Composition and Riemann Integrability

by Ng Tze Beng

It is natural to ask whether composition of Riemann integrable functions is again Riemann integrable. Unlike the case for differentiable functions, where composition of differentiable functions is again differentiable, composition of Riemann integrable functions need not be Riemann integrable. The starting point is of course to establish that non-Riemann integrable functions exist. The following example will be used in subsequent proceeding.

Example 1. The function h : [0, 1] ® R , defined by

,

is not Riemann integrable.

We shall prove that h is not Riemann integrable, using Theorem 1 of Riemann Integral and Bounded function or more precisely part 1 of this theorem.

Let P : 0 = x₀ < x₁ < x₂ < ¼ < x_n = 1 be any partition for the interval [0, 1]. Then by the density of the rational numbers and irrational numbers, in each of the subinterval [x_{i -} 1 , x_i ], (i = 1,¼, n) we can always find a rational number and an irrational number. Hence for i = 1,¼, n., {h(x): x Î[x_{i -} 1 , x_i ]} = {0, 1}. Therefore, for each i = 1,¼, n,

M_i (P, h) = sup{h(x): x Î[x_{i -} 1 , x_i ]} = sup{0, 1}= max{0, 1}= 1 and

m_i (P, h) = inf{h(x): x Î[x_{i -} 1 , x_i ]} = inf{0, 1}= min{0, 1}= 0.

Therefore, the upper Riemann sum with respect to the Partition P is

and the lower Riemann sum with respect to the Partition P is

.

The above is true for any partition P for [0, 1]. Hence the lower Riemann integral of h,

and the upper Riemann integral of h,

.

Therefore, the lower Riemann integral is not equal to the upper Riemann integral and so by Theorem 1 of Riemann Integral and Bounded Function, h is not Riemann integral over [0, 1].

There is a deeper theorem that we can invoke, namely the Theorem of Lebesgue.

Theorem 1 (Lebesgue). A real valued function f : [a, b] ® R from the closed and bounded interval [a, b] into R is Riemann integrable if and only if f is continuous except perhaps on a set of measure zero in [a, b].

Here is a new concept involved, more precisely, the concept of Lebesgue measure. This is a kind of generalisation of length. How can we use this theorem?. It is enough to know that the set of rational numbers is a set of measure zero. What we need to examine is the set of discontinuities of h.

Assertion 1. h is nowhere continuous on [0, 1].

Proof. Let x be an irrational number in [0, 1]. Then h(x) = 0. Take any e > 0 with e < 1, for any d > 0, there exists, by the density of rational numbers, a rational number p_d in (x - d, x + d)Ç[0,1] such that |h(p_d) - h(x)| = |1 - 0| = 1 > e. Hence Therefore, h is not continuous at x. Similarly let x be a rational number in [0, 1]. Then h(x) = 0. Take any e > 0 with e < 1, for any d > 0, there exists, by the density of the irrational numbers, an irrational number q_d in (x - d, x + d)Ç[0,1] such that |h(q_d) - h(x)| = |0 - 1| = 1 > e. Hence Therefore, h is not continuous at x.

For our purpose we only need to know that the function h is discontinuous on the setof irrational numbers in [0, 1] and that the measure of the irrational numbers in [0, 1] is not zero (in fact, the measure is equal to 1). We cannot easily establish the fact that the measure of the irrational numbers in [0, 1] is not zero without going into the theory of Lebesgue measure. Also the measure of [0, 1] is 1. For now we accept these facts.

Another proof of the non Riemnann integrability of h.

By assertion 1, h is discontinuous on a set of measure bigger than zero. By Theorem 1 h is not Riemann integrable on [0, 1].

How can we produce a counter example to the assertion that composite of Riemann integrable functions is Riemann integrable? If we factor h into a composite of Riemann integrable functions, then we are done. Indeed, we can do so. Let us first describe the factors.

Example 2. The function f : [0, 1] ® R defined by f (x) = 0 if x = 0 and f (x) = 1, for 0 < x £ 1 is Riemann integrable.

