7.3. THE RIEMANN-DARBOUX 239

7.3 The Riemann-Darboux Integral 7.3.1 Definition and Examples For various partitions P of [a, b] and a fixed bounded f, we set

S (P ) = S (f, P ) = S (f, P ): P is a partition of [a, b] and    S (P ) = S (f, P ) = S (f, P ): P is a partition of [a, b] { } { } { } It is important to understand that these sets are taken over all the possible par- titions P of [a, b]. As noticed above, they are both bounded above by M (b a) and below by m (b a). Thus, each set has a supremum and an infimum.− Therefore, we define:−

Definition 7.3.1 (Rieman-Darboux integral) Let f be bounded on [a, b].

b 1. The upper Darboux integral of f on [a, b], denoted af, is defined by b af = inf S (P ) . R R  b 2. The lower Darboux integral of f on [a, b], denoted a f, is defined by b f = sup S (P ) . R a { } R b b 3. If af = a f, the common value is called the Riemann-Darboux in- tegral (R-D integral for short) of the function f on the [a, b] R R b and is denoted a f. When this common value exists, we say that f is Riemann-Darboux integrable (or R-D integrable) on [a, b]. R Remark 7.3.2 From the definition, the following should be obvious.

b S (P ) f ≤ Za and b f S (P ) ≤ Z a We now need to understand what kind of functions will be R-D integrable. We also need to learn how to compute the Riemann-Darboux integral. The next results and examples will help us achieve that.

Theorem 7.3.3 If f is a bounded function on [a, b] then

b b f f ≤ Za Z a Proof. See exercises. Hint: use the previous theorem. 240 CHAPTER 7. RIEMANN INTEGRATION

This result is simply the analytic form of what is obvious geometrically. Though the results obtained above seem simple, we must not forget that they uses a very powerful principle: the supremum principle. Remark 7.3.4 Combining the inequality of the above theorem and those in remark 7.3.2, we have

b b S (P ) f f S (P ) ≤ ≤ ≤ Za Z a Example 7.3.5 Discuss the Riemann-Darboux integrability of f (x) = k on [a, b] where k is a constant. b b We need to prove that a f = af. To compute these , we need to find S (P ) and S (P ) forR any partitionR P . If P is any partition of [a, b], then m = mi = Mi = M = k (since the function is constant. Therefore, S (P ) = n n n n kdi = k di = k (b a). Similarly, S (P ) = kdi = k di = k (b a). i=1 i=1 − i=1 i=1 − Hence,P forP any partition P, S (P ) = k (b a) andP thereforeP does not depend − b on P . It follows that inf S (P ) = k (b a) and therefore f = k (b a). − a − Similarly, we see that bf = k (b a). So, the two integral are equal. It follows a  R − b that f is Riemann-DarbouxR integrable and f = k (b a). a − Example 7.3.6 Let c [a, b]. Discuss theR R-D integrability of f defined on [a, b] by ∈ k > 0 if x = c f (x) = 0 if x 6= c  From what we did above, we see that If P is any partition of [a, b], then the com- ponents of S (P ) will agree with the components of S (P ) except on the intervals which contain c (there can be at most two). Suppose that c [xi 1, xi]. The contribution from this subinterval to S (P ) will be zero, while∈ the− contribution to S (P ) will be kdi. It follows that S (P ) S (P ) + 2k P . In other words, ≤ k k S (P ) S (P ) 2k P − ≤ k k Since P can be made as small as we want, we conclude that for any  > 0, there isk ak partition P such that S (P ) S (P )  − ≤ Since b b f f S (P ) S (P ) a − a ≤ − Z Z it follows that for any  > 0,

b b f f  a − a ≤ Z Z

7.3. THE RIEMANN-DARBOUX INTEGRAL 241

b b From a theorem studied in class, it follows that a f = af thus f is R-D integrable. R R Remark 7.3.7 The function in the example above was identical to the function in the example before except at one point. They were both integrable. In fact, later on, we will prove a much more powerful result. If a function g is obtained from a function f by altering a finite number of values of f then if f is R-D integrable, so is g and the two integrals have the same value.

