What Substitution Requires
For a differentiable strictly increasing map, substitution is often introduced by writing \(y=\phi(x)\) and \(dy=\phi'(x)\,dx\). In measure theory, the formula remains valid when \(\phi\) is merely nondecreasing and absolutely continuous: strict increase is not needed, and intervals on which \(\phi\) is constant contribute zero. The measurability of the integrand, however, must be stated carefully.
Let \(a<b\), let \(\phi:[a,b]\to\mathbb{R}\) be nondecreasing and absolutely continuous, and set \(c=\phi(a)\) and \(d=\phi(b)\), so \(c\le d\). Absolute continuity implies that \(\phi'\) exists and is finite almost everywhere and that \(\phi(x)=\phi(a)+\int_a^x\phi'(t)\,dt\). Since \(\phi\) is nondecreasing, \(\phi'\ge0\) wherever the derivative exists. Choose a measurable version of \(\phi'\) by assigning value zero at points where the derivative does not exist.
The requirement that \(g\) be Borel measurable ensures that its composition with the continuous map \(\phi\) is measurable. The theorem does not assert this for every Lebesgue measurable \(g\); a concrete failure of that assertion is discussed below. The cases \(c=d\) and \(c<d\) are both covered: if \(c=d\), monotonicity forces \(\phi\) to be constant, so \(\phi'=0\) almost everywhere and both integrals are zero.
Proof by Identifying the Induced Measure
We prove the theorem by showing that weighting Lebesgue measure on \([a,b]\) by \(\phi'\), then mapping points through \(\phi\), gives Lebesgue measure on \([c,d]\). More precisely, for Borel \(E\subseteq[c,d]\), define $$ \nu(E)=\int_{\phi^{-1}(E)}\phi'(x)\,dx. $$ The preimage is Borel because \(\phi\) is continuous. Countable additivity follows from countable additivity of the integral on disjoint measurable sets, so \(\nu\) is a measure on the Borel subsets of \([c,d]\). It is finite, since the integral representation of \(\phi\) gives \(\nu([c,d])=\int_a^b\phi'(x)\,dx=d-c\).
Proof. First suppose \(c<d\) and fix \(y\in[c,d]\). The set \(\{x\in[a,b]:\phi(x)\le y\}\) is a nonempty closed initial interval, so it has the form \([a,r]\) for some \(r\in[a,b]\). If \(y<d\), continuity and maximality of \(r\) give \(\phi(r)=y\): if \(\phi(r)<y\), continuity would keep \(\phi\) below \(y\) just to the right of \(r\), contradicting maximality. For \(y=d\), \(r=b\) and \(\phi(r)=d=y\). Thus, by the integral representation of \(\phi\), $$ \nu([c,y])=\int_{\{x:\phi(x)\le y\}}\phi'(x)\,dx =\int_a^r\phi'(x)\,dx =\phi(r)-\phi(a)=y-c. $$ This computation includes \(y=c\), for which the value is zero. Lebesgue measure restricted to \([c,d]\) also assigns measure \(y-c\) to \([c,y]\). The sets \([c,y]\), for \(y\in[c,d]\), form a family closed under finite intersections and generate the Borel sigma-algebra of \([c,d]\). By the standard uniqueness theorem for finite measures agreeing on a generating pi-system, \(\nu\) is Lebesgue measure restricted to \([c,d]\). If \(c=d\), both measures are zero, as already noted, so the same conclusion holds.
For a Borel set \(E\subseteq[c,d]\), the equality of measures now gives \(\int_{\phi^{-1}(E)}\phi'\,dx=m(E)\). By finite additivity and homogeneity of the integral, it follows that $$ \int_a^b \psi(\phi(x))\phi'(x)\,dx=\int_c^d\psi(y)\,dy $$ for every nonnegative Borel simple function \(\psi\). Every nonnegative Borel measurable \(g\) is the increasing pointwise limit of nonnegative Borel simple functions. The Monotone Convergence Theorem applies on both measure spaces, so taking the limit in the simple-function identity proves the asserted formula, including when the value is infinite.
Finally, if \(g\) is real-valued or complex-valued and \(\int_c^d|g|\,dy<\infty\), then \(|g|\) is nonnegative and Borel measurable. The formula just proved gives $$ \int_a^b |g(\phi(x))|\phi'(x)\,dx=\int_c^d|g(y)|\,dy<\infty. $$ Thus the real and imaginary parts (or, in the real case, the positive and negative parts) are integrable against the weighted measure. Applying the nonnegative formula to those parts and subtracting, or applying it separately to real and imaginary parts, proves the signed or complex equality. \(\square\)
Worked Applications
Worked Example: A Power Map and an Indicator
Take \(\phi(x)=x^2\) on \([0,2]\) and \(g(y)=\chi_{[1,3]}(y)\) on \([0,4]\). The map is nondecreasing and continuously differentiable, hence absolutely continuous, and \(\phi'(x)=2x\). The integrand \(g\) is Borel measurable and nonnegative, so the substitution theorem applies. On the left, the interval \([1,3]\) has length \(3-1=2\), giving $$ \int_0^4\chi_{[1,3]}(y)\,dy=2. $$ On the right, \(g(x^2)=1\) exactly when \(1\le x^2\le3\). Since \(x\in[0,2]\), this is precisely \(1\le x\le\sqrt{3}\). Therefore $$ \int_0^2\chi_{[1,3]}(x^2)\,2x\,dx =\int_1^{\sqrt{3}}2x\,dx =\big[x^2\big]_1^{\sqrt{3}} =3-1=2. $$ Check: the preimage interval has weighted measure \(3-1=2\), exactly the length of the target interval \([1,3]\).
