partition already exists. Otherwise, choose \(c\in S\) with \(b-c<\gamma(b)\), which is possible because \(\sup S=b\), and append \([c,b]\) tagged by \(b\). This interval is contained in \((b-\gamma(b),b+\gamma(b))\), so the resulting partition is \(\gamma\)-fine. \(\square\)

We now prove the theorem. Fix a target \(\varepsilon>0\). The complement of the set where the derivative exists and is zero is contained in a null set; enlarge that null set to include the endpoints and call it \(N\). By the outer regularity of Lebesgue measure, there is an open set \(O\subseteq\mathbb{R}\) containing \(N\) with \(m(O)<\delta\), where \(\delta\) is the absolute-continuity threshold for \(\varepsilon/2\). For each \(x\in N\), choose \(\gamma(x)>0\) so that \((x-\gamma(x),x+\gamma(x))\subseteq O\).

At every \(x\in[a,b]\setminus N\), the derivative is zero. Set \(\alpha=\varepsilon/(2(b-a))\). By the definition of derivative, there is a positive \(\gamma(x)\) such that whenever \(y\in[a,b]\) and \(0<|y-x|<\gamma(x)\),

The inequality also holds when \(y=x\), since both sides then equal zero. Apply the lemma to obtain a finite \(\gamma\)-fine tagged partition. Call a partition interval exceptional if its tag lies in \(N\), and ordinary otherwise.

For an ordinary interval \([u,v]\) tagged by \(x\), the tag belongs to \([u,v]\). The derivative estimate applies at both endpoints, including an endpoint equal to \(x\). Thus

Summing over ordinary intervals gives a total contribution at most \(\alpha(b-a)=\varepsilon/2\). For an exceptional interval tagged by \(x\in N\), fineness ensures \([u,v]\subset O\). The exceptional intervals have disjoint interiors, so the sum of their lengths is at most \(m(O)<\delta\). Apply absolute continuity to this finite family: the sum of their increments is less than \(\varepsilon/2\). Finally, telescoping along the partition and using the triangle inequality yields

This holds for every \(\varepsilon>0\), so \(f(b)=f(a)\). The same argument applies on any subinterval \([u,v]\subseteq[a,b]\): the restriction of an absolutely continuous function is absolutely continuous, and the derivative condition still holds almost everywhere there. Consequently \(f(u)=f(v)\) for every \(u,v\in[a,b]\), proving that \(f\) is constant everywhere. \(\square\)

Worked Applications and the Role of the Hypotheses

What the Covering Argument Accomplishes

The derivative estimate alone is only local, and it holds only off a null set. The proof does not try to obtain one derivative radius that works uniformly at every ordinary point. Instead, the fine partition uses a radius chosen separately at each tag. On ordinary intervals, the tag lies between the endpoints, so the two estimates from the tag add to a bound proportional to the interval’s length. Exceptional intervals are handled collectively: their disjoint interiors lie inside an open set of small measure, and absolute continuity controls the sum of their increments.

A common mistake is to reason that a function with derivative zero almost everywhere must be constant, without checking absolute continuity. The step function \(H\) above disproves that statement: it has derivative zero except at one point but fails the interval-sum condition. Another mistake is to conclude that an almost-everywhere derivative condition gives equality only almost everywhere. Here the conclusion is everywhere because the proof applies on every subinterval and establishes equality between each pair of endpoints. Absolute continuity supplies the global control needed to turn the almost-everywhere local information into that pointwise conclusion.