Functions Having Finite Derivatives, Bounded Variation, Absolute Continuity and the Banach Zarecki Theorem

by Ng Tze Beng

We shall begin by examining the properties of the image under a function f of a set in which f has finite derivatives that are bounded by a constant. The first property we examine is the relation between the measure of such a set and the measure of its image. We state this property in the next theorem.

This result appears in Saks monograph on the theory of the integral and there are a number of proofs of the result. But I shall present a proof using some finiteness argument, a consequence of compactness and the triangle inequality.

Theorem 1. Suppose f : [a, b] ® R is a function. Suppose E is a subset of [a, b] such that at each point x of E, f is differentiable and | f ' (x)| £ K for some constant K ³ 0. Then if m denotes the Lebesgue outer measure,

m( f (E)) £ K m(E) ---------------------- (A)

Proof . Now E = { x Î [a, b] : | f ' (x)| £ K } Í [a, b] and so E has finite outer measure. If E is finite or denumerable, then the set f (E) is at most denumerable and so both m( f (E)) and m(E) are zero and we have nothing to prove since both sides of the inequality are zero. We shall now assume that E is uncountably infinite. We may assume that neither a nor b is in E since adding any finite number of points to E will not alter the inequality (A).

For any e > 0, by the definition of outer measure, there exists an open set U in [a, b] such that U Ê E and m(U) £ m(E) + e.

Since for each e in E, | f ' (x)| £ K , for e > o there exists a d _e > 0 such that

.

and so

.

Thus we have,

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

Since U is open, we may choose d _e > 0 such that the open interval (e - d _e , e + d _e) Í U. Denote (e - d _e , e + d _e) by I _e . Then inequality (1) says that

----------------- (2).

Then the collection C = { I_e : e Î E} covers E and the union W = È {V : V Î C} = È { I _e : e Î E} Í U. In particular the union W is open and so is a disjoint union of countable number of open intervals, i.e.,

,

where B the index set is a subset of the set N of natural numbers and each U_i is an open interval. We shall show next that for each i in B,

m( f (U_i Ç E)) £ (K+e) m(U_i).-------------------------- (3)

Note that U_i = È{ I _e : e Î U_i Ç E }. Observe that each U_i is a path component of W.

Plainly for e Î U_i Ç E , I _e Ç U_i ¹ Æ and since I _e Í W and U_i is a path component of W, I _e Í U_i. It follows that È{ I _e : e Î U_i Ç E } Í U_i . For any x in U_i , x Î I _e for some e in E, since W = È { I _e : e Î E} and so I _e Ç U_i ¹ Æ . It follows as in the above argument that I _e Í U_i and so e Î U_i Ç E . Thus, x Î I _e for some e Î U_i Ç E, that is, x Î È{ I _e : e Î U_i Ç E } and so U_i Í È{ I _e : e Î U_i Ç E }. This proves that

U_i = È{ I _e : e Î U_i Ç E }.

Now take any x < y in U_i . Since U_i is an open interval, the closed and bounded interval [x, y] is contained in U_i . Now plainly the collection B = { I _e : e Î U_i Ç E } is an open cover for [x, y]. Since [x, y] is compact, there exists a finite subcover say

I₁ , I₂ , ¼ , I_n

where I_i = (e_i - d(e_i), e_i + d(e_i), for some e_i in E and d(e_i) is as given in (1). We assume that the e_i 's are ordered in an increasing order. Hence

[x, y] Í I₁ È I₂ È ¼ ÈI_n

and e₁ < e₂ < ¼ < e_n.

We may assume that x Î I₁ . This is seen as follows. If x Ï I₁ x must belong to I _j for some 1 < j £ n and x Ï I_i for for 1 £ i < j. Then [x, y] Ç I_i = Æ for 1 £ i < j. It follows that [x, y] Í I _j È I_j+1 È ¼ ÈI_n and so we can rename if need be I _j to be I₁, I_j+1 to be I₂ and so on. By a similar argument we may assume that y Î I_n. We may also assume that I_i Ç I_{i +1} ¹ Æ for 1 £ i £ n -1 and that I_i Ë I_j for j ¹ i. We can deduce this as follows. If I_i Í I _j, then the collection of the I_k 's without I_i still covers [x, y] and so we can discard I_i and rename the I_j 's. Then starting with I₁ , suppose I₁ Ç I₂ = Æ. Then since [x, y] is path connected, I₁ Ç È{ I_j : 1< j £ n} ¹Æ implies for some 2 < j £ n, I₁ Ç I_j ¹ Æ . Then e_j - d(e_j) < e₂ - d(e₂) implies that d(e_j) > e_j - e₂ + d(e₂) > d(e₂) and so e_j + d(e_j) > e₂ + d(e₂) and so I₂ Í I_j . This contradicts that I₂ Ë I_j . We can repeat the same argument to show that I_i Ç I_{i +1} ¹ Æ for i > 1.

