Arc Length, Functions of Bounded Variation and Total Variation

by Ng Tze Beng

This article gives a formula for the length of a curve in Rⁿ . This is a general formula, not assuming differentiability and in its derivation some interesting results about function of bounded variation will be used. It gives a complete picture than otherwise given in most elementary text.

Definition 1. A curve in Rⁿ is a continuous function f : [a, b] ® Rⁿ , where the non-trivial interval [a, b] is called the parameter interval and f is called a parametrized curve.

We can consider an estimate of the length by taking points on this curve and take the length of an inscribed polygonal curve passing through theses points. To do this we take a partition D : a = x₀ < x₁ < x₂ < ¼ < x_n = b for the interval [a, b]. Let P₀ be the point f (x₀) and P_i = f (x_i). Then the length of the polygonal curve P₀P₁¼P_n is an approximation of the arc length P₀P_n .

We shall consider Rⁿ as the usual space with Euclidean metric. That is, the distance between two points is the usual Euclidean distance and we have the usual Euclidean norm function

|| || : Rⁿ ® R

defined by ||V || = Ö ( V₁² + V₂² + V₃² + ¼ + V_n² ) , where V_i is the i-th component of V. The distance between two points V and W in Rⁿ is then given by || V - W||.

The length of the polygonal curve P₀P₁¼P_n is given by

|P₀P₁| + |P₁P₂| + ­ + |P_n-1P_n| or .

Now the length of each line segment |P_i-1P_i| is the length of the line joining f (x_i-1) to f (x_i). Thus by the Euclidean distance

.

Therefore, the length of the polygonal curve P₀P₁¼P_n is given by

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

We define the arc length of the curve f (x), for a £ x £ b to be the least upper bound of all possible polygonal approximation as given by (1), if it exists. Therefore, the length of the curve f is just the total variation T_f [a, b]. If it exists, the curve is called a rectifiable curve, otherwise it is not.  It then follows form the definition of a function of bounded variation, f is rectifiable if and only if f is of bounded variation. Not all continuous curves on a closed and bounded domain are rectifiable.

Now we consider the component functions of f . First an easy observation.

Theorem 2. A curve f : [a, b] ® Rⁿ is rectifiable if and only if each component function is continuous of bounded variation.

Theorem 2 is an easy consequence of the definition of bounded variation and the following inequality,

|V_j |£ ||V || = Ö ( V₁² + V₂² + V₃² + ¼ + V_n² ) £ |V₁ | + |V₂ | + |V₃ |+ ¼ + |V_n|.

Using the first part of this inequality we can show that if f is of bounded variation, then each component function f _j is also of bounded variation. Using the second part of the above inequality we can show that if each f _j is of bounded variation, then f is of bounded variation. Also note that f is continuous if and only if each component of f is continuous. One may use the above inequality to prove this or simply observe that a map into a product space is continuous if and only if each component is continuous.

The next result is to show that there is a relation between the derivative of the bounded variation and the norm of the derived function of f . Note that if f is of bounded variation, then it is differentiable almost everywhere. In particular, if g: [a, b] ® R is of bounded variation, then g' is measurable and summable or integrable. This is a consequence of the fact that if g is of bounded variation, then it is the difference of two increasing functions. (See Theorem 13 in "Monotone Function, Function of Bounded Variation, Fundamental Theorem of Calculus.). Note that for any increasing function g on [a, b], g is differentiable almost everywhere and g' is measurable and Lebesgue integrable. (For a reference, see Proposition 24 in "Change of Variable or Substitution in Riemann and Lebesgue Integration".)

Lemma 3. Suppose g: [a, b] ® R is of bounded variation. Then the variation function V_g : [a, b] ® R, defined by V_g(x) = T_g[a, x] for a < x £ b and V_g(a) = 0, satisfies

(V_g)'(x) = |g'(x)| almost everywhere.

This result carries over to curve in Rⁿ.

Lemma 4. Suppose f : [a, b] ® Rⁿ is of bounded variation. Then the variation function T_f : [a, b] ® R, defined by T_f (x) = T_f [a, x] for a < x £ b and T_f (a) = 0, satisfies

(T_f )'(x) = || f '(x)|| almost everywhere on [a, b].

Lemma 3 is well known although not as well known as the weaker statement when g is also absolutely continuous.

To prove Lemma 4 we require several results.

