Change of Variable or Substitution in Riemann and Lebesgue Integration

By Ng Tze Beng

Because of the fact that not all derived functions are Riemann integrable (see Example 2.2.2.7 on page 139 of Theorems and Counterexamples in Mathematics by B.R. Gelbaum and JMH Olmsted), in applying the change of variable formula to Riemann integration we need to be a little careful.

Suppose F is a function which is differentiable and g is also a differentiable function. Suppose also that F and g can be composed to give the composite F ) g. Then F ) g is differentiable and by the Chain rule for differentiation,

(F ) g)' (x) = F' (g(x)) g' (x) = f (g (x)) g' (x),

where F' (x) = f (x). Thus F ) g is an antiderivative of f (g(x)) g' (x) and F is an antiderivative of f and so we can write

ò f (g (x)) g' (x)dx = F (g(x)) + C.

This is a statement about antiderivative and not about Riemann integration. For instance, f (g (x)) g' (x) need not be Riemann integrable. Here is an example when this is the case.

Example 1. Let F :[0, 1] ® R be the function as given in Example 2.2.2.7 on page 139 of Theorems and Counterexamples in Mathematics by B.R. Gelbaum and JMH Olmsted. F is differentiable on [0, 1] but the derivative F' = f is not continuous on a Cantor set of positive measure in [0, 1] and so f is not Riemann integrable. Now define g:[-1,0] ® [0,1] by g(x) = x + 1 for x in [-1, 0]. Then g is differentiable on [-1,0] and g' (x) = 1 for x in [-1,0]. We also have that f (g (x)) is not Riemann integrable since it is discontinuous on a set of positive measure which is a translation of the Cantor set of positive measure in [0,1]. This set is of positive measure because Lebesgue measure is translation invariant. Therefore f (g (x)) g' (x) has an antiderivative even though it is not Riemann integrable. Therefore, the formula      does not hold since the integrands on both sides are not Riemann integrable.

Example 2. Even if f is continuous the formula may not hold because f (g (x)) g' (x) may not be Riemann integrable simply because g' (x) is not Riemann integrable. We may take f (x) = exp(x) the exponential function and g :[0, 1] ® R be the function as given in Example 2.2.2.7 on page 139 of Theorems and Counterexamples in Mathematics by B.R. Gelbaum and JMH Olmsted. That is g' (x) is discontinuous on a Cantor set C of positive measure in [0,1]. Then it is easy to see that f (g (x)) g' (x) is also discontinuous at every point x in C using the fact that f ) g is continuous and non zero and g' (x) = 0 on C. Once again the formula

does not hold simply because the left hand side does not exist.

Now suppose g:[a, b] ® [c, d] is a differentiable function and f : [c, d] ® R is a bounded function. A necessary condition for the change of variable formula

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

to hold is that

1. f (g (x)) g' (x) is Riemann integrable on [a, b] and

2. f is Riemann integrable over a domain containing the range of g.

Theorem 1. If g: [a, b] ® [c, d] is a differentiable function and f : [c, d] ® R is Riemann integrable and has an antiderivative and if f (g (x)) g' (x) is Riemann integrable on [a, b], then formula (A) holds.

Proof. Let F be an antiderivative of f. Then F' = f . Also we have that F (g(x)) is an antiderivative of f (g (x)) g' (x) by the Chain rule. Then, since f (g (x)) g' (x) is Riemann integrable, by Darboux Theorem,

.

Since f is Riemann integrable with antiderivative F, again by Darboux Theorem,

Therefore, we have . This completes the proof.

Remark 1. 1. It is not necessary that the function g be differentiable at the end points a and b for the application of Darboux Theorem. We can replace the condition on g:[a, b] ® R by requiring that g be continuous on [a, b] and differentiable on the open interval (a, b). We then further require that the range of g be contained in the domain of f so that the composite function f Û g can be defined. Then the conclusion of Theorem 1 follows.

2. If the function f is continuous on its domain then f is Riemann integrable and has an antiderivative given by the Fundamental Theorem of Calculus. Thus we often replace the condition on f by continuity as in Theorem 2 below.

The following is the usual version of change of variable formula or substitution.

Theorem 2. If g: [a, b] ® [c, d] is a continuous function differentiable on the open interval (a, b) so that g' : (a, b) ® R is continuous and bounded and if f : [c, d] ® R is continuous, then formula (A) holds.

Proof. By assumption f ) g is continuous on [a, b] and so is bounded on (a, b). Since g' : (a, b) ® R is continuous and bounded, f (g (x)) g' (x) or ( f ) g) g' is continuous and bounded on (a, b). Therefore, ( f ) g) g' is bounded and continuous almost everywhere on [a, b] and so by Lebesgue Theorem ( f ) g) g' is Riemann integrable on [a, b]. Since f is continuous on [c, d], by the Fundamental theorem of Calculus , f has an antiderivative on [c, d], given by .. Therefore, by Theorem 1 and the remark following Theorem 1, formula (A) holds.

Now under what condition can we guarantee the Riemann integrability of ( f ) g) g' on [a, b]. Formula (A) requires that f be Riemann integrable on a domain containing the range of g. If we impose sufficient condition on g' we can deduce that ( f ) g) g' is Riemann integrable on [a, b]. It is not true in general that if f is Riemann integrable and g is continuous, then f ) g is Riemann integrable. For a counter example see Example 5 of Composition and Riemann Integrability.