Example 7.3.8 Discuss the R-D integrability of f defined on [a, b] by

1 if x is irrational f (x) = 0 if x is rational  Let P be any partition of [a, b]. Because any subinterval of [a, b] contains both irrational and rational numbers, it follows that mi = 0 and Mi = 1. Therefore, b S (P ) = b a and S (P ) = 0 for any partition P of [a, b]. It follows that a f = 0 b − and f = b a therefore f is not R-D integrable on [a, b]. R a − R Remark 7.3.9 This establishes the fact that not all bounded functions are R-D integrable. It also suggests that continuity will play an important role. There was an important difference between the last example and the one before. In the last example, no matter which partition we selected, we could not make S (P ) S (P ) as small as we wanted. We could do that in the previous one. This fact− is captured in the next result.

7.3.2 Integrability Theorems We now look at several theorem about integrability of functions. The first theorem, attributed to Riemann, gives a necessary and suffi cient condition for a function to be integrable.

Theorem 7.3.10 A bounded function f on [a, b] is R-D integrable on [a, b] if and only if for every  > 0, there is a partition P of [a, b] such that

S (P ) S (P ) <  − Proof. The proof is assigned as an exercise. We give an outline of the proof here. This is an "if and only if", so we must prove both directions.

( =) Use the definition of R-D integrability and the fact that S (P ) • ⇐ b ≤ bf f S (P ). a ≤ a ≤ R R (= ) we break the proof in several steps: • ⇒ 242 CHAPTER 7. RIEMANN INTEGRATION

1. Use the fact that if H is a non empty and bounded set of real numbers, then and upper bound b of H is the supremum of H if and only if for every positive  there exists x H such that b  < x b (this is another form of theorem 2.1.33∈ last semester)− to prove≤ that  bf < S (Q) for some partition Q a − 2 b  2. Similarly,R prove that S (R) < f + for some partition R. a 2 3. Apply both inequalities to theR refinement P = Q R and conclude using the fact that f is R-D integrable. ∪

The remaining theorems give either necessary conditions or suffi cient condi- tions.

Theorem 7.3.11 If f is continuous on [a, b] then f is R-D integrable on [a, b]. Proof. The proof is assigned as an exercise. We give an outline of the proof here. We break the proof in several steps.

1. Using the previous theorem, it is enough to show that S (P ) S (P ) < for some partition P of [a, b]. − 2. Use the fact that if f is continuous on [a, b], then it is uniformly continuous hence we can make f (s) f (t) as small as we want. Of course, you will have to determine how| small− we| want it to show that f is R-D integrable. 3. From the previous step and properties of continuous functions on a closed interval, conclude that we can make Mi mi as small as we want. Of course, you will have to determine how small− we want it to show that f is R-D integrable. Here, I am using the notation of the notes. 4. Using the estimate of the previous step, compute S (P ) S (P ) and show S (P ) S (P ) < . − −

Theorem 7.3.12 A monotonic function on [a, b] is R-D integrable. Proof. The proof is assigned as an exercise. We give an outline of the proof here in the case f is monotone increasing. The case f monotone decreasing is similar.

Case 1: f (a) = f (b). What does this imply? You should see that we’ve • already done it. Case 2: f (b) > f (a). We prove the result by proving that S (P ) S (P ) < • . We break the proof in several steps. −

n

1. Prove that (Mi mi) = f (b) f (a). − − i=1 X 7.3. THE RIEMANN-DARBOUX INTEGRAL 243

2. If P is a partition with norm less that a constant you will have to determine, prove that S (P ) S (P ) < . −

Theorem 7.3.13 If f is R-D integrable on [a, b] and g is obtained from f by altering the values of f at a finite number of points, then g is also R-D integrable. b b Furthermore, a f = a g Proof. The proof is assigned as an exercise. We give an outline of the proof here. We breakR the proofR in several cases and give an outline for each case.

