MACT 4127 Real Analysis II Spring 2025

The Riemann Methos of Integration

Contents
1 Properties of the Riemann Integral 1
1.1 The Fundamental Theorem of Calculus . . . . . . . . . . . . . . . . . . 5

1 Properties of the Riemann Integral

Our main aim in this section is to study the properties of the Riemann integral, this
means that we want to learn more about the behavior of the Riemann integral with
respect to different operations and relations on functions and on intervals.
1 Theorem. The Riemann integral on an interval [𝑎, 𝑏] is linear and monotone,
i.e. Given 𝑓, 𝑔 : [𝑎, 𝑏] −→ R two Riemann integrable functions, then

(a) For any 𝛼, 𝛽 ∈ R we have

∫︁ 𝑏 ∫︁ 𝑏 ∫︁ 𝑏
(𝛼𝑓 (𝑥) + 𝛽𝑔(𝑥)) 𝑑𝑥 = 𝛼. 𝑓 (𝑥) 𝑑𝑥 + 𝛽. 𝑔(𝑥) 𝑑𝑥.
𝑎 𝑎 𝑎

∫︀ 𝑏 ∫︀ 𝑏
(b) If 𝑓 (𝑥) ≤ 𝑔(𝑥) for all 𝑥 ∈ [𝑎, 𝑏], then 𝑎
𝑓 (𝑥) 𝑑𝑥 ≤ 𝑎
𝑔(𝑥) 𝑑𝑥.

Proof. — Let z = (𝑥0 , · · · , 𝑥𝑛 ) be a subdivision of [𝑎, 𝑏] and let 𝜉𝑘 ∈ 𝐼𝑘 , 𝑘 =

1, · · · , 𝑛. Then the corresponding Riemann sums satisfy:

𝑅(𝛼𝑓 + 𝛽𝑔; z) = 𝛼𝑅(𝑓 ; z) + 𝛽𝑅(𝑔; z),

and if 𝑓 (𝑥) ≤ 𝑔(𝑥)∀𝑥 ∈ [𝑎, 𝑏], then 𝑅(𝑓 ; z) ≤ 𝑅(𝑔; z).
∫︀ 𝑏 ∫︀ 𝑏
Now, let 𝐴 = 𝑎 𝑓 (𝑥) 𝑑𝑥, 𝐵 = 𝑎 𝑔(𝑥) 𝑑𝑥 and put 𝐶 = 𝛼𝐴 + 𝛽𝐵. For any 𝜀 > 0 there
exist subdivisions z1 , z2 such that
𝜀
z finer than z1 =⇒ |𝑅(𝑓 ; z) − 𝐴| < ,
2 |𝛼|
𝜀
and z finer than z2 =⇒ |𝑅(𝑔; z) − 𝐵| <
2 |𝛽|

Let z0 be a subdivision finer than both z1 , z2 , then

z finer than z0 =⇒ |𝑅(𝛼𝑓 + 𝛽𝑔; z) − 𝐶| ≤ |𝛼| |𝑅(𝑓 ; z) − 𝐴| + |𝛽| |𝑅(𝑔; z) − 𝐵|

𝜀 𝜀
< |𝛼| + |𝛽| = 𝜀.
2 |𝛼| 2 |𝛽|
∫︀ 𝑏
This shows that 𝛼𝑓 + 𝛽𝑔 is Riemann integrable and that 𝑎
(𝛼𝑓 (𝑥) + 𝛽𝑔(𝑥)) 𝑑𝑥 =
∫︀ 𝑏 ∫︀ 𝑏
𝛼. 𝑎 𝑓 (𝑥) 𝑑𝑥 + 𝛽. 𝑎 𝑔(𝑥) 𝑑𝑥.