Theorem 3. Suppose f : [c, d] ® R is Riemann integrable and g: [a, b] ® [c, d] is a continuously differentiable strictly increasing function mapping [a, b] onto [c, d]. Then ( f ) g) g' is Riemann integrable on [a, b] and formula (A) holds, that is,

.

Remark 2. In theorem 3 we do not require that f has an antiderivative. This includes simple step functions.

Proof of Theorem 3. Since f is Riemann integrable on [c, d], given any e > 0, there exists a partition W for [c, d],

W: c = l₀ < l₁< ... < l_n = d

such that

U(W, f ) - L(W, f ) < e/2 --------------------------- (1)

(See Theorem 1 of Riemann integral and bounded function.)

Now since g: [a, b] ® [c, d] is a strictly monotonic increasing bijective map, its inverse is also a strictly monotonic increasing bijection and so

g ^-1W: a = z₀ < z₁< ... < z_n = b

where z_i = g ^-1 l_i , i =1,2, ¼, n, is a partition for [a, b].

Because f is Riemann integrable, f is bounded on [c, d]. That is, there exists a real number M > 0 such that | f (x) | < M for all x in [c, d]. Since g' :[a, b] ®R is continuous on [a, b], g' is uniformly continuous. Therefore, given e > 0, there exists d > 0 such that

|x - y | < d Þ |g' (x) - g' (y) | < e/(2(3M+1)(b-a)) ------------ (2)

Now refine the partition g ^-1W, by adding points if need be, to get a partition Q

Q: a = x₀ < x₁< ... < x_k = b , where k ³ n,

with || Q || = max {x_i - x_i-1 : i =1, 2,¼, k} < d. This means g ^-1W Í Q. Then gQ = P is a refinement of W since W Í g(Q) = P. Let

P: c = y₀ < y₁< ... < y_k = d ,

then y_i = g (x_i), i =1, 2,¼, k. Therefore, by the definition of upper and lower Riemann sums, (see for example Riemann integral and bounded function),

L(W, f ) £ L(P, f ) £ U(P, f ) £ U(W, f )

and so we have, U(P, f ) - L(P, f ) £ U(W, f ) - L(W, f ) £ e/2, that is,

U(P, f ) - L(P, f ) < e/2. ---------------------- (3)

This means,

. ------------ (4).

Since g is differentiable, by the Mean Value Theorem, there exists x_i Î [x_i-1, x_i] such that

D y_i = y_i - y_i-1 = g (x_i ) - g (x_i-1 ) = g' (x_i )( x_i - x_i-1) = g' (x_i )D x_i ------ (5)

, g' (x_i )> 0 for i =1, 2,¼, k .

We shall now show that ( f ) g) g' satisfies Riemann's condition to conclude that it is Riemann integrable.

U(Q, ( f ) g) g' ) - L(Q, ( f ) g) g' )

=.----------------(6)

But for x , y in [x_i-1, x_i],

| f ) g (x) g' (x) - f ) g (y) g' (y)|

=|[ f ) g (x)- f ) g (y)] [g' (x) - g' (x_i )] + [ f ) g (x)- f ) g (y)] g' (x_i ) + f ) g (y)] [g' (x) - g' (y )]|

£ | f ( g (x))- f (g (y))| |g' (x) - g' (x_i )| + | f ( g (x))- f (g (y))|g' (x_i ) + | f (g (y))| |g' (x) - g' (y )|

£ 2M e/(2(3M+1)(b-a)) + | f ( g (x))- f (g (y))|g' (x_i ) + M e/(2(3M+1)(b-a))< e/(2(b-a)) + | f ( g (x))- f (g ( y))|g' (x_i ).

Therefore, for each i =1, 2,¼, k.

sup{| f ) g (x) g' (x) - f ) g (y) g' (y)|: x, y Î [x_i-1, x_i]}

< e/(2(b-a)) + sup{ | f ( g (x))- f (g ( y))|: x, y Î [x_i-1, x_i]}g' (x_i )

= e/(2(b-a)) + sup{ | f ( x)- f ( y)|: x, y Î [y_i-1, y_i]}g' (x_i )

since g is a bijection and g([x_i-1, x_i]) = [y_i-1, y_i] , where y_i = g (x_i).

Therefore,

<

= by (5)

< e/2 + e/2 = e by (4).

Therefore, U(Q, ( f ) g) g' ) - L(Q, ( f ) g) g' ) < e and so ( f ) g) g' satisfies the Riemann's condition and so it is Riemann integrable on [a, b]. (See Theorem 1 of Riemann integral and bounded function.)

We can also show that the upper and lower Riemann integrals of ( f ) g) g' are the same to conclude that ( f ) g) g' is Riemann integrable. (See Theorem 1 of Riemann integral and bounded function.) This will turn out to be a more efficient way to prove the theorem.

 

We shall use what we have just proved. Note that the upper Riemann sum of ( f ) g) g' with respect to the partition Q (as given before with ||Q|| < d) is defined to be

U(Q, ( f ) g) g' ) = .

But sup{ f ) g (x) g' (x): x Î [x_i-1, x_i]}

= sup{ f ) g (x) (g' (x) - g'(x_i))+ f ) g (x) g' (x_i): x Î [x_i-1, x_i]}

£ sup{ M e/(2(3M+1)(b-a)) + f ) g (x) g' (x_i): x Î [x_i-1, x_i]} by (2)

< e/(6(b - a)) + sup{ f ) g (x) :x Î [ x_i-1, x_i]}g' (x_i) since g' (x_i) > 0

= e/(6(b - a)) + sup{ f (x) :x Î [y_i-1, y_i]}g' (x_i) since g is a bijection.