Proof. We shall use the method of Example 9.2.6 of Calculus, An Introduction to show this. Given any e > 0, there exists a positive integer m such that 1/m < e. Take any partition P : 0 = x₀ < x₁ < x₂ < ¼ < x_n = 1 such that the norm of P, ||P|| = max{x_i - x_{i -} 1 : i = 1, ¼, n} < 1/m. Then the difference of the upper and lower Riemann sum,

U(P, f ) - L(P, f ) = ,

where M_i (P, f ) = sup{ f (x): x Î[x_{i -} 1 , x_i ]} and

m_i (P, f ) = inf{ f (x): x Î[x_{i -} 1 , x_i ]}

= (M₁ (P, f ) - m₁ (P, f ))(x₁ - x₀ ),

since M_i (P, f ) = m_i (P, f ) = 1 for i > 1

= (x₁ - x₀), since M₁ (P, f ) = 1 and m₁ (P, f ) = 0

£ ||P|| < 1/m < e.

Hence by the Riemann's condition (Theorem 1, Riemann Integral and Bounded Function) f is Riemann integrable.

We consider now the next example.

Example 3. The real valued function g : [0, 1] ® [0, 1] defined by

.

is Riemann integrable.

Before we show that g is Riemann integrable we observe that f and g are the required factors for our counter example.

Example 4. The functions f, g and h as defined in example 1, 2 and 3 satisfy

h = f Û g . The functions f and g are Riemann integrable but h is not Riemann integrable.

Proof. If x is an irrational number in [0, 1], then f Û g (x) = f (g(x)) = f (0) = 0 = h(x).

Suppose now that x is a rational number in [0, 1]. Then by the definition of g, 0 < g(x) £ 1. Hence by the definition of f, f Û g (x) = f (g(x))= 1 = h(x) for any rational number in [0, 1]. Thus for all x in [0, 1], f Û g (x) = h(x). Therefore, h = f Û g.

Assertion 2. The function g: [0, 1] ® [0, 1] as defined in Example 3 is Riemann integrable.

We shall use Lebesgue theorem. This requires us to show that g is continuous at every irrational point in [0, 1]. Thus since the rational numbers in [0, 1] is of measure zero, by the Lebesgue Theorem, g is Riemann integrable. Since the proof of the continuity at irrational points is of some interest, especially the simple logic involved, we shall present the proof along this line. We shall do more, we shall prove that g is discontinuous at every rational point in [0,1]. Given any e > 0, by the Archimedean property of R, there exists a positive integer m > 1 such that 1/m < e. Next we observe that there can only be a finite number of reciprocals of integers that are greater than or equal to 1/m: For any rational number p/q with p/q in its lowest terms and 0 < p/q £ 1, g(p/q) ³ 1/m if and only if 1/q ³ 1/m if and only if 1£ q £ m . This means that 1 £ p £ q and the greatest common divisor of p and q is 1. Thus the number of rational numbers in [0, 1] that have values greater than or equal to 1/m is finite and this set includes the point 0 since g(0) = 1. That is the finite set S_m = { p/q: q = 1, ¼, m; p = 1, ¼, q}È{0} is precisely the set on which the values of g are greater than or equal to 1/m. Take an irrational number x in [0, 1]. Then let d = min{|x - y| : y Î S_m }. Then d > 0 since x is irrational. Obviously then the open interval (x - d, x + d) do not meet S_m . This is because if (x - d, x + d)Ç S_m ¹ Æ, then there exists p/q in S_m such that |x - p/q| < d contradicting that |x - p/q| ³ d. That means for all y in (x - d, x + d)Ç [0, 1], y Ï S_m and so consequently, g(y) < 1/m < e. Therefore, |g(y) - g(x)| = |g(y) - 0| = g(y) < e. Hence, for all y in [0, 1] such that |y - x| < d we have |g(y) - g(x)| < e. Thus g is continuous at x. Now we shall show that g is discontinuous at rational point x. Now for any rational point x in [0, 1], g(x) > 0. Let e = g(x)/2 > 0. Then for any d > 0, by the density of the irrational numbers in any interval, there exists an irrational number y_d in (x - d, x + d)Ç [0, 1]. Then |g( y_d) - g(x)| = |0 - g(x)| = g(x) > g(x)/2 = e. That means g is not continuous at x. Hence g is not continuous at any rational number in [0,1]. This completes the proof of the assertion.

Another proof of assertion 2.

We shall show that g satisfies the Riemann's condition. For any e > 0, take any positive integer m > 1 such that 1/m < e/2. Recall that the finite set S_m is precisely the set {y Î[0, 1]: g(y) ³ 1/m}. Note that 0 ,1 Î S_m. Let the number of points of S_m be k +1. Order the elements y₀ , y₁ , y₂ , ¼, y_k of S_m.as follows:

0 = y₀ < y₁ < y₂ < ¼ y_k -1 < y_k = 1.