Thus, in this way we may assume that we have a sequence of points x₁ , x₂, ¼, x_n - 1 such that

e₁ < x₁ < e₂ < x₂ < ¼ < e_n - 1 < x_n - 1 < e_n

and x_i Î I_i Ç I_{i +1} for 1 £ i £ n -1. Therefore, by (2) and the triangle inequality.

£ (K+e) m( I₁ È I₂ È ¼ ÈI_n ) £ (K+e) m( U_i).

Hence the diameter of f (U_i) £ (K+e) m( U_i). It follows that m( f (U_i Ç E)) £ (K+e) m(U_i). This proves (3).

Then using (3), we see that

.

Since e is arbitrary, we conclude that .

Theorem 2. Suppose f : [a, b] ® R is a measurable function. Suppose E is any

measurable subset such that f ' (x) exists finitely for every x in E. Then

,

where m is the Lebesgue outer measure.

Proof. Since f is measurable and finite on [a, b], its Dini derivatives are measurable. (Banach Theorem). Consequently, f ' is measurable on E and so | f ' | is measurable on E. Suppose now g = | f ' | is bounded on E, by a positive integer K, i.e., | f ' (x)| < K for each x in E. For any positive integer n and integer i =1,2, ¼,2ⁿ K, let . Define for each positive integer n. Then ( g_n ) is a sequence of simple functions converging pointwise to g on E. In particular,

.

By Theorem 1, for integer i =1,2, ¼,2ⁿ K, Thus,

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

Therefore, . Since, and , we conclude that

.

We now consider the case when g is unbounded. For each integer k > 1, let Then it is obvious that E is a disjoint union of the E_k's. Note that on E _k , g is bounded by k. Hence, by what we have just shown, for each integer k > 0, . Therefore,

.

This completes the proof of Theorem 2.

We have some easy consequences of the above theorems.

Theorem 3. Suppose f is defined and finite on [a, b]. Suppose E = {x Î [a, b]: f is differentiable at x and f ' (x) = 0}. Then m( f (E)) = 0.

Proof. By Theorem 1 m( f (E)) £ 1/n m(E) for any positive integer n. Therefore, m( f (E)) = 0.

Recall a set is called a null set if its measure is zero.

Theorem 4. Suppose f : [a, b] ® R has a finite derivative at every point of [a, b]. Then f maps null sets onto null sets.

Proof. Suppose E is a null set in [a, b]. Then by Theorem 2,

.

Hence m( f (E)) = 0. This proves the theorem.

Theorem 5. Suppose f : [a, b] ® R has a finite derivative at every point of [a, b] and f ' is Lebesgue integrable on [a, b]. Then for every closed and bounded interval [c, d] in [a, b],

.

Proof. Since f is continuous on [a, b], | f (d) - f (c)| £ m( f ([c, d])). Since f is differentiable at every point of [c, d], by Theorem 2,

.

It follows that .

We can apply Theorem 5 to the next result.

Theorem 6. Suppose f : [a, b] ® R has a finite derivative at every point of [a, b] and f ' is Lebesgue integrable on [a, b]. Then f is absolutely continuous.

Proof. Since f ' is Lebesgue integrable, | f ' | is also Lebesgue integrable on [a, b]. For each positive integer n, let g_n = min( | f ' |, n). Then each g_n is Lebesgue integrable on [a, b] and the sequence ( g_n ) converges pointwise to | f ' |. In particular, for each n, |g_n |= g_n £ | f ' | and so by the Lebesgue Dominated Convergence Theorem,

.

Hence, given any e > 0, there exists a positive integer N such that

.

It follows that

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

Now take . Suppose [a_i , b_i], i = 1, 2, ¼, k are non-overlapping intervals in [a, b]. If , then

by Theorem 5,

by (1) ,

This shows that f is absolutely continuous on [a, b].

Remark. Theorem 6 is Theorem 8.21 in Rudin's Real and Complex Analysis in an equivalent formulation.

More generally we may relax the requirement of everywhere differentiability but we need to impose the requirement that f maps null sets to null sets. This is a necessary condition for absolute continuity.

Theorem 7. Suppose f : [a, b] ® R is continuous and f ' exists almost everywhere and is Lebesgue integrable on [a, b]. Suppose f maps null sets to null sets. Then f is absolutely continuous.

Proof. Let E Í [a, b] be the subset where f ' exists at each point so that the measure of [a, b] - E is zero. Then | f ' | = g almost everywhere, where g(x) = | f ' (x)| for x in E and g(x) = 0 for x outside E. Then there exists an increasing sequence of simple functions (g_n) converging pointwise to g almost everywhere and

.