Lemma 5. Suppose f : [a, b] ® Rⁿ is such that each component is Lebesgue integrable. Then .

Proof. , where < , > is the inner product for Rⁿ .

Let . Then

, where a _i and f _i are respectively the i-th components of a and f ,

by the Schwarz inequality

Hence . This proves the required inequality.

Now we shall state a weaker version of Lemma 4.

Lemma 6. Suppose f : [a, b] ® Rⁿ is absolutely continuous. Then the variation function T_f : [a, b] ® R, defined by T_f (x) = T_f [a, x] for a < x £ b and T_f (a) = 0, satisfies

(T_f )'(x) = || f '(x)|| almost everywhere on [a, b].

Proof. Recall that f : [a, b] ® Rⁿ 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 each component of f is also absolutely continuous.

First observe that if f is of bounded variation, then

|| f (x) - f (y) || £ T_f (x) - T_f (y) for x > y.

From this we easily deduce that for almost all x in [a, b],

|| f ' (x) || £ (T_f )'(x) ------------------------------------- (2)

Now fixed a x in [a, b]. For any partition D : a = x₀ < x₁ < x₂ < ¼ < x_k = x for the interval [a, x],

because for each j as f _j is absolutely continuous

by Lemma 5

.

Thus, T_f (x) being the supremum for such sum for any partition D : a = x₀ < x₁ < x₂ < ¼ < x_k = x,

we have

Therefore, for any y > x we get by taking the interval [x, y] in place of [a, x] above

Thus . Taking limits as y tends to x from above we have whenever the limits exist.

Similarly interchanging the role of x and y for x > y we get

whenever the limits exist.

Therefore, (T_f )'(x) £ || f '(x)|| almost everywhere. Thus (T_f )'(x) = || f '(x)|| almost everywhere.

This completes the proof of Lemma 4.

Note that since f is of bounded variation on [a, b], f _i ' is measurable and integrable and so

|| f ' (x)|| is integrable since it is dominated by the integrable function because of the inequality .

Now we shall prove Lemma 4. Our strategy is to write f as the sum of an absolutely continuous function and a singular function. Indeed for any function f : [a, b] ® Rⁿ of bounded variation, f = g + h where g is absolutely continuous and h is a singular function such that h'(x) = 0 almost everywhere. We can let g be defined by and h(x) = f (x) - g(x).

Proof of Lemma 4. We shall prove first for the case when f is singular, that is, f '(x) = 0 almost everywhere. That means for each i =1,2, ¼, n, f _i '(x) = 0 almost everywhere.

For any y > x we have

where f = ( f ₁, f ₂, ¼, f _n ). Therefore, for y > x,

Therefore, for almost all x in [a, b],

Similarly by interchanging the role of x and y, we have for almost all x in [a, b],

.

Thus almost everywhere by Lemma 3.

Hence, (T_f )'(x) = || f '(x)|| = 0 almost everywhere.

Now we proceed to consider the general case when f is of bounded variation. Write f = g + h , where g is absolutely continuous and h is a singular function. Then for y > x,

Therefore, for y > x,

.

Thus, as before, for almost all x,

.

This means, for almost all x,

(T_f )'(x) £ (T_g)'(x) + (T_h )'(x) = (T_g)'(x) + 0 since h is singular

= || g '(x)|| by Lemma 6 since g is absolutely continuous

= || f '(x)|| since f '(x) = g'(x) almost everywhere.

On the other hand, by (2), for almost all x in [a, b]

|| f ' (x) || £ (T_f )'(x).

Therefore, (T_f )'(x) = || f ' (x) || almost everywhere.

Now it remains to look at the arc length of the curve f : [a, b] ® Rⁿ itself. Note that this is just the total variation T_f [a,b] of the function. By the definition of a curve, f is continuous and so uniformly continuous since the domain [a, b] is compact. Thus there is actually a practical way to obtain an approximation of the arc length. That is for each n, we consider a regular partition of [a, b], D_n : a = x₀ < x₁ < x₂ < ¼ < x_n = b such that the norm of the partition ||D_n|| = (b - a)/n. Then the arc length L_n of the polygonal curve defined by the partition D_n tends to the arc length of f as n tends to infinity. We could not obtain totally an integral formula for the arc length of f in general. There will be an error term, which vanishes if f is absolutely continuous. The integral part of the formula will actually come from the absolutely continuous part of f . The next theorem will imply that the total variation of the function f is the sum of the total variation of its absolutely continuous part and that of the singular part.