Choose k-1 pair of points, each pair constitute an interval containing each y_i in its interior and of length < e/2 for i =1 1, 2, ¼, k-1 and such that they are all mutually disjoint. That is, we choose x₀ < x₁ < x₂ < ¼ < x_2k-2 such that

0 = y₀ < x₁ < y₁ < x₂ < x₃ < y₂ < x₄ < x₅ < ¼ x_2k - 4 < x_2k - 3 < y_k -1 < x_2k - 2 < y_k =1

We choose further two more points, x₀ and x_2k - 1 and name y₀ as x - 1 , y_k as x_2k such that 0 = x - 1 = y₀ < x₀ < x₁ and x_2k - 2 < x_2k - 1 < x_2k = y_k =1. We further require that

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

Obviously,

P: 0 = x_-1 < x₀ < x₁ < x₂ < x₃ < x₄ < x₅ < ¼ x_2k - 4 < x_2k - 3 < x_2k - 2 < x_2k - 1 < x_2k =1

forms a partition for [0, 1].

Now by the density of the irrational numbers in any interval, for i = 0, 1, ¼, k

m _i (P, g ) = inf{ g (x): x Î[x_{i -} 1 , x_i ]}= 0. ---------------- (2)

Now since for each j = 0, 1, 2, ¼, k, y_j Î [x_{2 j} -1 , x_{2 j} ] and so by the definition of S_m

M _{2 j} (P, g ) = sup{ g (x): x Î[x_{2 j - 1} , x_{2 j} ]}= g (y_j ) ------------------- (3).

for j = 0, 1, 2, ¼, k .

Now because for j = 1, 2, ¼, k , [x_{2 j - 2} , x_{2 j} - 1 ]Ç S_m = Æ,

M _{2 j -}1 (P, g ) = sup{ g (x): x Î[x_{2 j - 2} , x_{2 j} - 1 ]}< 1/m ---------------- (4).

j = 1, 2, ¼, k

Hence U(P, g ) - L(P, g ) = ,

= by (2)

=

£ by (3) and (4)

£ since g (y_j ) £ 1

< e/2 + by (1)

< e/2 + 1/m since

< e/2 + e/2 = e.

Thus g satisfies the Riemann's condition and so by Theorem 1 of Riemann Integral and Bounded Function g is Riemann integrable. This completes the proof.

This wraps up our demonstration of a counter example. It is then natural to ask: Under what condition can we deduce that the composite of two Riemann integrable functions is Riemann integrable? In view of Lebesgue's Theorem, it is necessary to examine the set of discontinuities of the composite function. If the set of discontinuities of the composite function is contained within the set of discontinuities of the first function, then it becomes an easy matter to make a deduction. Note that it is the set of discontinuities that determines the Riemann integrability of a function. We have the following theorem.

Theorem 2. Suppose g: [a, b] ® R is a Riemann integrable function and f : [c, d] ® R is a continuous function such that the range of g, g( [a, b] ) is contained in [c, d]. Then the composite function f Û g :[a, b] ® R is Riemann integrable on [a, b].

Proof. We shall prove this using Lebesgue's theorem. Note that if g is continuous at a point x in [a, b], then since f is continuous at g(x) (because f is a continuous function), f Û g is continuous at x. This means that whenever g is continuous at x the composite f Û g is also continuous at x. Hence the set of discontinuities of the composite function f Û g, D _f Û g is contained in the set of discontinuities of g, D _g . Since g is Riemann integrable, D _g is of measure zero and since D _f Û g Í D _g , D _f Û g too is of measure zero. Therefore, by Lebesgue's theorem f Û g is Riemann integrable.

This proof demonstrates the power of Lebesgue's theorem. But one need not use Lebesgue's theorem. One can make use of the uniform continuity of g too. It will involve a clever manipulation of the Riemann sums.

Another Proof of Theorem 2. f : [c, d] ® R is a continuous function, by Theorem 1 of The Boundedness Theorem, Extreme Value theorem and Intermediate Value Theorem, f is bounded. Therefore, there exists a real number M > 0, such that | f (x) | < M for all x in [c, d]. Also by Theorem 9 of Closed and bounded sets, Heine Borel Theorem, etc, f is uniformly continuous. Therefore, given any e > 0, there exists 0 < d < e such that for all x, y in [c, d],

|x - y| < d implies that | f (x) - f ( y)| < ------------------ (5).

