** Functions Having Finite Derivatives, Bounded Variation, Absolute Continuity, the Banach Zarecki Theorem and de La Vall****é****e Poussin's Theorem**

**by Ng Tze Beng**

- We shall begin by examining the properties of the image under a function

- 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.

- For any e > 0, by the definition of outer measure, there exists an open set

- Since for each

- .

- and so

- .

- Thus we have,

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

- Since

- Then the collection C = {

- ,

- where

- Note that

- Plainly for

- Now take any

- where

- [

- and

- We may assume that

- Thus, in this way we may assume that we have a sequence of points

- and

- £ (

- Hence the diameter of

- Then using (3), we see that

- .

- Since e is arbitrary, we conclude that .

- measurable subset such that

- where

- .

- By Theorem 1, for integer

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

- Therefore, . Since, and , we conclude that

- .

- We now consider the case when g is unbounded. For each integer

- .

- This completes the proof of Theorem 2.

- We have some easy consequences of the above theorems.

- Recall a set is called a

- .

- Hence

- .

- .

- It follows that .

- We can apply Theorem 5 to the next result.

- .

- Hence, given any e > 0, there exists a positive integer

- .

- It follows that

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

- Now take . Suppose [

- by Theorem 5,

- by (1) ,

- This shows that

- More generally we may relax the requirement of everywhere differentiability but we need to impose the requirement that

- .

- Thus, given any e > 0, there exists a positive integer

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

- Suppose [

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

- Since

- since

- where

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

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

- .

- This shows that

- As a corollary we have the Banach Zarecki Theorem.

- With a little thought we shall derive the following theorem.

- The next result is a consequence of a function having the property of being a continuous

- ,

- where each

- Each

- Therefore,

- being a countable union of measurable sets is measurable.

- Since

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

- For functions that are strictly increasing (or strictly dedcreasing) we have the following useful result for absolute continuity.

- (a)

- (b) The inverse function

- (a) By Theorem 8,

- .

- Hence . Since

- .

- It follows that

- (b) Suppose

- As in part (a), note that

- Now assume that

- Let

- By Theorem 15 of "Functions of Bounded Variation and Johnson's Indicatrix",

- Write

- But

- This completes the proof.

- The proof of Theorem 12 (a) word for word with minor modification changing "increasing" to "of bounded variation" and "¥" to "± ¥" gives the following theorem.

- Then

- We shall now give a proof of the Theorem of De La Vallée Poussin.

- where

- The following elementary proof is due to F. S. Cater.

- The following technical lemma is the key to the proof.

- Then

- The first step is to choose some anchor partition for [

- .

- Then given any e > 0, there exists a partition such that

- . i.e.,

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

- Then for any partition containing all the points of the partition

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

- Let

- We may assume also that

- Let

- Thus we can find arbitrary small non trivial intervals [

- (Note that since

- such that and for each

- Therefore, the intervals are also mutually disjoint and

- and so

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

- Without loss of generality we may assume that the points of the partition
does not contain any points of

If *P* contains a point in *E* we may just remove this point
from *E*. We may thus remove all the points in *P* that are in *
E* from *E* without affecting the conclusion of the lemma as only a
finite number of points is removed from *E*. We may take
e = 1/*N* then by passing to the limit as *
N* tends to infinity only at most a denumerable number of points are
removed from *E*. Consequently as the measure of a set of denumerable
number of points and its image under* f* or *v _{ f}* is of
measure zero, the conclusion of the lemma remains valid.

- Now since at each point

- ,

- and that

- for each

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

- But by (1) using the fact that for any

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

- (Show this for finite number of the intervals and pass to the limit.)

- Since ,

- . ------------------------- (5)

- Then from (3) and (4), we arrive at

- Thus,

- and using (2) we get

- .

- Since e = 1/

- Now we proceed to show that

- and . Hence we may cover

- , ----------------- (6)

- where the last inequality is deduced using (1).

- Similarly as before using the negative derived number of

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

- Since ,

- ,

- ,

- where

- .

- This combining with (6) and (7) yields,

- and so

- .

- Hence, .
,Since e = 1/

- Therefore,

- and so .

- Now for any e > 0, take an open set

- .

- We have used the fact that for each

- It remains now to show that

- Since

- Let

- Let Let

- By Lemma 15,

- sets of measure zero and so

- since the set is a countable union of sets

- We now prove the property of

- Therefore, for any derived number

*DV* £
inf{ | *D f* | : *D f* is a derived number of *
f * at *x* .}.

- .

- Therefore, the sequence is bounded. Since we have for each

- is a derived number of

- |

- But

- |

- and

- Suppose now

- We have thus proved that

- Ng Tze Beng