 

Therefore,

U(Q, ( f ) g) g' ) <

= = e/6 + U(P, f )

< e/6 + L(P, f ) + e/2 by (3).

Thus using the fact the upper Riemann integral of ( f ) g) g' £ U(Q, ( f ) g) g' ) and that L(P, f ) £ lower Riemann intergral of f , we have

.

where is the upper Riemann integral of ( f ) g) g' and is the lower Riemann integral of f .

Since this is true for arbitrary e > 0, we have that

. --------------- (7)

(More precisely the upper Riemann integral of ( f ) g) g' on [a, b] £ lower Riemann integral of f on [c, d].)

Similarly, L(Q, ( f ) g) g' ) = .

But inf{ f ) g (x) g' (x): x Î [x_i-1, x_i]}

= inf{ f ) g (x) (g' (x) - g'(x_i))+ f ) g (x) g' (x_i): x Î [x_i-1, x_i]}

³ inf{- M e/(2(3M+1)(b-a)) + f ) g (x) g' (x_i): x Î [x_i-1, x_i]}

> -e/(6(b - a)) + inf{ f ) g (x) :x Î [x_i-1, x_i]}g' (x_i) since g' (x_i) > 0

= -e/(6(b - a)) + inf{ f (x) :x Î [y_i-1, y_i]}g' (x_i) since g is a bijection.

Therefore,

L(Q, ( f ) g) g' ) >

= = - e/6 + L(P, f ) --------- (8)

Hence,

by (3)

< L(Q, ( f ) g) g' ) + e/6 +e/2 by (8) above

<

using the fact that the upper Riemann integral of f £ U(P, f ) and that L(Q, ( f ) g)g') £ the lower Riemann integral of ( f ) g) g' .

Since this is true for any e > 0, . (More precisely, the upper Riemann integral of f on [c, d] £ lower Riemann integral of ( f ) g) g' on [a, b].)

Therefore, because lower Riemann integral is always less than or equal to the upper Riemann integral, we have . This then with (7) and the above inequality gives us

.

Since f is Riemann integrable on [c, d], and so

.

Hence ( f ) g) g' is Riemann integrable on [a, b] and

This completes the proof.

 

The case when g is strictly decreasing is given as follows.

Theorem 4. Suppose f : [c, d] ® R is Riemann integrable and g: [a, b] ® [c, d] is a continuously differentiable strictly decreasing function mapping [a, b] onto [c, d]. Then ( f ) g) g' is Riemann integrable on [a, b] and

The proof is exactly the same except that we use the function - f instead of f and noting that this time round, g' (x_i ) < 0 and in the equation D y_i = g' (x_i )D x_i

the points y_i are oriented in the opposite direction, that is

d = y₀ > y₁> ... > y_k = c ,

where y_i = g (x_i), i =1, 2,¼, k. Care should be exercised when taking supremum or infimum, use - g' (x_i ) > 0, for instance

sup{ f ) g (x) g' (x): x Î [x_i-1, x_i]}

= sup{ f ) g (x) (g' (x) - g'(x_i)+ f ) g (x) g' (x_i): x Î [x_i-1, x_i]}

£ sup{ M e/(2(3M+1)(b-a)) + f ) g (x) g' (x_i): x Î [x_i-1, x_i]}

< e/(6(b - a)) + sup{- f ) g (x) :x Î [x_i-1, x_i]}(-g' (x_i)) since - g' (x_i) > 0.

There is a version of Theorem 1 for Lebesgue integrals. We shall state the result as follows. Note that it is a consequence of the following theorem.

 

Theorem 5. Suppose f : [a, b] ® R is differentiable on [a, b] and its derivative f ' is bounded, that is, there exists a number K ³ 0 such that for all x in [a, b], | f ' (x)| £ K.

Then (1) f is absolutely continuous and

(2) the derived function f ': [a, b] ® R is Lebesgue integrable, and for any x in [a, b], the Lebesgue integral,

.

This is a well known result from Lebesgue integration theory. Part of the theorem is a consequence of the characterisation of functions satisfying the conclusion of Darboux Theorem or "Fundamental Theorem of Calculus" with Riemann integral replaced by Lebesgue integral and Riemann integrability replaced by Lebesgue integrability in terms of absolute continuity. This is a new concept which says that f : [a, b] ® R is absolutely continuous if for any e > 0, there exists some d > 0 such that for any finite disjoint open intervals (a₁, b₁), (a₂, b₂),¼, (a_n b_n) in [a, b] such that , then we have . Obviously absolute continuity implies continuity. If f is differentiable and has a bounded derivative it follows from the Mean Value Theorem that f is absolutely continuous on [a, b]. Thus Theorem 5 is just the characterization of functions satisfying the conclusion of the "Fundamental Theorem of Calculus" for Lebesgue integrals, with bounded derivative guaranteeing the absolute continuity of f .

  Theorem 6. If g: [a, b] ® [c, d] is a differentiable function such that its derivative is bounded and if f : [c, d] ® R is a bounded function which has an antiderivative, then we have the following equality for Lebesgue integrals.

.

 

Proof. Suppose F is an antiderivative of f . Then since its derivative f is bounded on [c, d], F is absolutely continuous. By Theorem 5, F' = f is Lebesgue integrable. Also the composite function F) g is differentiable and its derivative by the Chain Rule is given by ( f ) g)g' , which is bounded since both f ) g and g' are bounded. Therefore by Theorem 5, . This proves the theorem.