Next, since g is Riemann integrable by Theorem 1 of Riemann Integral and Bounded Function, we can find a partition P : a = x₀ < x₁ < x₂ < ¼ < x_n = b for [a, b] such that U(P, g ) - L(P, g ) < d ² /(4M),

where , for i = 1, ¼, n,

M _i (P, g ) = sup{ g (x): x Î[x_{i -} 1 , x_i ]}, m _i (P, g ) = inf{ g (x): x Î[x_{i -} 1 , x_i ]}.

Note that U(P, g ) - L(P, g ) = < d ² /(4M) ---- (6)

Now M _i (P, g ) - m _i (P, g ) = sup{ g (x): x Î[x_{i -} 1 , x_i ]}- inf{ g (x): x Î[x_{i -} 1 , x_i ]}

= sup{g (x) - g(y): x, y Î[x_{i -} 1 , x_i ]}.

Similarly, M _i (P, f Û g ) - m _i (P, f Û g ) = sup{f Û g (x) - f Û g(y): x, y Î[x_{i -} 1 , x_i ]}. Therefore, using the same partition P for the composite f Û g, the difference of the upper and lower Riemann sum with respect to P for f Û g is

U(P, f Û g ) - L(P, f Û g ) =

Let J = {i : 1 £ i £ n, sup{g (x) - g(y): x, y Î[x_{i -} 1 , x_i ]}< d}. So we can rewrite the above difference as

U(P, f Û g ) - L(P, f Û g ) =

+ ----- (7).

If J ¹ Æ, then for i Î J ,

sup{f Û g (x) - f Û g(y): x, y Î[x_{i -} 1 , x_i ]}

= sup{| f (g(x)) - f (g(y))|: x, y Î[x_{i -} 1 , x_i ]} £ by (5) because for any x, y Î[x_{i -} 1 , x_i ] , | g(x) - g(y)| £ sup{g (x) - g(y): x, y Î[x_{i -} 1 , x_i ]}< d . Hence

------------------------ (8)

If {1, ¼,n} - J ¹ Æ , then i Î {1, ¼,n} - J implies that M _i (P, g ) - m _i (P, g ) ³ d. Therefore, . Hence by (6),

• and so -------------------------------- (9)

Therefore, using (9),

.

---------------------------- (10)

Therefore, by (7), if J = Æ using (10), U(P, f Û g ) - L(P, f Û g ) < e/2 < e . If J = {1, ¼,n}, by (8) U(P, f Û g ) - L(P, f Û g ) < e/2 < e. If J ¹ Æ and J ¹ {1, ¼,n}, by (7), (8) and (10), U(P, f Û g ) - L(P, f Û g ) < e/2 + e/2 = e. Thus by the Riemann's conndition (Theorem 1 of Riemann Integral and Bounded Function ), f Û g is Riemann integrable on [a, b]. This completes the proof.

Now we ask the question: If g is Riemann integrable and f is a continuous function, is it true that the composite g Û f is Riemann integrable? In fact it need not be the case. More specifically, if f : [a, b] ® [c, d] is continuous and g : [c, d] ® R is Riemann integrable, the composite function g Û f : [a, b] ® R need not necessarily be Riemann integrable, even if f is a monotonically increasing bijective continuous function. Our counter example will involve two types of Cantor set, one of measure zero and another of positive measure. We shall not go into the construction of these Cantor sets. But we shall state the properties these Cantor sets enjoy.

Example 5 Cantor sets. There are Cantor sets in [0, 1] with measure k with 0 £ k < 1. Let C_k denote the Cantor set of measure k. Thus C₀ is the Cantor set of measure zero. This is the usual cantor set we meet. The construction of these Cantor sets is similar. They satisfy the following properties:

1. Any Cantor set C is closed in [0,1].

2. Any Cantor set C is nowhere dense; that is to say C does not contain any interval, meaning C has empty interior.

3. Any Cantor set is uncountable, more precisely, it has the cardinality of the real numbers R.

4. Any Cantor set is perfect, that is, it is its own accumulation points.

5. At the n-th stage of the construction of C_k for 1 > k > 0 there is left a disjoint union of 2ⁿ closed intervals. The (n+1)-stage is obtained by deleting from the centre of each of these disjoint intervals, an open interval of length 2 ^{- 2n -1}(1- k).