Theorem 7. Suppose g : [a, b] ® Rⁿ is absolutely continuous and h : [a, b] ® Rⁿ is such that each component of h is of bounded variation and that h'(x) = 0 almost everywhere in [a, b]. Then T_{g + h}[a, b] = T_g[a, b] + T_h[a, b].

Proof. First note that g + h is of bounded variation. It then follows that for any y > x in [a, b],

T_{g + h}[x, y] £ T_g[x, y] + T_h[x, y] ------------------------------- (3)

We shall show that T_{g + h}(x) = T_g(x) + T_h(x) for all x in [a, b]. To do this it is sufficient to show that T_{g + h}(x) - T_h(x) is absolutely continuous.

Let f (x) = T_{g + h}(x) - T_h(x). Then for any y > x,

f (y) - f (x) = T_{g + h}(y) - T_{g + h}(x) - (T_h(y) - T_h(x))

= T_{g + h}[x, y] - T_h[x, y]

£ T_g[ x, y]= T_g(y) - T_g(x) ---------------------- (4)

by inequality (2).

Since g is absolutely continuous, it follows that the variation function T_g is also absolutely continuous. Therefore, using (4) and the definition of absolute continuity, we can easily show that f is absolutely continuous.

Then, for almost all x in [a, b],

f '(x) = (T_{g + h})'(x) - (T_h)'(x).

= || (g + h)'(x)|| - ||h'(x)|| by Lemma 4

= || g'(x)|| since h'(x) = 0 almost everywhere.

= (T_g )'(x) by Lemma 6.

Now since f (x) - T_g (x) is absolutely continuous since both f (x) and T_g (x) are absolutely continuous and that its derivative f '(x) - (T_g )'(x) = 0 almost everywhere on [a, b], f (x) - T_g (x) is a constant. Because f (a) = T_g (a) = 0, we conclude that f (x) = T_g (x) for all x in [a, b]. This means T_{g + h}(x) - T_h(x) = T_g (x) and so T_{g + h}(x) = T_g (x) + T_h(x) for all x in [a, b]. In particular,

T_{g + h}(b) = T_g (b) + T_h(b). i.e., T_{g + h}[a, b] = T_g[a, b] + T_h[a, b]. This completes the proof.

Theorem 8. Suppose f : [a, b] ® Rⁿ is a rectifiable curve . Then the arc length of f , L_f is given by the following formula,

.

for some singular function h such that f = g + h, where g is absolutely continuous and h'(x) = 0 for almost all x in [a, b]. We may take . In particular, f is absolutely continuous if and only if .

Proof. Let . Then g is absolutely continuous since it is an indefinite integral. Let . Then f = g + h. Therefore, by Theorem 7

L_f = T_f[a, b] = T_{g + h}[a, b] = T_g[a, b] + T_h[a, b]

= T_g(b) + T_h[a, b]

since T_g is absolutely continuous

since (T_g)'(t) = || g'(t)|| almost everywhere

since g'(t) = f '(t) almost everywhere.

If f is absolutely continuous, i.e., each component of f is absolutely continuous, then is a constant function and so T_h[a, b] = 0. Conversely suppose T_h[a, b] = 0, then T_h[a, x] = 0 for all x in [a, b]. Therefore, by Theorem 7, T_f (x) =T_{g + h}(x) = T_{g + h}[a, x] = T_g[a, x] + T_h[a, x] = T_g[a, x] + 0 =T_g(x). Since g is absolutely continuous, T_g(x) is also absolutely continuous and so T_f (x) =T_g(x) is absolutely continuous. (See Theorem 23 of "Change of Variable or Substitution in Riemann and Lebesgue Integration" or Theorem 29.14 and its proof in "Principles of Real Analysis" by C.D. Aliprantis and Owen Burkinshaw.) It follows easily that f is absolutely continuous. This completes the proof.