Worked Example: A Plateau Does Not Add Mass
Define \(\phi:[0,3]\to[0,2]\) by $$ \phi(x)= \begin{cases} x,&0\le x\le1,\\ 1,&1\le x\le2,\\ x-1,&2\le x\le3. \end{cases} $$ The definitions agree at the join points, and the map is nondecreasing and piecewise linear, hence absolutely continuous. Its derivative is \(1\) on \((0,1)\), \(0\) on \((1,2)\), and \(1\) on \((2,3)\); its values at the two join points do not affect the integral. For the Borel function \(g(y)=y\), the target integral is $$ \int_0^2 y\,dy=\left[\frac{y^2}{2}\right]_0^2=2. $$ On the domain, the plateau contributes zero because \(\phi'=0\) there. The other two intervals give $$ \int_0^3\phi(x)\phi'(x)\,dx =\int_0^1x\,dx+\int_1^2 0\,dx+\int_2^3(x-1)\,dx =\frac12+0+\left[\frac{(x-1)^2}{2}\right]_2^3 =\frac12+\frac{4-1}{2}=2. $$ Check: the two nonconstant pieces map onto \([0,1]\) and \([1,2]\); their target integrals are \(\frac12\) and \(\frac{4-1}{2}=\frac32\), summing to \(2\). The interval on which \(\phi\) is constant contributes no weighted mass.
Worked Example: An Integrable Function Under an Affine Map
Let \(\phi(x)=3x+1\) on \([0,2]\), and let \(g(y)=1/y\) on \([1,7]\). Here \(\phi\) is nondecreasing and absolutely continuous, \(\phi'=3\), and \(g\) is Borel measurable with \(\int_1^7|g(y)|\,dy=\ln 7<\infty\). Directly on the target interval, $$ \int_1^7\frac{1}{y}\,dy=[\ln y]_1^7=\ln7-\ln1=\ln7. $$ The transformed integral is $$ \int_0^2\frac{1}{3x+1}\,3\,dx =\left[\ln(3x+1)\right]_0^2 =\ln7-\ln1=\ln7. $$ Check: the transformed endpoints are \(\phi(0)=1\) and \(\phi(2)=7\), the same endpoints used in the target integral. Both computations give \(\ln7\).
Why Borel Measurability Matters
It is a common mistake to replace “Borel measurable” in the theorem by “Lebesgue measurable” and assume continuity of \(\phi\) still makes \(g\circ\phi\) Lebesgue measurable. Continuity guarantees that the preimage of a Borel set is Borel; it does not guarantee that the preimage of every Lebesgue measurable set is Lebesgue measurable.
Here is a concrete reason. Let \(K\subset[0,1]\) be a nowhere-dense closed set of positive Lebesgue measure whose complement has positive measure in every nonempty interval, and define \(\phi(x)=\int_0^x\chi_{K^c}(t)\,dt\). This map is absolutely continuous and strictly increasing: every nonempty interval contains a positive-measure part of \(K^c\), so the integral increases across it. Its image \(\phi(K)\) is null. Indeed, each complementary interval of \(K\) maps to an interval of the same length, and these image intervals are disjoint; their lengths sum to \(m(K^c)=\phi(1)\), the length of the entire range. Their complement in that range, which contains \(\phi(K)\), therefore has measure zero.
Choose a non-Lebesgue-measurable subset \(A\subset K\), and set \(E=\phi(A)\). Since \(E\subset\phi(K)\) and Lebesgue measure is complete, \(E\) is Lebesgue measurable. But strict increase gives \(\phi^{-1}(E)=A\), which is not Lebesgue measurable. Consequently \(g=\chi_E\) is Lebesgue measurable while \(g\circ\phi=\chi_A\) is not. The point is the distinction between a Borel set and a Lebesgue measurable set, not a failure of the substitution theorem for Borel integrands.
If an application begins with a Lebesgue measurable \(g\), do not automatically claim that \(g\circ\phi\) is measurable. One safe formulation is to use a Borel representative when the integral is taken with respect to the measure induced by \(\phi'\); the substitution theorem identifies that measure with Lebesgue measure on the target. If the pointwise composition is separately known to be measurable, it can also be used: a Lebesgue measurable \(g\) agrees almost everywhere with a Borel measurable representative, and the set where the representatives differ has zero induced measure by the measure identity proved above. The Borel hypothesis in the stated theorem avoids needing this additional measurability condition.
Check Your Understanding
For each question, identify the hypotheses that make substitution applicable before calculating.
- For \(\phi(x)=x^3\) on \([0,1]\) and \(g(y)=y^2\), compute both integrals in the substitution formula.
- For the piecewise map in the plateau example, compute \(\int_0^2\chi_{[0,1]}(y)\,dy\) and its transformed integral. Which part of the domain contributes zero?
- State why the substitution theorem permits a nonnegative Borel \(g\) whose integral is infinite, but requires \(\int|g|\,dy<\infty\) for its signed or complex-valued version.
- Explain why continuity of \(\phi\) ensures measurability of \(g\circ\phi\) when \(g\) is Borel measurable, and why that argument does not cover every Lebesgue measurable \(g\).