6. At the n-th stage of the construction of C₀ , we have in a disjoint union of 2ⁿ closed intervals. The (n+1)-stage is obtained by deleting from the centre of each of these disjoint intervals, the middle open interval.

7. The complement of any of these Cantor sets is a disjoint union of a countable set of open intervals.

Our example for the continuous function f mentioned above will be given by the following lemma.

Lemma 1. For each k with 0 < k < 1, there exists a function f : [0, 1] ® [0, 1] such that

1. f is monotonically increasing and onto and so f is continuous. (Refer to Theorem 2 of Monotonicity and Continuity of Inverse Function.)

2. f maps the Cantor set C_k onto the Cantor set C₀ .

Proof. We shall look at the complement of the Cantor set C_k and C₀ . They are disjoint unions of open intervals. Let {I₁ , I ₂, I₃, ¼, I_n , ¼} be the open intervals of the complement of C_k , listed from left to right following the construction of C_k in the order of deletion and the natural ordering of the open intervals in each stage of the construction . Similarly let {J₁ , J ₂, J₃, ¼, J_n , ¼} be the open intervals of the complement of C₀ , listed from left to right following the construction of C₀ . For each n denote I_n by (a_n , b_n) and J_n by (c_n , d_n). Now we shall define our function f : [0, 1] ® [0, 1] as follows.

1. f (0) = 0.
2. For x in I_n = (a_n , b_n), f (x) = . This maps I_n bijectively onto J_n.
3. For x ¹ 0 and x in C_k ,

   .

   This is well defined by the completeness property of R, since the set

   is bounded above by 1.

   The function f is increasing on [0,1]-C_k

   First we shall show that f is increasing on the complement of C_k and maps the complement of C_k bijectively onto the complement of C₀ in [0, 1].

   By definition f is increasing on each I_n and maps I_n bijectively onto J_n . At the n-th stage of the construction of the Cantor set we obtained 2 ⁿ - 1 disjoint open intervals, that have been deleted from [0, 1], I₁ , I ₂, I₃, ¼, . The ordering of these intervals is in the order of the deletion starting from the left to the right. The natural ordering defined as follows follows a very simple rule. I_k < I_l if and only if there exists some x in I_k such that x < y for some y in I_l if and only if for any x in I_k

   x < y for all y in I_l . At the n-th stage we can map {I₁ , I ₂, I₃, ¼, I_n , ¼} onto the set { j / 2 ⁿ : j = 1, 2, ¼, 2 ⁿ - 1} according to the order of dissection of [0, 1] into 2 ⁿ parts: I₁ corresponds to 2 ^{n - 1} /2 ⁿ = 1/2, I₂ corresponds to 2 ^{n - 2} /2 ⁿ = 1/4,

   I₃ corresponds to 3´2 ^{n - 2} /2 ⁿ = 3/4, I₄ corresponds to 2 ^{n - 3} /2 ⁿ = 1/8, I₅ corresponds to 3´2 ^{n - 3} /2 ⁿ = 3/8 and so on. This map g(n) : {I₁ , I ₂, I₃, ¼, I_n , ¼} ® { j / 2 ⁿ : j = 1, 2, ¼, 2 ⁿ - 1} is canonical, meaning that it is exactly the same for any of the Cantor set. Hence the natural ordering follows the simple rule I_k < I_l if and only if the corresponding image g(n)(I_k ) < g(n) (I_l ). Thus we can conclude that {J₁ , J ₂, J₃, ¼, J_n , ¼} is ordered in exactly the same way and so I_j < I_k if and only if J_j < J_k . We now claim that f is increasing on the complement of C_k in [0, 1] = . Let x, y be in [0, 1] - C_k be such that x < y. If x and y are in some I_k , then since f is by definition increasing on I_k , f (x) < f (y). Suppose now x is in I_j and y is in I_l . Then x < y implies that I_j < I_l . This is easily seen by taking a positive integer m such that max(j, l ) £ 2 ^m - 1 and consider the ordering map g(m) at the m-th stage of the construction of the Cantor set.. Hence J_j < J_l . Therefore, f (x) < f (y) since f (x) Î J_j and f (y) Î J_l . We have thus shown that f is increasing on the complement of C_k in [0, 1]. Now for any y in [0, 1] - C₀ = , y Î J_k for some k. Since f maps I_k onto J_k , there exists x in I_k such that f (x) = y. Hence f maps the the complement of C_k in [0, 1] onto the complement of C₀ in [0, 1].