Now we specialize to the graph of a function, the curve given by a continuous function

f : [a, b] ® R . Then the arc length of the curve in R² is the arc length of the function g : [a, b] ® R² given by g(x) = (x, f (x)). Thus g is rectifiable if and only if f is of bounded variation. The arc length of the curve is the total variation of g. Thus applying Theorem 8 to g we get:

Theorem 9. Suppose f : [a, b] ® R is continuous and the graph of f is rectifiable (hence f is of bounded variation). Then the arc length L of the curve is given by

for some singular function h such that h'(x) = 0 almost everywhere ,. In particular f is absolutely continuous if and only if

Proof. The graph of f is the curve in question. Let g : [a, b] ® R² be defined by g(x) = (x, f (x)) for each x in [a, b]. Then g is of bounded variation. Then we can take a decomposition of g as follows

g (x) = (x, F(x)) + (0, h(x))

where and .

Then G(x) = (x, F(x)) is absolutely continuous and H(x) = (0, h(x)) is singular since h'(x) = 0 almost everywhere. Therefore, by Theorem 8, the arc length L is given by the arc length of g,

L=L _g =

since F'(x) = f ' (x) almost everywhere and the total variation T_H[a, b] =T_(0,h) [a, b] = T_h[a, b]. f is absolutely continuous if and only if h(x) is a constant function if and only if

L=L _g .

This completes the proof.

Now we proceed to show that a sequence of polygonal length tends to the arc length. This is a consequence of the following result.

Theorem 10. Suppose f : [a, b] ® Rⁿ is a rectifiable curve. Then for any e > o there exists a d > 0 such that if P: a = x₀ < x₁ < x₂ < ¼ < x_n = b is a partition for [a, b] such that ||P|| < d, then

Proof. Since

is a partition for [a, b]}

given any e > 0 there exists a partition Q : a= T₀ < T₁ < T₂ < ¼ < T_N = b such that

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

Let . Now f is continuous on [a, b] implies that f is uniformly continuous on [a, b]. Therefore there exists a d > 0 such that whenever |x - y| < d we have

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

Suppose now P: a = t₀ < t₁ < t₂ < ¼ < t_m = b is any partition such that the norm of the partition ||P|| < d. Then the length of the polygonal curve defined by this partition is given by

Now let S= Q ÈP . Now we write the partition as

S: a= s₀ < s₁ < s₂ < ¼ < s_M = b

Then the length of the polygonal curve defined by S is given by

Obviously, by the triangle inequality, L_P £ L_S and L_Q £ L_S . Now we shall see that our partition P satisfies the conclusion of the theorem.

Note that S is formed by adding points in Q to P. Adding a point T_i to P will increase the polygonal length of P by at most e/(2N). This is seen as follows.

Suppose for some j, t_j-1 < T_i < t_j ,

Then,

by (6).

Therefore, since Q has at most N-1 points different from points in P, we conclude that

Hence, by (5) Thus, the partition P: a = t₀ < t₁ < t₂ < ¼ < t_m = b satisfies . This completes the proof.

Now it remains to investigate how Lemma 3 can be proved. We shall use the Lebesgue decomposition of an increasing function to prove the assertion of the Lemma. We shall recall some interesting properties of increasing function. We shall define the saltus function of an increasing function. This is made possible because the set of points of discontinuity of an increasing function is at most countable.

Theorem 11. Suppose f : [a, b] ® R is a monotone function. Then the set of discontinuity of f is countable.

This is Theorem 3 in my article "Monotone Function, Function of Bounded Variation and Fundamental Theorem of Calculus".

we shall decompose an increasing function as a sum of a continuous function and a 'jump' function.

Definition 12. Suppose f : [a, b] ® R is an increasing function. Then the discontinuity of f can only be jump discontinuity. By Theorem 11, the set of discontinuity of f is countable. So let the set of discontinuity be {x₁, x₂, ¼ , x_n , ¼} where x₁ < x₂ < ¼ < x_n < ¼, i.e., the x_i 's is ordered in an increasing order. We define the saltus function (or jump function) s : [a, b] ® R of f by

s (a) = 0,

, for a < x £ b,

where are the respective right and left limits at x.

Plainly s(x) is an increasing function.

Let f(x) = f (x) - s(x) for x in [a, b]. Then we shall show that f : [a, b] ® R is an increasing, continuous function.

Theorem 13. Suppose f : [a, b] ® R is an increasing function. Let s be the saltus function of f . Then the function f : [a, b] ® R defined by f(x) = f (x) - s(x) for x in [a, b] is increasing and continuous on [a, b]

Proof. For a £ x < y £ b,

£ f (y) - f (x) by Theorem 2 of "Monotone Function, Function of Bounded Variation and Fundamental Theorem of Calculus".

