Composition and the Chain Rule for Absolutely Continuous Functions

Composition: Controlling Increments on the Range

Composition combines two different kinds of control. Absolute continuity of the inner function controls the total change in its values over short collections of intervals. A bound on the outer function’s derivative turns those changes into bounds on the changes of the composition. Together, these facts show that composition preserves absolute continuity. Differentiability then gives the chain rule wherever the inner function has a derivative; absolute continuity ensures that this is almost everywhere.

Throughout, \([a,b]\) is a compact interval with \(a<b\), and the functions considered are real-valued. A family of intervals \([u_j,v_j]\subseteq[a,b]\) is pairwise nonoverlapping if their interiors are disjoint. The definition below uses finite families of such intervals; allowing shared endpoints does not affect the sums.

The interval-sum condition is stronger than continuity: it controls the combined increments across many intervals, not just the change across one small interval. In particular, an absolutely continuous function is continuous. To check this, fix \(\varepsilon>0\) and choose \(\delta\) from the definition. Whenever \(x,y\in[a,b]\) and \(|x-y|<\delta\), apply the condition to the single interval with endpoints \(x,y\); it gives \(|f(y)-f(x)|<\varepsilon\).

Proof. Set \(K=f([a,b])\). The continuity of \(f\) implies that \(K\) is compact, and the intermediate value theorem implies that it is an interval. For any \(s,t\in K\), the whole interval between \(s\) and \(t\) lies in \(K\). The mean value theorem, applied to \(g\) on that interval, therefore gives \(|g(s)-g(t)|\le M|s-t|\). If \(M=0\), this inequality says that \(g\) is constant on \(K\), so \(g\circ f\) is constant and hence absolutely continuous. If \(M>0\), fix \(\varepsilon>0\) and use the absolute continuity of \(f\) with tolerance \(\varepsilon/M\). It supplies \(\delta>0\). For any finite pairwise nonoverlapping intervals \([u_j,v_j]\) with total length less than \(\delta\), the inequality on \(K\) gives \[ \sum_j|g(f(v_j))-g(f(u_j))| \le M\sum_j|f(v_j)-f(u_j)| < M\frac{\varepsilon}{M} =\varepsilon. \] This is precisely the definition of absolute continuity for \(g\circ f\). \(\square\)

Only the bound on the range of \(f\) is used: the derivative of \(g\) need not be bounded everywhere on the real line. In this theorem the bound is automatic if \(g'\) is continuous on an open interval containing the compact set \(f([a,b])\), since a continuous function is bounded on a compact set. It is useful to state the bound explicitly because the increment estimate depends on it.

Why the Chain Rule Holds Almost Everywhere

To pass from a pointwise calculation to an almost-everywhere statement, we need to know that \(f\) is differentiable almost everywhere. Here is the needed link to the earlier material on bounded variation. For a real-valued function on \([a,b]\), its total variation is the supremum of \(\sum_{i=1}^n|f(x_i)-f(x_{i-1})|\) over all partitions \(a=x_0<x_1<\cdots<x_n=b\). A function has bounded variation if this supremum is finite. The earlier result that a function of bounded variation on a compact interval is differentiable almost everywhere will be used below.

Proof. Apply absolute continuity with tolerance \(1\), obtaining \(\delta>0\). Choose a finite subdivision \(a=c_0<c_1<\cdots<c_N=b\) whose subintervals have length less than \(\delta\). Consider any partition \(a=x_0<x_1<\cdots<x_n=b\). Insert all the subdivision points \(c_i\) into this partition. The triangle inequality shows that inserting points cannot decrease the sum of absolute increments. Within each subinterval \([c_{i-1},c_i]\), the resulting partition intervals have total length \(c_i-c_{i-1}<\delta\). By absolute continuity, the sum of the increments over those intervals is less than \(1\). Summing over the \(N\) subintervals shows that the original partition sum is at most \(N\). This bound does not depend on the original partition. Taking the supremum over all partitions proves that \(f\) has bounded variation. \(\square\)

Combining the lemma with the earlier almost-everywhere differentiability result for functions of bounded variation, \(f\) is differentiable at Lebesgue-almost every point of \((a,b)\). This is the only almost-everywhere differentiability fact needed for the chain rule here; no claim is made about differentiability at every point.

Proof. By the lemma and the earlier almost-everywhere differentiability result for functions of bounded variation, there is a Lebesgue-null set \(N\subseteq(a,b)\) outside of which \(f\) is differentiable. Fix \(x\in(a,b)\setminus N\), and put \(y=f(x)\). For nonzero \(h\) sufficiently small that \(x+h\in[a,b]\), write \(k_h=f(x+h)-f(x)\). Differentiability of \(f\) at \(x\) means \[ k_h=f'(x)h+r_h,\qquad \frac{r_h}{h}\longrightarrow0\quad\text{as }h\to0. \] In particular, \(k_h\to0\) and \(k_h/h\to f'(x)\). Differentiability of \(g\) at \(y\) gives \[ g(y+k_h)-g(y)=g'(y)k_h+s_h, \] where \(s_h/k_h\to0\) whenever \(k_h\ne0\). If \(k_h=0\), then \(g(y+k_h)-g(y)=0\), so the same expansion holds with \(s_h=0\). Also, \(s_h/h\to0\): when \(k_h\ne0\), write \(s_h/h=(s_h/k_h)(k_h/h)\), whose first factor tends to \(0\) and whose second factor tends to the finite value \(f'(x)\); when \(k_h=0\), the quotient is zero. Dividing the expansion by \(h\) and taking the limit now yields \[ \lim_{h\to0}\frac{g(f(x+h))-g(f(x))}{h} =g'(y)\lim_{h\to0}\frac{k_h}{h} =g'(f(x))f'(x). \] Thus the chain-rule identity holds at each \(x\in(a,b)\setminus N\), and hence almost everywhere. \(\square\)

A Common Pitfall

The composition theorem establishes absolute continuity everywhere on the interval, but the chain-rule identity is asserted only where \(f\) is differentiable. A common error is to apply the pointwise chain rule at an exceptional point anyway. In the preceding example, one might write at \(x=0\) that \((\sin|x|)'=\cos(|x|)(|x|)'=1\). This calculation is invalid: \((|x|)'(0)\) does not exist. Indeed, the right and left difference quotients of \(\sin|x|\) at \(0\) are \(1\) and \(-1\), respectively. The correct statement is that the chain rule holds for every \(x\ne0\), and hence almost everywhere; a null exceptional set is permitted, but it cannot be ignored when claiming an everywhere identity.

Two distinct hypotheses play distinct roles. Absolute continuity of \(f\) supplies differentiability almost everywhere and the interval-sum control needed to prove absolute continuity of the composition. The derivative bound for \(g\) on the range supplies the inequality \(|g(s)-g(t)|\le M|s-t|\). A derivative bound outside the range is not needed, while a pointwise chain-rule conclusion at points where \(f\) is not differentiable is not justified by the almost-everywhere theorem.