   The function f maps C_k into C₀.

   We shall next show that f maps C_k into C₀ . For x = 0 , f (x) = 0 by definition. We now assume x ¹ 0 and x Î C_k . Recall then that

   .

   Suppose that f (x) Ï C₀ . Then for some integer l f (x) Î J_l and since J_l = f ( I_l ) there exists x₀ in I_l such that f (x₀) = f (x) . Then since I_l is open there exists y₀ in I_l with y₀ < x₀ such that f (y₀) < f (x₀) = f (x). Thus by the definition of supremum, there exists y' in [0, 1] - C_k with y' < x and f (y₀) < f (y' ) £ f (x) = f (x₀). Since f is increasing on [0, 1] - C_k , y₀ < y' < x. Then since y₀ Î I_l and so for all y in I_l y < x for otherwise if there exists z in I_l with z > x , then x would belong to ( y₀ , z) Í I_l Í [0, 1] - C_k , contradicting x Î C_k . Now since I_l is open, there exists x' in I_l

   such that x' > x₀ . Thus f (x' ) > f (x₀) = f (x). Also since x' < x , f (x') £ sup{ f (y): y < x and y Î [0, 1] - C_k} = f (x), contradicting f (x' ) > f (x). We can thus deduce that f (x) is in C₀. Therefore, this shows that f maps C_k into C₀.

   The function f is strictly increasing on [0, 1].

   Next we shall show that f is (strictly) increasing on [0, 1]. We have already shown that f is increasing on [0, 1] - C_k. Thus if x, y are in [0, 1] - C_k. and x < y, then f (x) < f (y ). Suppose now x Î C_k and y Ï C_k and x < y. Then for any z Î , z < x implies that z < y. Therefore, since c and z are in [0, 1] - C_k , f (z) < f (y ). Hence f (x) = sup{ f (z): z < x and z Î }£ f (y ). Now since f (x) Î C₀ =[0, 1] - , f (x) ¹ f (y ) and so f (x) < f (y ).

   Suppose now x Î C_k and y Ï C_k and x > y. Then f (y) £ sup{ f (z): z < x and z Î [0, 1] - C_k}= f (x ). Again since f (x) ¹ f (y ), we must have f (x) > f (y ).

   Suppose both x and y are in C_k and x < y. This time we shall use the property of the Cantor set here. Because C_k is nowhere dense, the intersection (x, y)Ç([0, 1]-C_k) ¹ Æ. Therefore, there exists z Î [0, 1] - C_k such that x < z < y. By what we have just proved f (x) < f (z) and f (z) < f (y ). Therefore, we can conclude that f (x) < f (y ). Hence we have shown that f is strictly increasing on [0, 1].

   The function f is onto and maps C_k onto C₀ .

   Now we shall show that f is onto. Since f maps the complement of C_k in [0, 1] onto the complement of C₀ in [0, 1], it is sufficient to show that f maps C_k onto C₀. By examining the definition of f we can consider a similar function mapping C₀ into C_k which is the inverse of f . We are going to use this inverse function to construct a pre image of y in C₀ under f . For y = 0 in C₀ , by definition of f , f (0) = 0 and 0 is also in C_k . For a fixed y ¹ 0 in C₀ , define the following

   x = .

   Note that this is well defined because is in the range of f , the set is non-empty and bounded above by 1 and so the supremum exists by the completeness property of R. Note that we also have

   x = .

   Essentially the same argument for showing that for any l ¹ 0 in C_k , f ( l ) is in C₀ , applies here to conclude that x Î C_k . Now we claim that f (x) = y.

   Note that

   

   .

   Therefore, . We now claim that

   for any z' in ,

   z' < x Û f (z' ) < y. -------------------------- (*)

   This is deduced as follows. Let z' in .

   z' <

   Û there exists z₀ in such that z' < z₀ £ x

   Þ there exists z₀ in such that f (z') < f (z₀ ) < y

   Þ f (z') < y.