Suppose that g differs from f at one point that is let g be such that g (x) = • f (x) if x = c [a, b] . We outline the proof in the case y > f (c). y = f (c) if6 x =∈ c The other6 case is similar.

1. First, we prove that g is R-D integrable by proving that S (g, P ) S (g, P ) < . For this, using the notation in the notes, prove that S (f, P ) −S (g, P ) | − | ≤ 2 y mi di and S (f, P ) S (g, P ) 2 Mi y di. Then, explain why| − one can| find partitions−Q and R such≤ that| −S (|f, Q) S (g, Q) <   | − | , S (f, Q) S (g, Q) < and S ( f, R) S (f, R) < . Conclude 3 − 3 − 3 one can find a partition P such that S (g, P ) S (g, P ) < . − b b 2. Next, we prove that f = g. For this explain why S (f, R) a a b b b b ≤ a f S (g, R) a gR af R S (f, P ) ag S (g, P ) and using ≤ ≤ ≤ ≤ ≤ ≤b b Rthe fact that f andR g areR R-D integrable, showR a f = a g.

Suppose g differs from f at k points. Build a sequenceR of functionsR fi • { } where f0 = f, fk = g such that each two consecutive function differ by only one point and apply the result to the sequence.

Corollary 7.3.14 A bounded function on a closed interval [a, b] whose set of discontinuities is finite is R-D integrable on [a, b]. Proof. The proof is assigned as an exercise.

We finish with a slightly more general version of the previous theorem and corollary. We will state the result without proof. This result involves the concept of measure, more specifically what a set of measure zero is. Measure will be treated a few chapters ahead. Here, we will only define what a set of measure zero is.

Definition 7.3.15 (set of measure zero) Let S R. S is said to have ⊆ measure zero if for every  > 0, there exists a countable collection In of open intervals such that { }

S In ⊆ n [ 244 CHAPTER 7. RIEMANN INTEGRATION and l (In)  n ≤ X where l (In) denotes the width of In.

Example 7.3.16 Every finite set has measure zero. Let S = x1, x2, ..., xn .   { } Let  > 0 be given. Define Ii = xi 2n , xi + 2n . Then, clearly S In and − ⊆ n  [ n

l (Ii) =  i=1 X Example 7.3.17 Every countable subset of R has measure zero. If S is count- able, then we can write S = xi ∞ . Let  > 0 be given. For each i N, define { }i=1 ∈    ∞ ∞  ∞ 1 Ii = xi i+1 , xi + i+1 . Then l (Ii) = 1 and l (Ii) = i =  i = − 2 2 2 2 2 i=1 i=1 i=1 .  X X X

Remark 7.3.18 The above example implies that Q has measure zero. Remark 7.3.19 When we say that a property holds almost everywhere, we mean the property holds everywhere except on a set of measure zero.

Theorem 7.3.20 (Lebesgue) Let f be a bounded function on [a, b]. f is R-D integrable if and only if its set of discontinuities has measure zero.

7.3.3 Exercises 1. Prove theorem 7.3.3. b 2. Prove that bf f S (P ) S (P ) for any partition P . a − a ≤ −

3. Prove theorem R 7.3.10R following the outline in the notes.

4. Prove theorem 7.3.11 following the outline in the notes. 5. Prove theorem 7.3.12 following the outline in the notes. 6. Prove theorem 7.3.13 following the outline in the notes. 7. Prove theorem 7.3.14 following the outline in the notes. 8. Discuss the R-D integrability of the functions below in the given interval.