Therefore, f(x) = f (x) - s(x) £ f (y) - s(y)= f(y) and so f is an increasing function.

Now if x is a point of continuity of f , then obviously, for x < y

f (x+) - f (x) £ s(y) - s(x)

and if x is a point of discontinuity of f, i.e., x = x_j for some j, then

³ f (x+) - f (x)

Therefore, f (x+) - f (x) £ s(y) - s(x) for x < y.

Hence, . Therefore,

f(x+) = f (x+) - s(x+) £ f (x) - s(x)= f(x)

Since f is increasing, f(x) £ f(x+). It follows that f(x) = f(x+).

Now, for x < y if x is a point of discontinuity of f ,

³ f ( y) - f ( y-)

Plainly, if x is a point of continuity, then and so

³ f ( y) - f ( y-).

Therefore, . I.e., s( y) - s( y-) ³ f ( y) - f ( y-).

Thus, f( y-) = f ( y-)- s( y-) ³ f (y) - s(y)= f(y). Once again since f is increasing, f( y-) £ f(y). Thus for any y in [a, b]. It then follows that f(x) = f(x+)=f(x-). Thus, f is continuous at x for any x in [a, b]. This completes the proof.

Now we summarize the above in the following:

Theorem 14. Suppose f : [a, b] ® R is an increasing function. Then f can be decomposed as a sum of a continuous function and a saltus type function as follows:.

f (x) = f(x) + s(x) for all x in [a, b],

where f(x) = f (x) - s(x) is increasing and continuous and s(x) is the saltus function of f .

Obviously s'(x) = 0 almost everywhere on [a, b] and f '(x) = f '(x) almost everywhere.

We shall seek for a better decomposition of f . We would like to decompose the function into a sum of an absolutely continuous function and a singular function. This is the Lebesgue decomposition of a monotone function.

Theorem 15. Suppose f : [a, b] ® R is an increasing function. Then

f (x) = F(x) + g(x) + s(x) for all x in [a, b] ------------------- (7)

where F(x) is an increasing, absolutely continuous function, g(x) is a continuous increasing function which is singular, i.e., g'(x) = 0 almost everywhere on [a, b] and s(x) is a saltus function of f .

Proof. Take the decomposition f (x) = f(x) + s(x) given by Theorem 14.

Let . Then F(x) is absolutely continuous being the indefinite integral of an integrable function. Then F'(x) = f '(x) = f'(x) almost everywhere on [a, b]. Let g(x) = f(x)-F(x). Then g(x) is continuous since both F and g are continuous on [a, b]. Also g'(x) = f'(x)- F'(x) = f '(x) - f '(x) = 0 almost everywhere on [a, b]. Note that g is also increasing. For a £ x < y £ b,

because for increasing function f

(see for example Proposition 24 in "Change of Variable or substitution in Riemann and Lebesgue Integration")

Therefore, g(x) £ g(y) for x < y. Hence g is increasing.

Thus f (x) = f(x) + s(x) = F(x) + g(x) + s(x) is the desired decomposition.

Next we shall consider for a function of bounded variation f : [a, b] ® R the most efficient way of decomposing f (x) - f (a) as the difference of two increasing functions.

Let Q: a = x₀ < x₁ < x₂ < ¼ < x_n = b is a partition for [a, b]. Define

,

,

.

Note that p(Q) is just the sum over the terms for which f ( x_i) - f (x_i-1) ³ 0, -n(Q) is the sum over the terms for which f ( x_i) - f (x_i-1) £ 0 and so

Define for a function f : [a, b] ® R of bounded variation

the positive variation,

P_f [a,b] = sup { p(Q): Q a partition of [a, b] }

the negative variation,

N_f [a,b] = sup { n(Q): Q a partition of [a, b] }

and the total variation,

T_f [a,b] = sup { t(Q): Q a partition of [a, b] }.

Plainly, all three variations exist since f is of bounded variation.

Since p(Q) + n(Q) = t(Q),

P_f [a,b] + N_f [a,b] = T_f [a,b]. ------------------------- (8)

We deduce this as follows. Plainly, p(Q) + n(Q) £ T_f [a,b]. It follows that P_f [a,b] + N_f [a,b] £ T_f [a,b]. On the other hand, t(Q)= p(Q) + n(Q)£ P_f [a,b] + N_f [a,b], consequently T_f [a,b] £ P_f [a,b] + N_f [a,b]. Thus we have P_f [a,b] + N_f [a,b] = T_f [a,b].