 

Remark 3. For the functions f and g as given in Example 2, by Theorem 6 formula (A) for Lebesgue integrals holds. Though ( f ) g)g' there is not Riemann integrable, it is nevertheless Lebesgue integrable.

There is a version of Theorem 3 in terms of Lebesgue integral. The difficulty with the proof lies mainly in showing that the Chain Rule is true almost everywhere unlike Theorem 6 when the Chain Rule does apply easily.

We state the theorem and then we shall describe the problem in detail.

Theorem 7. If g: [a, b] ® [c, d] is a monotone increasing (not necessarily strictly increasing) absolutely continuous function mapping [a, b] onto [c, d] and f : [c, d] ® R is a Lebesgue integrable function, then we have the following equality for Lebesgue integrals.

.

Now we set up the functions involved in the Theorem. Since f is Lebesgue integrable on [c, d], we can define the indefinite integral on [c, d] by

For any x in [c, d]. Then F: [c, d] ® R is an absolutely continuous function which is differentiable almost everywhere on [c, d] and F ' = f almost everywhere on [c, d]. Since g :[a, b] ® [c, d] is monotonic and absolutely continuous the composite function F ) g :[a, b] ® R is also absolutely continuous. We shall phrase the result in the following proposition.

 

Proposition 8. Suppose g :[a, b] ® [c, d] is monotonic and absolutely continuous and F: [c, d] ® R is absolutely continuous. Then

F ) g :[a, b] ® R is also absolutely continuous.

Proof. We shall assume that g is monotonic increasing and absolutely continuous. Since F is absolutely continuous given any e > 0, there exists d₁ > 0 such that for any finite disjoint open intervals (c₁, d₁), (c₂, d₂),¼, (c_n, d_n) in [c, d] such that

Þ . ---------------------------- (9)

Thus for such a d₁ > 0, since g is absolutely continuous, there exists d₂ > 0 such that

for any finite disjoint open intervals (a₁, b₁), (a₂, b₂),¼, (a_k , b_k) in [a, b] with , then we have . But then since g is monotonic increasing the interval (g(a_i), g(b_i)) is either empty when g(a_i) = g(b_i) or a nontrivial open interval and the intervals (g(a₁), g(b₁)), (g(a₂), g(b₂)),¼, (g(a_k), g(b_k)) are then pairwise disjoint in [c, d]. Thus by (9) .

If g is monotonic decreasing, then (g(b₁), g(a₁)), (g(b₂), g(a₂)),¼, (g(b_k), g(a_k)), are then pairwise disjoint open intervals in [c, d]. Therefore, again by (9) we can also conclude that given any e > 0, there exists d₂ > 0 such that

. Therefore, F ) g is absolutely continuous. This completes the proof.

 

Before we embark on the proof of Theorem 7, we shall establish some results that we shall use in the proof. In the following we shall be using the Vitali Covering Theorem, a very useful theorem for our purpose. The theorem is stated later for reference.

From now on we shall use the term "measure" interchangeably with "Lebesgue outer measure".

Proposition 9. Suppose g :[a, b] ® R is an absolutely continuous function. If E is a subset of [a, b] of measure zero, then its image g(E) is also of measure zero.

 

Proof. Since g is absolutely continuous given any e > 0, there exists some d > 0 such that for any finite disjoint open intervals (a₁, b₁), (a₂, b₂),¼, (a_n, b_n) with , then we have

  --------------------- (10).

At this point, it is useful to note that when proving statement about set being of measure zero, it is equivalent to proving the same of the same set minus a set of countable number of points, since such a set is of measure zero. Very often for technical reason, we may also add countable or finite number of points to it. One of the use of this device involves the fact that any open set in R is at most the union of countable open intervals, which are pairwise disjoint. What we may actually need is union of countable closed intervals which are non overlapping in the sense that any two sets in this collection can have at most one point in common. Since the measure of {g(a), g(b)} is zero, we may assume that E Í (a, b). Therefore, because E is of measure zero, for this d > 0, there exists open set I such that E Í I Í (a, b) such that the measure m(I - E) < d. Therefore, since m(E) = 0, m(I) < d. Therefore, I is the union of countable disjoint open intervals { I_i , i =1, ¼}, each of finite length. The number of open intervals I_i may be finite. Let I_i = (a_i , b_i) and its closure J_i = [a_i , b_i]. Then E Í J Í [a, b], where J = È J_i . Then g(E) Í g(J) = . Since g is continuous, and each J_i is closed and bounded, by the Extreme Value Theorem, each g(J_i) = [c_i, d_i] is also a closed and bounded interval, where c_i is the absolute minimum of g on J_i and d_i is the absolute maximum of g on J_i and there exist x_i , y_i in J_i such that g(x_i) = c_i, g(y_i) = d_i , either x_i £ y_i or y_i £ x_i. Denote the interval [x_i, y_i] or [y_i , x_i] by K_i . Then K_i Í J_i and g(K_i) = g(J_i ) for each i. Thus for any integer n,

.

Thus this implies by (10) that

Since this is true for any integer n, Thus  Since this is true for any e > 0, m(g(E)) = 0. This completes the proof.

 

The next proposition will be useful for redefining the function f.

Proposition 10. Suppose g :[a, b] ® R is absolutely continuous and monotonic increasing. If E is the set { x Î [a, b] : g'(x) = 0}, then its image g(E) is of measure zero.