(a) 1 if 0 x < 1 f (x) = 2 if ≤x = 1  1 on [0, 1]. In addition, find 0 f if it exists. R 7.3. THE RIEMANN-DARBOUX INTEGRAL 245

(b) 1 1 1 0 if x = 1, , , , ... f (x) = 2 3 4 ( 1 otherwise 1 on [0, 1]. In addition, find 0 f if it exists. (c) R 1 if x Q f (x) = ∈ 1 if x R Q  − ∈ \ on [a, b] for any a < b. (d) x if x Q f (x) = ∈ 0 if x R Q  ∈ \ on [0, 1] (e) 0 if x is irrational f (x) = 1 p  if x = Q in lowest terms  q q ∈ on [0, 1]  246 CHAPTER 7. RIEMANN INTEGRATION

7.3.4 Hints for the Exercises 1. Prove theorem 7.3.3. Hint: Using the previous theorems, for a fixed partition P of [a, b], show that S (Q) S (P ) for every partition Q of [a, b]. Taking the supremum ≤ over all the partitions Q, show that bf S (P ). Now, since P was a ≤ arbitrary, this result is true for every partition P of [a, b]. Use that to R conclude.

b 2. Prove that bf f S (P ) S (P ) for any partition P . a − a ≤ − b Hint: From R the aboveR theorems, explain how we have S (P ) f a b ≤ ≤ f S (P ) then use it to derive the result. R a ≤ 3. RProve theorem 7.3.10 following the outline in the notes. Hint: none other than what is already given.

4. Prove theorem 7.3.11 following the outline in the notes. Hint: none other than what is already given.

5. Prove theorem 7.3.12 following the outline in the notes. Hint: none other than what is already given.

6. Prove theorem 7.3.13 following the outline in the notes. Hint: none other than what is already given.

7. Prove theorem 7.3.14 following the outline in the notes. Hint: none other than what is already given.

8. Discuss the R-D integrability of the functions below in the given interval.

(a) 1 if 0 x < 1 f (x) = 2 if ≤x = 1  1 on [0, 1]. In addition, find 0 f if it exists. Hint: Using the results of this section, compare this function with g (x) = 1 on [0, 1]. R (b) 1 1 1 0 if x = 1, , , , ... f (x) = 2 3 4 ( 1 otherwise

1 on [0, 1]. In addition, find 0 f if it exists. Hint: Using the results of this section, compare this function with g (x) = 1 on [0, 1]. R 7.3. THE RIEMANN-DARBOUX INTEGRAL 247

(c) 1 if x Q f (x) = ∈ 1 if x R Q  − ∈ \ on [a, b] for any a < b. b Hint: Compute S (P ) and S (P ). From this, deduce the value of af and b f and conclude. a R (d) R x if x Q f (x) = ∈ 0 if x R Q  ∈ \ on [0, 1] Hint: Show that S (P ) S (P ) cannot be made as small as we want (less than ) and conclude.− To do this, show that S (P ) = 0 and 1 S (P ) for some partition P . ≥ 4 (e) 0 if x is irrational f (x) = 1 p  if x = Q in lowest terms  q q ∈ on [0, 1]  Hint: Prove that given  > 0, one can find a partition P such that S (P ) S (P ) < . − Bibliography

[B] Bartle, G. Robert, The elements of , Second Edition, John Wiley & Sons, 1976.

[C1] Courant, R., Differential and Integral , Second Edition, Volume 1, Wiley, 1937. [C2] Courant, R., Differential and Integral Calculus, Second Edition, Volume 2, Wiley, 1937.

[DS] Dangello, Frank & Seyfried Michael, Introductory Real Analysis, Houghton Miffl in, 2000. [F] Fulks, Watson, Advanced Calculus, an Introduction to Analysis, Third Edi- tion, John Wiley & Sons, 1978.

[GN] Gaskill, F. Herbert & Narayanaswami P. P., Elements of Real Analysis, Prentice Hall, 1998. [LL] Lewin, J. & Lewin, M., An Introduction to Mathematical Analysis, Second Edition, McGraw-Hill, 1993. [MS] Stoll, Manfred, Introduction to Real Analysis, Second Edition, Addison- Wesley Higher Mathematics, 2001.

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