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 To prove the monotonicity of the Riemann integral, lets asuume that 𝑓 (𝑥) ≤ 𝑔(𝑥)
∫︀ 𝑏 ∫︀ 𝑏
for all 𝑥 ∈ [𝑎, 𝑏] and suppose that 𝐴 = 𝐴 = 𝑎 𝑓 (𝑥) 𝑑𝑥 > 𝐵 = 𝑎 𝑔(𝑥) 𝑑𝑥. Take
𝜀 = (𝐴 − 𝐵)/2. Then there exist subdivisions z1 , z2 such that
z finer than z1 =⇒ |𝑅(𝑓 ; z) − 𝐴| < 𝜀,
and z finer than z2 =⇒ |𝑅(𝑔; z) − 𝐵| < 𝜀.
Again, take z to be a subdivision finer than both z1 , z2 , then
𝑅(𝑔; z) < 𝐵 + 𝜀 = 𝐴 − 2𝜀 + 𝜀 = 𝐴 − 𝜀 < 𝑅(𝑓 ; z)
which contradicts our first deduction that 𝑅(𝑓 ; z) ≤ 𝑅(𝑔; z).
The class of all Riemann integrable functions on [𝑎, 𝑏] is usually denoted by ℛ([𝑎, 𝑏]).
The previous theorem and results from the previous section tell us that
2 Theorem.
(a) ℛ([𝑎, 𝑏]) is a vector space and the Riemann integral is a linear monotone operator
from ℛ([𝑎, 𝑏]) into R.
(b) ℛ([𝑎, 𝑏]) ⊂ B([𝑎, 𝑏]).
With the aid of the Cauchy criterion we will be able to prove:
3 Proposition. Let 𝑓 : [𝑎, 𝑏] −→ R be a Riemann integrable function, then for any
𝑐, 𝑑 ∈ [𝑎, 𝑏] the function 𝑓 is Riemann integarble on [𝑐, 𝑑].
4 Theorem. [Adition Property of the Riemann Integral]
Let 𝑓 : [𝑎, 𝑏] −→ R, then for any 𝑐 ∈ [𝑎, 𝑏]:
∫︁ 𝑏 ∫︁ 𝑐 ∫︁ 𝑏
𝑓 (𝑥) 𝑑𝑥 = 𝑓 (𝑥) 𝑑𝑥 + 𝑓 (𝑥) 𝑑𝑥,
𝑎 𝑎 𝑐

where the equality is understood in the sense that if one of the two sides exists, then
the other side exists also and the two sides are equal.
Sketch of Proof. — The first step is to prove that given a subdivision z of [𝑎, 𝑏]
we can get two subdivisions: z1 of [𝑎, 𝑐] and z2 of [𝑐, 𝑏], such that z1 ∪ z2 is finer than
z.
On the otherhand, it is clear that given any subdivisions z1 of [𝑎, 𝑐], and 𝑧2 of [𝑐, 𝑏],
then their union is a subdivision of [𝑎, 𝑏].
The second step is to prove that |𝑅(𝑓, z) − (𝑅(𝑓, z1 ) + 𝑅(𝑓, z2 ))| can be made as
small as we like by choosing a sufficiently fine subdivisions z, z1 , z2 .
Another useful fact about Riemann integrable functions is
5 Proposition. If 𝑓, 𝑔 are Riemann integrable on [𝑎, 𝑏], then the functions 𝑢 =
max(𝑓, 𝑔) and ℓ = min(𝑓, 𝑔) are also Riemann integrable on [𝑎, 𝑏].
Let’s now see some negative aspects of the Riemann integral:
6 Example.
(a) There exists a bounded function 𝑓 : [0, 1] −→ R such that |𝑓 | is Riemann inte-
grable while 𝑓 is not Riemann integrable.
{︃
1 if 𝑥 ∈ Q,
Let 𝑓 (𝑥) = Then |𝑓 (𝑥)| ≡ 1, hence is integrable while 𝑓 is not
−1 otherwise.
because any lower Darboux sum of 𝑓 equals −1 and any upper Darboux sum
equals 1.

2
 (b) There exists a positive Riemann integrable function 𝑔 on [0, 1] and a function
𝑓 : [0, 1] −→ R such that 0 ≤ 𝑓 (𝑥) ≤ 𝑔(𝑥) for all 𝑥 ∈ [0, 1], but 𝑓 is not Riemann
integrable.
let 𝑔(𝑥) = 1 and take 𝑓 to be the Dirichlet function. Then 𝑔 is Riemann inte-
grable, 𝑓 is not Riemann integrable and 𝑓 (𝑥) ≤ 𝑔(𝑥) for all 𝑥.

Finally, lets study the relation of Riemann integrability to limits: a negative result
is:
7 Example. There exists a sequence (𝑓𝑛 )𝑛∈N* of Riemann integrable functions on
[0, 1], such that
(i) 𝑓𝑛 (𝑥) ≤ 𝑓𝑛+1 (𝑥) for all 𝑥 ∈ [0, 1],
∫︀ 1
(ii) lim𝑛→∞ 0 𝑓𝑛 (𝑥) 𝑑𝑥 exists,

(iii) The function 𝑓 (𝑥) = lim𝑛→∞ 𝑓𝑛 (𝑥) is not Riemann integrable.