Thus, given any e > 0, there exists a positive integer N such that

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

Suppose [a_i , b_i], i = 1, 2, ¼, k are non-overlapping intervals in [a, b]. Let

E_i ={x Î [a_i , b_i]: f ' (x) exists.}. Then since f maps null sets to null sets and m([a_i , b_i]-E_i) = 0, m( f ([a_i , b_i]) = m( f (E_i)). By Theorem 2, and so for each i,

. ------------------------ (2)

Since f is continuous, f is also continuous on [a_i , b_i] and so by continuity, | f (b_i ) - f (a_i )| £ m( f ([a_i , b_i]) )for each i = 1, 2, ¼, k. Therefore, by (2) we have

since m([a_i , b_i]-E_i) = 0

where K > 0 is an upper bound for g _N..

------------------------------- (3).

Take , It follows from (3) that if , then

.

This shows that f is absolutely continuous.

As a corollary we have the Banach Zarecki Theorem.

Theorem 8 (Banach Zarecki) . Suppose f : [a, b] ® R is continuous and is a function of bounded variation. Suppose f maps null sets to null sets. Then f is absolutely continuous.

Proof. Since f is of bounded variation, f is differentiable almost everywhere and f ' is Lebesgue integrable. Therefore, by Theorem 7, f is absolutely continuous.

Remark. It is easy to see that if f is absolutely continuous on [a, b], then f is continuous and of bounded variation on [a, b]. Any function of bounded variation on [a, b] is the difference of two increasing functions (see for instance Theorem 13 of "Monotone functions, function of bounded variation , fundamental theorem of Calculus"). Since any increasing function on [a, b] is differentiable almost everywhere on [a, b] and its derived function is Lebesgue integrable on [a, b], any function of bounded variation is therefore, differentiable almost everywhere on [a, b] and its derivative is Lebesgue integrable on [a, b]. So if f is absolutely continuous on [a, b], then f is differentiable almost everywhere on [a, b] and f ' is Lebesgue integrable on [a, b]. If f is absolutely continuous on [a, b], then f maps null sets in [a, b] to null sets (see for instance Proposition 9 of my article "Change of variable or substitution in Riemann and Lebesgue Integration"). Thus the converse of Theorem 7 and Theorem 8 are also true.

With a little thought we shall derive the following theorem.

Theorem 9. Suppose f : [a, b] ® R is absolutely continuous and f ' (x) = 0 almost everywhere on [a, b]. Then f is a constant function.

Proof. It is enough to show that the range of f has measure zero. Let E = {x Î [a, b] : f ' (x) = 0}. Then m( [a, b] - E) = 0. By Theorem 3, m( f (E)) = 0. Since f is absolutely continuous, it maps null sets to null sets (see Proposition 9 of my article "Change of variable or substitution in Riemann and Lebesgue Integration"). It follows that m( f ([a, b] - E)) = 0. Therefore, m ( f ([a, b])) £ m( f (E)) + m( f ([a, b] - E)) = 0. It follows that m ( f ([a, b])) = 0. Since f is continuous and [a, b] is compact and connected, f ([a, b])) is compact and connected and so is either a non-trivial closed and bounded interval or a single point. Since a non-trivial closed and bounded interval has non-zero measure, f ([a, b])) must be a single point, consequently f is a constant function.

The next result is a consequence of a function having the property of being a continuous N function. In particular the result applies to an absolutely continuous function on [a, b].

Theorem 10. Suppose f : [a, b] ® R is continuous and maps null sets to null sets, i.e., f is a continuous N function. Then f maps measurable sets to measurable sets.

Proof. Since the Lebesgue measure is a regular measure, for any measurable set E there is a subset, a F_s set, K in [a, b] such that K Í E and m(E - K) = 0. By a F_s set K, we mean K is a countable union of closed sets in [a, b]. Thus

,

where each K_n is a closed subset in [a, b].

Each K_n is closed and bounded and so by the Heine Borel Theorem, is compact. Since f is continuous, each f (K_n ) is compact and so is closed and bounded by the Heine Borel Theorem. Since f (K_n ) is closed, it is measurable.

Therefore,

being a countable union of measurable sets is measurable.

Since f maps null sets to null sets, m( f (E - K)) = 0. It then follows that f (E - K) is measurable. Hence,

f (E ) = f (K ) È f (E - K))

is a union of two measurable sets and so is measurable.

Corollary 11. Suppose f : [a, b] ® R is absolutely continuous. Then f maps measurable sets to measurable sets.

Proof. Since f is absolutely continuous on [a, b], f maps null sets in [a, b] to null sets (see for instance Proposition 9 of my article "Change of variable or substitution in Riemann and Lebesgue Integration"). Thus f is a continuous N function and so by Theorem 10, f maps measurable sets to measurable sets.