Change of Variables Theorems

by Ng Tze Beng

Suppose g: [a, b] ® R is a function and f : [c, d] ® R is another function. Suppose the range of g is contained in [c, d]. The change of variable formula is a formula of the following form

--------------------------------- (A)

for any x in the interval [a, b]. This formula is usually used to find the integral whenever it is possible to express the integrand H(t) in the form of f ( g(t) )g' (t), where f is a function whose primitive is known or f is equal almost everywhere to the derivative of an absolutely continuous function. The change of variable formula (A) requires that f be Lebesgue integrable on the domain containing the range of g and that the function f ( g(t) )g' (t) be Lebesgue integrable on [a, x] for every x in [a, b]. Now assume that f is Lebesgue integrable on [c, d] and define F : [c, d] ® R by . Then we can write (A) as

--------------------------- (B)

for every x in [a, b]. Thus (B) is equivalent to (A). Assume the function f ( g(t) )g' (t) is Lebesgue integrable on [a, b], hence on [a, x] for every x in [a, b]. By virtue of the left hand side of (B) being an indefinite integral of a Lebesgue integrable function, the function F ) g is absolutely continuous on [a, b]. Conversely suppose that F ) g is absolutely continuous on [a, b]. Then F ) g is of bounded variation and so has finite derivatives almost everywhere on [a, b]. F being an indefinite integral of a Lebesgue integrable function is absolutely continuous and so is an N function and has finite derivatives almost everywhere on [c, d]. If g has finite derivative almost everywhere on [a, b], then by Theorem 3 below (F ) g )' (x) = f (g(x)) g'(x) almost every where on [a, b]. Consequently since F ) g is absolutely continuous on [a, b], f (g(x))g'(x) is Lebesgue integrable on [a, b] and (B) holds.

Thus we have deduced the following theorem.

Theorem 1. Suppose g: [a, b] ® R is a function having finite derivatives almost everywhere on [a, b] and f : [c, d] ® R is a Lebesgue integrable function such that the range of g is contained in [c, d]. Let F : [c, d] ® R be defined by . Then f (g(x)) g'(x) is Lebesgue integrable on [a, b] and if and only if F ) g is absolutely continuous on [a, b].

Remark. In the proof of the converse of Theorem 1, we make use of the fact that g has finite derivatives almost everywhere on [a, b]. Question arises if there are functions f and g not having finite derivatives almost everywhere on [a, b] such that F ) g is absolutely continuous on [a, b] but (F ) g )' (x) ¹ f (g(x)) g'(x) almost every where on [a, b].

We shall need the next technical result regarding critical points in the inverse image of a set of measure zero.

Theorem 2. Suppose g has derivatives (finite or infinite) on a set E with m(g(E)) = 0. Then g' = 0 almost everywhere on E.

Proof.

Let B = { t Î E : |g'(t)| > 0}. Let C_n ={ t Î B : |g'(t)| ³ 1/n } and

B_n ={ t Î B : |g(s) - g (t)| ³ |s - t|/n , |s - t| < 1/n }for positive integers n.

It is easy to see that B = È C_n. Note that for each x in C_n , there exists k such that x Î B_k . This is because if x is in C_n  ,then either |g'(t)| ³ 1/n when g'(t) is finite, or g'(t) is infinite. If g'(t) is finite, then there exists d > 0 such that

for 0 < |s - x| < d. Take any integer k such that k > 2n and 1/k < d. Then we have

0 < |s - x| < 1/k Þ .

This means that x Î B_k . If |g'(t)| is infinite then there exists d > 0 such that

for 0 < |s - x| < d. In this case, just take any positive integer k such that 1/k < d. Then we have

0 < |s - x| < 1/k Þ .

Hence x Î B_k . This implies that B = È C_n Í È B_n Í B and so B = È B_n. . We shall show that the measure of B is 0 by showing that the measure of B_n is zero. Fix an integer n and consider any interval I of length 1/n and its intersection with B_n , A = IÇ B_n . We claim that the measure of A is zero. Since m(g(E)) = 0, m(g(B)) = 0 and so m(g(A)) = 0. Given any e > 0, cover g(A) by a countable union of disjoint interval I_k such that g(A) =È I_k and å m(I_k) < e. Let A_k =g ^-1(I_k)ÇA . Then A = È A_k and

--------------------------- (1)

Note here that exists because A_k is bounded as A is bounded. Observe that A_k Í IÇ B_n Í I and I is an interval of length less than 1/n and so for any s, t in A_k , |s - t| < 1/n . Thus by the definition of B_n , for s, t in A_k , |s - t| £ n |g(s) - g(t)|. Thus

as g(A_k) Í I_k . It then follows from (1) that

.