   Conversely, if z' in and f (z') < y , then by definition of x, z' £ x and so since z' Î [0, 1] - C_k , z' < x. This proves our claim.

   Therefore,

   { f (z' ): z' < x and z' Î } = { f (z' ): f (z' ) < y and z' Î }

   ={ y' : y' < y and y' Î }.

   Thus f (x) = sup{ f (z' ): z' < x and z Î }

   = sup{ y' : y' < y and y' Î }= y . This is seen as follows, obviously f (x) £ y.

   If f (x) < y, then there exists y₀ in such that f (x) < y₀ < y since both f (x) and y are in C₀ and C₀ is nowhere dense. Therefore, there exists x₀ in with

   f (x) < y₀ = f (x₀) < y. Since f is increasing , x < x₀. But by (*) f (x₀) < y implies that x₀ < x contradicting x < x₀. Therefore, f (x) = y. Thus we have shown that f maps C_k onto C₀ and consequently f is onto..

   We have now thus shown that f is a strictly increasing function mapping the closed and bounded interval [0, 1] onto itself and so by Theorem 3 of Inverse Function and Continuity, f is continuous on [0, 1]. This completes the proof of Lemma 1.

   Example 5. Riemann integrable function of a continuous function need not necessarily be Riemann Integrable.

   Let g: [0, 1] ® R be defined by g(x) = 0 if x Ï C₀ and g(x) = 1 if x Î C₀ . Let f : [0, 1] ® [0,1] be the continuous strictly increasing bijective function mapping a Cantor set of positive measure C_k onto the Cantor set of measure zero C₀ given by Lemma 1. Then g is Riemann integrable, f is continuous but g Û f is not Riemann integrable.

   Proof. The function g is obviously bounded. Since [0, 1] - C₀ = is a countable disjoint union of open interval and g is zero on each of these open intervals, g is continuous on [0, 1] - C₀. The function g is discontinuous at every point in C₀. This is seen as follows. Take e = 1/2. For any d > 0 and any x in C₀, (x - d, x + d)Ç [0,1] M C₀ because C₀ is nowhere dense and so there exists x_d Î (x - d, x + d)Ç [0,1] - C₀ . Hence | f (x_d) - f (x)| = |0 - 1| = 1 > e = 1/2. This implies that f is discontinuous at x in C₀ . Since C₀ is of measure zero, by Lebesgue theorem, g is Riemann integrable. Take any k such that 0 < k <1, for instance, k = 1/2 . Then C_k is of positive measure. The function f defined by Lemma 1 is a continuous bijection of [0, 1] onto [0, 1] and maps the Cantor set of measure k, C_k bijectively onto C₀. Then

   .

   Thus g Û f is constant on the complement of C_k in [0, 1] which is a disjoint union of open intervals and so g Û f is continuous on each of these intervals and so g Û f is continuous on = [0, 1] - C_k . As before we can check that g Û f is discontinuous at any point x in C_k . Take again e = 1/2. Then for any d > 0 and any x in C_k , (x - d, x + d)Ç [0,1] contains a point x_d not in C_k because C_k is nowhere dense. Thus | g Û f (x_d) - g Û f (x)| = |0 - 1| = 1 > e = 1/2. This means g Û f is discontinuous at any point in C_k . Therefore, by Lebesgue's Theorem, since C_k is of positive measure , g Û f is not Riemann integrable.

   There are other examples: one example would be to take g to be the function defined on [0, 1] such that g(x) = 0 for all x, such that 0 £ x < 1 and g(x) =1 when x = 1 and f to be a function on [0, 1] such that on each of the disjoint interval I_n the graph of f is a 'U' - shape graph with limit tending to 1 at both end points. Then the composite g Û f would be the same as the composite in the above example.

   Example 5 inspires the next example.

   Example 6. Let g: [0, 1] ® R be defined by g(x) = 0 if x Ï C_k for k = 1/2 and g(x) = 1 if x Î C_k. Let f ^-1: [0, 1] ® [0,1] be the inverse of the continuous strictly increasing bijective function f , given by Lemma 1, which maps a Cantor set of positive measure C_k onto the Cantor set of measure zero C₀. Then g is not Riemann integrable, f ^-1 is continuous but g Û f ^-1 is Riemann integrable.

   The next question that we would ask is: What about composite of Lebesgue integrable functions? This will be discussed next.

ã Ng Tze Beng 2001