Proof. If E is of measure zero, then the measure of g(E) is zero by Proposition 9. So we now consider the case where the measure of E is greater than 0. Since the measure of {g(a), g(b)} is zero, we may assume that E Í (a, b). (If need be, just remove both points a, b from E.) Let g([a, b]) = [c, d].

For each x in E, g'(x) = 0 and so given any e > 0, there are arbitrary small intervals [x, x + h] such that where h > 0. We may assume without loss of generality that each [x, x + h] is contained in (a, b). Then these arbitrary small intervals form a Vitali covering of E. Choose d > 0 for the given e > 0 satisfying (10) in the definition of absolute continuity since g is absolutely continuous. Then by the Vitali Covering Theorem, there is a finite disjoint collection of these intervals I_i = [x_i, y_i], i = 1,2,¼,n, in [a, b] such that the measure of E - is less than d. That is I_i , i = 1,2,¼, n covers all of E except for the subset of E, E - of measure less than d. We order these intervals so that

a < x₁ < y₁ £ x₂ < y₂ £ x₃ < ¼ £ x_n < y_n < b.

Then the measure

Now (a, b) - is an open set containing E' = E - . Let k = m(E'). Then since k = there exists open set J containing E' such that the measure m(J - E') < . Thus m(J) = m(J-E')+m(JÇE') < We may assume that J Í (a, b) - . Since J is open it is a union of countable pairwise disjoint intervals J_i = (a_i, b_i) i = 1,2,¼. Then Hence . This gives us

Since e is arbitrary this implies that m(g(E)) = 0.

We shall prove the assertion we make previously, namely the following proposition.

Proposition 11. Suppose g :[a, b] ® R is an absolutely continuous and monotonic increasing function. Suppose the range of g is [c, d] and E is a subset of [c, d] of measure zero. Let H = { x Î [a, b] : g'(x) ¹ 0}. Then the measure of g^-1(E) Ç H is zero.

Proof. Let H_n = { x Î [a, b] : |g'(x)| > 1/n}. Then . We shall show that the measure of g^-1(E) Ç H_n = E_n is zero for each positive integer n.

Suppose on the contrary that the measure m(g^-1(E) Ç H_n) = m(E_n) = k > 0. Now since g(E_n) Í E and E is of measure zero, the measure m(g(E_n)) is 0. Thus given any e' > 0, there exists an open set G' containing g(E_n) such that m(G') < e' . Then G = g^-1(G') is an open set containing E_n . For each x in E_n there exists arbitrary small interval [x, x+h] such that or

, ----------------------- (11)

where h > 0.

We may assume that these intervals are in G. Therefore, this collection forms a Vitali covering for E_n . Then by the Vitali Covering Theorem, given any e > 0 there exists a finite disjoint intervals I_i , i = 1,2,¼, N so that I = I₁ È I₂ È ¼ È I_N covers a subset of E_n of outer measure > k - e. But if we write I_i = [x_i, x_i + h_i] , then (11) implies that

But since g(I ) Í G' and g is monotonic increasing ,

. So if we choose e' = k/(2n) and e = k/2, we would obtain a contradiction. Hence m(E_n) = 0. Since g^-1(E) Ç H_n = E_n so that g^-1(E) Ç H = and so the measure since m(E_n) = 0. That means m(g^-1(E) Ç H)=0. This completes the proof.

Similar result when g is a decreasing function also hold.

 Proposition 12. Suppose g :[a, b] ® R is an absolutely continuous and monotonic decreasing function. If E = { x Î [a, b] : g'(x) = 0}, then its image g(E) is of measure zero.

Proof. If g is decreasing, then - g is increasing. Then Proposition 10 says that - g(E) is of measure zero. Since measure is invariant under reflection, g(E) is also of zero measure.

Proposition 13. Suppose g :[a, b] ® R is an absolutely continuous and monotonic decreasing function. Suppose the range of g is [c, d] and E is a subset of [c, d] of measure zero. Let H = { x Î [a, b] : g'(x) ¹ 0}. Then the measure of g^-1(E) Ç H is zero.

Proof. If g maps onto [c, d], then -g maps onto [-d, -c]. Suppose E is a subset of [c, d] of zero measure, then -E is a subset of [-d, -c] of zero measure. Then by Proposition 11 (-g)^-1 (-E)Ç H' , where H' = { x Î [a, b] : (-g)'(x) ¹ 0}, is of measure zero. But H' = H and (-g)^-1 (-E) = g^-1(E) and so the measure of g^-1(E) Ç H is zero.

Proposition 14. Suppose g :[a, b] ® R is absolutely continuous and monotonic. Suppose the range of g is [c, d] and F: [c, d] ® R is an absolutely continuous function. Let E = { x Î [a, b] : (F ) g)'(x) ¹ 0 and g'(x) = 0}. Then the measure of E is zero.

Proof. Let E_n = { x Î [a, b] : |(F ) g)'(x)| > 1/n and g'(x) = 0 }. Then . We shall show that the measure of E_n is zero for each positive integer n. We may assume as usual that E_n Í (a, b).

Suppose on the contrary that the measure m(E_n) = k > 0.

Now we use the condition that the derivative is zero at every point of E_n . For any x in E_n , since g'(x) = 0, given any K > 0 there exists arbitrary small h > 0 such that

------------------- (12)

We are going to choose a suitable K to give a contradiction. The choice of K will depend on k, n and the absolute continuity of F.

Given any e' > 0, since m(E_n) = k , there exists open set G containing E_n such that the measure of G , m(G) < k + e' . We shall choose our e' carefully.