Since e is arbitrary, we conclude that m(A) = 0. We can cover B_n by a countable non-overlapping intervals I of length < 1/n. Thus by the above argument the measure of B_n is less than the sum of measure of sets of of measure zero and so is of measure zero. It follows that the measure of B is 0. Therefore, g' = 0 almost everywhere on E.

Theorem 3. Suppose F has finite derivatives almost everywhere on [c, d] and g and F ) g have finite derivatives almost everywhere on [a, b]. It is assumed that the range of g is contained in [c, d]. Suppose F is an N-function, i.e., F maps sets of measure zero to sets of measure zero. Then (F ) g )' = ( f )g ) g' almost everywhere on [a, b], where F' = f almost everywhere on [c, d], that is to say, the chain rule holds almost everywhere on [a, b].

Proof. Let Z = {x Î [c, d]: F'(x) does not exist or F'(x) = ± ¥ or F'(x) ¹ f (x)}.

Since F has finite derivative almost everywhere on [c, d], the set {x Î [c, d]: F'(x) does not exist or F'(x) = ± ¥} has measure zero. Also as F' = f almost everywhere on [c, d], the set {x Î [c, d]: F'(x) ¹ f (x)} has measure zero consequently, the measure of Z m(Z) = 0.

Consider the preimage S = g ^-1(Z) of Z. Let T =[a, b] - S be the complement of S in [a, b] Then for any t in T, g(t) Ï Z and so F' (g(t)) exists, is finite and F'(g(t) = f (g(t)).

Now since g has finite derivatives almost everywhere on [a, b], for any t in [a, b], either g'(t) exists finitely or t belongs to a set of measure zero. Thus we need to consider only those t in [a, b] where g'(t) exists finitely.

Suppose t is in T and g is differentiable at t (finitely). Then F is differentiable at g(t) and so by the usual chain rule, (F ) g )' (t) = F' (g(t)) g' (t) = f (g(t)) g' (t). This means (F ) g )' = ( f )g ) g' almost everywhere on T.

Now we consider the chain rule on S. Since g(S) Í Z , m(g(S)) = 0. Then since g is differentiable (finitely) almost everywhere on S by considering points in S where g is differentiable finitely, by Theorem 2, g' = 0 almost everywhere on S. Hence ( f )g ) g' = 0 almost everywhere on S. Since F is an N function m(F ) g(S) ) = 0. Since F ) g has finite derivatives on [a, b], F ) g is differentiable almost everywhere on S. Therefore, by Theorem 2, (F ) g)' = 0 almost everywhere on S. Hence (F ) g )' = ( f )g ) g' (=0) almost everywhere on S.

Thus we have shown that (F ) g )' = ( f )g ) g' almost everywhere on S and on T and so (F ) g )' = ( f )g ) g' almost everywhere on [a, b]. This completes the proof.

Suppose f : [c, d] ® R is Lebesgue integrable and the range of g is contained in [c, d]. Let F : [c, d] ® R be defined by . Then F is absolutely continuous and so is an N function and also a function of bounded variation and thus has finite derivatives almost everywhere on [c, d]. If g and F ) g are of bounded variation, then (F ) g )' = ( f )g ) g' almost everywhere on [a, b]. We record this conclusion below.

Corollary 4. Suppose f : [c, d] ® R is Lebesgue integrable, g is a function of bounded variation whose range is contained in [c, d]. Let F : [c, d] ® R defined by . If F ) g is of bounded variation then (F ) g )' = ( f )g ) g' almost everywhere on [a, b].

Note that if f : [c, d] ® R is Lebesgue integrable, then is an N function and has finite derivatives almost everywhere on [c, d]. Thus by Theorem 3 we have the following simple deduction of the chain rule holding almost everywhere on [a, b].

Corollary 5. If g and F ) g have finite derivatives almost everywhere on [a, b] and F is absolutely continuous, then the chain rule holds almost everywhere on [a, b].