Similarly, since p(Q)- n(Q) = f (b) - f (a),

P_f [a,b] - N_f [a,b] = f (b) - f (a). ---------------------- (9)

Now we define the positive variation function P_f : [a, b] ® R of f by

P_f (a)= 0, P_f (x) = P_f [a,x] for a < x £ b.

The negative variation function N_f : [a, b] ® R of f is defined similarly by

N_f (a)= 0, N_f (x) = N_f [a,x] for a < x £ b.

Finally the total variation function of is of course defined by

T_f (a)= 0, T_f (x) = T_f [a,x] for a < x £ b.

Thus it follows from (8), that

T_f (x) = P_f (x) + N_f (x) --------------------- (10)

and from (9)

f (x) - f (a) = P_f (x) - N_f (x) ------------------- (11).

Plainly these three variation functions are increasing functions.

We state the above conclusion as follows:

Theorem 16. Suppose f : [a, b] ® R is a function of bounded variation. Let T_f (x), P_f (x) and N_f (x) be respectively the total, positive and negative variation functions of f . Then f (x) - f (a) = P_f (x) - N_f (x) and T_f (x) = P_f (x) + N_f (x).

The decomposition (11) is the most efficient in the sense of the following theorem.

Theorem 17. Suppose f : [a, b] ® R is a function of bounded variation. Suppose we have another decomposition of f (x) - f (a),

f (x) - f (a) = g(x) - h(x)

where g and h are increasing functions with g(a)=h(a) =0. Then P_f (x) £ g(x) and N_f (x) £ h(x) on [a, b].

Proof. Let Q: a = x₀ < x₁ < x₂ < ¼ < x_n = b is a partition for [a, b]. Then

f ( x_i) - f (x_i-1) = g( x_i) - g(x_i-1) - [h( x_i) - h(x_i-1)] £ g( x_i) - g(x_i-1).

Therefore, p(Q) being the sum over the terms for which f ( x_i) - f (x_i-1) ³ 0

.

Thus P_f [a,b] £ g(b). It follows from this that P_f [a,x] £ g(x), for any a< x £ b. This means

P_f (x) £ g(x) for all x in [a, b]. Therefore, P_f (x) - N_f (x) =f (x) - f (a)= g(x) - h(x) £ g(x) - N_f (x) and so N_f (x) £ h(x).

Proof of Lemma 3. If f : [a, b] ® R is a function of bounded variation, then (T_f )'(x) = | f '(x)| almost everywhere on [a, b]

Let P(x) = P_f (x) and N (x) = N_f (x) be the positive and negative variation functions of f . Take their respective Lebesgue decomposition as given by Theorem 15,

and

where r₁'(x) = r₂'(x) = 0 almost everywhere on [a, b] and both r₁(x) and r₂(x) are increasing functions satisfying r₁(a) = r₂(a) = 0.

Let h(x) = min( P'(x), N'(x)). This is defined almost everywhere on [a, b]. Now define

and

.

Then obviously both w₁(x) and w₂(x) are increasing functions since both P'(x)-h(x) and N'(x)-h(x) are greater or equal to 0 almost everywhere on [a, b]. Also we have w₁(a) = w₂(a) = 0.

In particular,

w₁(x) - w₂(x) = P(x) - N(x) =f (x) - f (a) -------------------- (12)

Therefore, by Theorem 17, P (x) £ w₁(x).

Thus and so . Hence

. But since h(t) = min( P'(t), N'(t)) ³ 0 almost everywhere. Therefore, for all x and consequently h(x) = 0 almost everywhere on [a, b]. That is to say, min( P'(x), N'(x)) = 0 almost everywhere on [a, b].

Now by (12) f '(x) = P'(x) - N'(x) almost everywhere on [a, b] and so

[P'(x) - h(x)] + [N'(x) - h(x)] = |P'(x) - N'(x)| = | f '(x)|

almost everywhere.

Hence, | f '(x)| = P'(x) + N'(x) -2h(x) = P'(x) + N'(x) since h(x) = 0 almost everywhere. But because T_f (x) = P(x) + N(x), (T_f )'(x) = P'(x) + N'(x) and so (T_f )'(x) = | f '(x)| almost everywhere on [a, b]. This completes the proof of Lemma 3.

Ng Tze Beng

May 2006