For each x in E_n, since (F ) g)'(x) ¹ 0, there exists arbitrary small interval [x, x+h] such that

          or

, ------------------------------ (13)

where h > 0.

We may assume that these intervals are in G. Thus (12) and (13) hold simultaneously for these arbitrary small intervals [x, x+h]. Therefore, this collection forms a Vitali covering for E_n . Then by the Vitali Covering Theorem, there exists a finite number of pairwise disjoint intervals I_i , i = 1,2,¼, N so that I = I₁ È I₂ È ¼ È I_N covers a subset of E_n of outer measure > k - e' But if we write I_i = [x_i, x_i + h_i] , then (12) implies that

------------------ (14)

and (5) gives

--------------------- (15)

Take e' = k/2, Then .

By definition of the absolute continuity of F, there exists d > 0 such that for any finite disjoint number of open intervals (a₁, b₁), (a₂, b₂),¼, (a_N , b_N) in [c, d]

----------------------------- (16)

Now choose K > 0 so that K(k + e') = K(3k/2) < d. Then (14) implies that . Therefore, since g is monotonic, if g is increasing (g(x_i), g(x_i+h_i)), i =1 ,2, ¼,n are pairwise disjoint or degenerate (i.e., g(x_i) = g(x_i+h_i)), and if g is decreasing (g(x_i+h_i), g(x_i)), i =1 ,2, ¼,n are pairwise disjoint or degenerate. Therefore, by (16) we have

But this contradicts (15). Hence, m(E_n) = 0 for each positive integer n. Consequently

the measure . That means m(E) = 0. This completes the proof.

Before we proceed with the proof of Theorem 7, we state the Vitali Covering Theorem.

Definition 15. For a subset E of R, a collection C of intervals is said to cover E in the sense of Vitali if given x in E, and any e > 0, there is an interval J in C with xÎ J and 0 < m(J) < e, where m(J) = m(J) is the length of J.

Theorem 16. Vitali Covering Theorem.

Suppose E Í R has finite Lebesgue outer measure and is covered in the sense of Vitali by a class C of intervals. Then there is a countable disjoint subclass J Í C such that

the outer measure m( E - È{I: I Î J}) = 0.

The following more useful form of the theorem provides a finite covering that covers enough of the set E for use.

Corollary 17. Suppose E Í R has finite Lebesgue outer measure and is covered in the sense of Vitali by a class C of intervals. Then given any e > 0, there is a finite set J₁ , J₂, ¼, J_n of disjoint intervals of C such that

For the proof of Theorem 16 see page 225 of "Introduction to Measure and Integration" by S.J. Taylor or page 98 of "Real Analysis" by H.L. Royden.

We now summarize the idea of the proof of Theorem 7 in the following Venn diagram.

Proof of Theorem 7.

 Now if F: [c, d] ® R is is given by and g :[a, b] ® [c, d] is monotonic increasing, surjective and absolutely continuous, then the composite F ) g :[a, b] ® R by Proposition 8 is also absolutely continuous and consequently the Lebesgue integral

,, since g(a) = c and g(b) = d. Then we may want to ask the question: When is

(F) g)' (x) = F'(g(x)) g'(x) = f (g(x))g'(x) almost everywhere on [a, b]? It turns out that this is not an easy question. To prove the theorem, we need to answer this question in the affirmative and the Venn diagram above illustrates the situation as we proceed.

 Because F) g is absolutely continuous and hence of bounded variation on [a, b], F) g is differentiable almost everywhere on [a, b]. Since g is monotonic increasing, g is also differentiable almost everywhere on [a, b]. Thus it is enough to consider subset K of [a, b], where both (F) g)'(x) and g'(x) exist for every x in K, because the complement of K in [a, b] is of measure zero. If x is in K , either g'(x) = 0 or g'(x) ¹ 0. Now suppose x is in K and g'(x) ¹ 0. Then since (F) g)' (x) exists, F'(g(x)) exists. This is seen as follows. For h ¹ 0, if g(x + h) - g(x) ¹ 0, then

Consider the function Then since , there exists d > 0 such that 0 < |h| < d implies G(h) ¹ 0 and so k(h) = g(x + h) - g(x) ¹ 0. Thus for 0 < |h| < d, we have Hence exists say equals to L. That means given e > 0, there exists d > 0 such that 0 < |h| < d Þ k(h) ¹ 0 and . Hence on the interval (- d, d), k(h) is monotonic increasing and since k(x) = 0 if and only if x = 0, k((- d, d)) contains an open interval containing 0. Thus, there exists d' > 0 such that (-d', d') Í k((- d, d)). Therefore, this implies that for every p in (-d', d') there exists a h in (- d, d) such that k(h) = p. Therefore, 0 < |p| < d' implies that