Corollary 6. If g is monotone, F ) g has finite derivatives almost everywhere on [a, b] and F is absolutely continuous, then the chain rule holds almost everywhere on [a, b].

We have relied on some results concerning absolutely continuous functions. For convenience we state the results here as the next theorem.

Theorem 7. Suppose F is an absolutely continuous function on [a, b].

Then (i) F is a continuous function of bounded variation,

(ii) F is an N function, i.e., it maps sets of measure zero to sets of measure zero,

(iii) F is differentiable (finitely) almost everywhere and F' is Lebesgue integrable.

Moreover, (i) and (ii) implies that F is absolutely continuous (Banach-Zarecki).

If F is continuous and satisfies (ii) and (iii), then F is absolutely continuous.

In particular F is absolutely continuous if and only if it is the indefinite integral of its derivative. Thus the indefinite integral of a Lebesgue integrable function is absolutely continuous.

A very good reference for this theorem is the article "On Absolutely Continuous Functions" , (The American Mathematical Monthly vol 72 (1963), pp. 831-841) by Dale Varberg.

Now we can prove easily

Theorem 8. Suppose g: [a, b] ® R is an absolutely continuous function and let f : [c, d] ® R be a bounded Lebesgue integrable function such that the range of g is contained in [c, d]. Then we have the following equality for Lebesgue integrals.

.

Proof. F : [c, d] ® R defined by is absolutely continuous. Since f is bounded F is also Lipschitz. Therefore, F Û g is absolutely continuous on [a, b] (see Proposition 21 of my article "Change of variable ot substitution in Riemann and Lebesgue integration). Then Theorem 8 follows from Theorem 1.

Remark. Theorem 8 is Theorem 31 of "Change of Variable or Substitution in Riemann and Lebesgue Integration", where I have proved this using a weaker version of Theorem 2 and Theorem 3 for absolutely continuous functions instead of functions having only finite derivatives almost everywhere.

If we drop the condition that f be bounded, we then have the following theorem.

Theorem 9. Suppose g: [a, b] ® R is an absolutely continuous function and f : [c, d] ® R is a Lebesgue integrable function such that the range of g is contained in [c, d] and ( f Û g ) g ' is Lebesgue integrable on [a, b]. Then we have the change of variable formula for Lebesgue integral.

.

Proof. Write f as f ⁺ - f ^- , where f ⁺ and f ^- are the positive and negative parts of f , i.e., f ⁺(x) = max{0, f (x)} and f ^- (x) = - min{0, f (x)}. Then f is Lebesgue integrable if and only if f ⁺ and f ^- are Lebesgue integrable. Then for each positive integer n , f ⁺_n = min {n, f ⁺} is Lebesgue integrable and f ⁺_n converges pointwise to f ⁺ on [c, d]. Obviously each f ⁺_n is bounded. Similarly f ^-_n = min {n, f ^-} is Lebesgue integrable on [c, d] and converges pointwise to f ^- on [c, d]. Then h_n = f ⁺_n - f ^-_n is bounded, Lebesgue integrable and converges pointwise to f ⁺ - f ^- = f . Then by Theorem 8, ( h_n )g ) g' is Lebesgue integrable on [a, b] and

---------------------- (1)

Since ( h_n )g ) g' converges pointwise to ( f )g ) g' which is Lebesgue integrable and so by the Lebesgue Dominated convergnce Theorem, the left hand side of (1) converges to . Also since h_n converges pointwise to f and f is Lebesgue integrable, by the Lebesgue Dominated Convergence Theorem again, converges to . Consequently, .

Corollary 10. Suppose g: [a, b] ® R is an absolutely continuous function and f : [c, d] ® R is a Lebesgue integrable function such that the range of g is contained in [c, d] and ( f Û g ) g ' is Lebesgue integrable on [a, b]. Then F Û g is absolutely continuous on [a, b]

Theorem 11. Suppose g: [a, b] ® R is of bounded variation on [a, b] and f : [c, d] ® R is a Lebesgue integrable function such that the range of g is contained in [c, d] and F Û g is absolutely continuous on [a, b], where F is defined as above. Then we have the change of variable formula for Lebesgue integral.

.

Proof. Since g is of bounded variation on [a, b], g has finite derivatives almost everywhere on [a, b]. The theorem then follows from Theorem 1.