. This shows that exists and so F is differentiable at g(x). This means that if x is in K and g'(x) ¹ 0, then the chain rule (F) g)' (x) = F'(g(x)) g'(x) holds. Now if for x in K, if g'(x) = 0 and if F'(g(x)) exists, then we have the chain rule and so (F) g)' (x) = F'(g(x)) g'(x) = 0. On the other hand if g'(x) = 0 and (F) g)' (x) = 0, the equality still holds and there is no contribution to the Lebesgue integrals on both sides. It remains to check the case when g'(x) = 0 and (F) g)' (x) ¹ 0. Of course this means g'(x) = 0 and (F) g)' (x) ¹ 0 and F'(g(x)) does not exist for its existence would imply that (F) g)' (x) = 0. By Proposition 14 the set { x Î [a, b]: g'(x) = 0, (F) g)' (x) ¹ 0 (therefore F'(g(x)) does not exist)} has measure zero. We shall also show that the set {x Î [a, b]: g'(x) ¹ 0, and F'(g(x)) ¹ f (g(x)) } has measure zero. If F'(g(x)) ¹ f (g(x)), then g(x) belongs to a set of measure zero since F' = f almost everywhere on [c, d]. Therefore, the set {x Î [a, b]: g'(x) ¹ 0, and F'(g(x)) ¹ f (g(x))}Í g^-1(E) Ç H, where E is a set of measure zero and H={x Î [a, b]: g'(x) ¹ 0}. Thus, by Proposition 11, {x Î [a, b]: g'(x) ¹ 0, and F'(g(x)) ¹ f (g(x))} has measure zero. Thus we have

almost everywhere on [a, b]

almost everywhere on [a, b]

almost everywhere on [a, b]

almost everywhere on [a, b]

because the measure of {x Î[a, b]: g' (x) ¹ 0}Çg^-1(E) is zero where E is a set of measure zero (by Proposition 11)

Therefore,

.

From the above proof itself, we can easily construct a similar proof for the case when g is an absolutely continuous decreasing function. Hence, we state the following theorem for record.

Theorem 18. If g: [a, b] ® [c, d] is a monotone decreasing (not necessarily strictly decreasing) absolutely continuous function mapping [a, b] onto [c, d] and f : [c, d] ® R is a Lebesgue integrable function, then we have the following equality for Lebesgue integrals.

.

 

Remark 4. It is clear that since any Riemann integrable function is also Lebesgue integrable and the two integrals are the same, Theorem 3 is a special case of Theorem 7 because any g: [a, b] ® [c, d] which is monotonic and continuosly differentiable on [c, d] has bounded derivative and so is absolutely continuous. Indeed we may replace the requirement that g be strictly increasing by just increasing, courtesy of Theorem 7.

For simplicity and for the reader who may not be familiar with measure theory, we choose to present the two proofs. It is of course possible to modify the proof of Theorem 3 by not requiring the strict monotonicity on g. Likewise Theorem 4 is a special case of Theorem 18.

 

Remark 5. It is also possible to replace the function f by another function equal to f almost everywhere. We assume the hypothesis of Theorem 7, that is, g: [a, b] ® [c, d] is a monotone increasing (not necessarily strictly increasing) absolutely continuous function mapping [a, b] onto [c, d] and f : [c, d] ® R is a Lebesgue integrable function. By Proposition 12, the measure m(E) is zero, where E = { x Î [a, b] : g'(x) = 0}. Now we can define f _E : [c, d] ® R by f _E(y) = y if y is not in g(E) and f _E(y) = 0 if y is in g(E). Then f _E = f almost everywhere on [c, d] and we have

.

The conclusion of Theorem 7 need not be true if we drop the monotonicity condition on g as illustrated in the following example.

Example 3. Let g:[0, 1] ® R be defined by and Let f :[0, 1] ® R be defined by . Then f is not bounded on [0, 1]. The function f is Lebesgue integrable and the integral is given by the improper Riemann integral F :[0, 1] ® R defined by . Then the composite function F ) g is not absolutely continuous even though both F and g are. This is because and is obviously not of bounded variation and so cannot be absolutely continuous. But of course since F is differentiable on (0, 1] and g is differentiable on [0, 1], (F ) g)'(x) = f (g(x))g'(x) almost everywhere on [0, 1]. Thus . But since F ) g is not absolutely continuous, therefore there exists x such that

.

Hence the change of variable formula does not hold.

Thus, in view of the fact that f is unbounded in the above example, we consider function f which is bounded and Lebesgue integrable on the interval [a, b]. We can now obtain the change of variable formula even if g is not monotonic but still absolutely continuous. The full generality is not so easy to observe and will involve some subtle technical results in measure theory and also of functions of bounded variation or having absolute continuity. We shall state a weaker version of the theorem, which is mostly what we need.

Theorem 19. Suppose g: [a, b] ® R is an absolutely continuous function and f : [c, d] ® R is a continuous function such that the range of g is contained in [c, d]. Then we have the following equality for Lebesgue integrals.

.

We shall need the following technical result.

Proposition 20. Suppose f : [c, d] ® R is a bounded and Lebesgue integrable function. Define F : [c, d] ® R by . Then F satisfies a Lipschitz condition, that is, there is a constant K such that |F(y) - F(x)| £ K| y - x| for all x and y in [c, d]. Equivalently, if F is absolutely continuous and F' is bounded, then F satisfies a Lipschitz condition.

 

Proof. If F is absolutely continuous on [c, d], then for all x in [c, d]. Since F' is bounded almost everywhere in [c, d], there exists a K such that |F'(x)| < K almost everywhere on [c, d]. Then for any y > x in [c, d] Thus for any y > x,

|F(y) -F(x)| = .

Therefore, F satisfies a Lipschitz condition with constant K.

A crucial step in proving Theorem 19 is to show that the composite function F) g is absolutely continuous when g is absolutely continuous and F is also absolutely continuous but having bounded derivative. Indeed this is a consequence of the following proposition.

Proposition 21. Suppose g: [a, b] ® R is an absolutely continuous function and F : [c, d] ® R is an absolutely continuous function satisfying a Lipschitz condition with constant K, such that the range of g is contained in [c, d]. Then F) g is absolutely continuous.

Proof. Since g is absolutely continuous, given any e > 0, there exists d > 0 such that for any disjoint open intervals (a₁, b₁), (a₂, b₂),¼, (a_n b_n) in [a, b],

Then for each i =1, ¼, n, since F satisfies a Lipschitz condition with constant K,

|F(g(b_i)) - F(g(a_i))| £ K|g(b_i) - g(a_i)|.

Therefore, taking summation,

.

Hence, . Therefore, F) g is absolutely continuous.

 

Proof of Theorem 19. Since f : [c, d] ® R is continuous, it is therefore bounded and Lebesgue integrable. Therefore, the function F : [c, d] ® R defined by is absolutely continuous and satisfies a Lipschitz condition. It follows by Proposition 21, that F) g is absolutely continuous. Hence we have

.

Now F is differentiable everywhere on [c, d] by the Fundamental Theorem of calculus since f is continuous on [c, d]. Note that g is differentiable almost every where on [a, b], since g is absolutely continuous. Therefore, F) g is differentiable almost everywhere on [a, b] because if g is differentiable at x and since F is differentiable at g(x), F) g is differentiable at x. In particular, (F) g)'(x) = F'(g(x)) g'(x) = f (g(x)) g'(x) almost everywhere on [a, b]. Therefore, . This completes the proof.

Now if f : [c, d] ® R is not continuous but just bounded and Lebesgue integrable. Then the question remains if (F) g)'(x) = f (g(x)) g'(x) almost everywhere.

We shall need the analogue of Proposition 10, Proposition 11 and Proposition 14.

First we recall the definition of the total variation function of a function of bounded variation. A function g is said to be of bounded variation on [a, b] if the total variation

V_g[a, b] =

exists or equivalently there exists a constant K > 0 such that for any partition If g: [a, b] ® R is an absolutely continuous function, then the total variation function V_g : [a, b] ® R is defined by V_g(x) = 0 for x = a and V_g(x) = V_g[a, x], the total variation of the function g on [a, x] for x in (a, x]. This is well defined because for any partition a = x₀ < x₁ < ¼ < x_n = x, so that

exists.

The following results concerning the properties of function of bounded variation and of absolutely continuous functions can be found on page 267 to page 269 in "Principles of Real Analysis" by C.D. Aliprantis and Owen Burkinshaw or in my article "Monotone Function, Function of Bounded Variation, Fundamental Theorem of Calculus" .

 

Theorem 22.

1. g: [a, b] ® R is a function of bounded variation if and only if g is the difference of two increasing functions.

2. Any function g: [a, b] ® R of bounded variation is differentiable almost everywhere on [a, b].

3. If g: [a, b] ® R is of bounded variation, then the variation function V_g : [a, b] ® R and the function V_g - g are both increasing functions.

4. If g: [a, b] ® R is of bounded variation, then for any x and y in [a, b] with

a £ x £ y £ b we have |g(y) - g(x)| £V_g (y) -V_g (x) . In particular, |g(x) - g(a)| £ V_g (x).

The next result concerns absolutely continuous functions.

Theorem 23. Suppose g: [a, b] ® R is absolutely continuous on [a, b]. Then the following statement holds.

1. g is of bounded variation.

2. The variation function V_g : [a, b] ® R is also absolutely continuous and so g is the difference of two increasing absolutely continuous functions.

We shall need the following technical result about increasing function.

 

Proposition 24. Suppose g: [a, b] ® R is an increasing function.

Then g is differentiable almost everywhere and the derivative g' is Lebesgue integrable (therefore measurable) and for any x in [a, b].

The proof of Proposition 24 can be found on page 100 of Royden's "Real Analysis".

 

We shall need the following consequence of absolute continuity.

 

Proposition 25. Suppose g: [a, b] ® R is absolutely continuous on [a, b]. Then for any x in [a, b],

 

Proof. By Theorem 22 part 4, for any y ¹ x in [a, b], (V_g(y)-V_g(x))/(y-x) ³ (g(y) - g(x))/(y-x). Consequently, V_g'(x) ³ g'(x) almost everywhere on [a, b]. We also have , for any y ¹ x in [a, b], (V_g(y)-V_g(x))/(y-x) ³ -(g(y) - g(x))/(y-x) and so we have V_g'(x) ³ - g'(x) almost everywhere on [a, b]. Therefore, V_g'(x) ³ |g'(x)| almost everywhere on [a, b]. Then

The last inequality is by Proposition 24 since the function V_g is an increasing function.

On the other hand for any x in (a, b] and for any partition of [a, x], a = x₀ < x₁ < ¼ < x_n = x, since g is absolutely continuous (and therefore the fundamental theorem for Lebesgue integral holds),

.

Therefore,

.

It follows that . Therefore, .

Corollary 26. Suppose g: [a, b] ® R is absolutely continuous on [a, b]. Then

almost everywhere on [a, b].

 

Proof, Since , therefore almost eveywhere on [a, b]. The last equality is the differentiation of an indefinite integral. See for example Theorem 10 on page 107 in Royden's "Real Analysis" or see the reference to Theorem 2 of my article "Integration by Parts".

 

Remark 6. The conclusion of Corollary 26 is actually true just for function of bounded variation defined on [a, b]. But the proof is much more delicate, it requires a careful handling of how the finite difference g(x_i) - g(x_i-1) changes sign. One proof of this is to choose a dissection of [a, b] so that the summation differs from the total variation by say 1/2ⁿ. The choice is such that the series converges. Define a function g_n