We know that {︃ Q ∩ [0, 1] is countable, so we may write Q ∩ [0, 1] = (𝑟𝑘 )𝑘∈N* . De-
1 if 𝑥 = 𝑟𝑘 , 1 ≤ 𝑘 ≤ 𝑛,
fine 𝑓𝑛 (𝑥) = . The sequence (𝑓𝑛 )𝑛∈N* satisfies all the above
0 otherwise.
conditions.
But we have a positive result
8 Theorem. If 𝑓 is the uniform limit on [𝑎, 𝑏] of a sequence (𝑓𝑛 )𝑛∈N* of Riemann
integrable functions, then 𝑓 is Riemann integrable and
∫︁ 𝑏 ∫︁ 𝑏
𝑓 (𝑥) 𝑑𝑥 = lim 𝑓𝑛 (𝑥) 𝑑𝑥.
𝑎 𝑛→∞ 𝑎

Proof. — Let (𝑓𝑛 )𝑛∈N* be a sequence of Riemann integrable functions on [𝑎, 𝑏],
∫︀ 𝑏
which converges uniformly on [𝑎, 𝑏] to 𝑓 . Let’s denote the integral 𝑎 𝑓𝑛 (𝑥) 𝑑𝑥 by 𝐼𝑛 .
Claim (1): (𝐼𝑛 )𝑛∈N* converges to a real number 𝐼.
Proof of Claim (1): First we note that the sequence (𝑓𝑛 )𝑛∈N* is uniformly Cauchy,
so, given any 𝜖 > 0, we may choose 𝑛0 so that

𝑛, 𝑚 > 𝑛0 =⇒ sup |𝑓𝑛 (𝑥) − 𝑓𝑚 (𝑥)| < 𝜖/𝑏 − 𝑎.

𝑥∈[𝑎,𝑏]

Hence,
⃒∫︁ 𝑏 ⃒ ∫︁ 𝑏
⃒ ⃒
𝑛, 𝑚 > 𝑛0 =⇒ |𝐼𝑛 − 𝐼𝑚 |] = ⃒⃒ (𝑓𝑛 (𝑥) − 𝑓𝑚 (𝑥))𝑑𝑥⃒⃒ ≤ |𝑓𝑛 (𝑥) − 𝑓𝑚 (𝑥)| 𝑑𝑥
𝑎 𝑎
∫︁ 𝑏
≤ sup |𝑓𝑛 (𝑥) − 𝑓𝑚 (𝑥)| 𝑑𝑥 = sup |𝑓𝑛 (𝑥) − 𝑓𝑚 (𝑥)| (𝑏 − 𝑎) < 𝜖.
𝑎 𝑥∈[𝑎,𝑏] 𝑥∈[𝑎,𝑏]

So, (𝐼𝑛 )𝑛∈N* is a Cauchy sequence of real numbers. By completeness of R it converges

to a real number 𝐼. ∫︀ 𝑏
Claim (2): 𝑓 is Riemann integrable with 𝑎 𝑓 (𝑥) 𝑑𝑥 = 𝐼.
Proof of Claim (2): Let 𝜖 > 0 be given, there exists 𝑛1 such that

𝑛 > 𝑛1 =⇒ sup |𝑓 (𝑥) − 𝑓𝑛 (𝑥)| < 𝜖/4(𝑏 − 𝑎).

𝑥∈[𝑎,𝑏]

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Also, there exists 𝑛2 for which

𝑛 > 𝑛2 =⇒ |𝐼𝑛 − 𝐼| < 𝜖/2.

Let 𝑛 > 𝑚𝑎𝑥(𝑛1 , 𝑛2 ). The function 𝑓𝑛 is Riemann integrable on [𝑎, 𝑏], so there exists
a subdivision z0 ∑︀
of [𝑎, 𝑏] so that for any subdivision w finer than z0 and any Riemann
sum 𝑅(𝑓𝑛 , w) = 𝑘 𝑓𝑛 (𝑡𝑘 )(𝑥𝑘 − 𝑥𝑘−1 , we have: |𝑅(𝑓𝑛 , 𝑤) − 𝐼𝑛 | < 𝜖/4.
For this subdivision w we find
⃒ ⃒
⃒∑︁ ∑︁ ⃒
|𝑅(𝑓, w) − 𝑅(𝑓𝑛 , w)| = ⃒ 𝑓 (𝑡𝑘 )(𝑥𝑘 − 𝑥𝑘−1 ) − 𝑓𝑛 (𝑡𝑘 )(𝑥𝑘 − 𝑥𝑘−1 )⃒
⃒ ⃒
⃒ ⃒
⃒ 𝑘 𝑘
⃒
⃒∑︁ ⃒ ∑︁
= ⃒ (𝑓 (𝑡𝑘 − 𝑓𝑛 (𝑡𝑘 ))(𝑥𝑘 − 𝑥𝑘−1 )⃒ ≤ |𝑓 (𝑡𝑘 ) − 𝑓𝑛 (𝑡𝑘 )| (𝑥𝑘 − 𝑥𝑘−1 )
⃒ ⃒
⃒ ⃒
𝑘 𝑘
𝜖 ∑︁
< (𝑥𝑘 − 𝑥𝑘−1 ) = 𝜖/4.
4(𝑏 − 𝑎) 𝑘

Now, consider a Riemann sum of 𝑓 with respect to w, we see that

|𝑅(𝑓, w) − 𝐼| ≤ |𝑅(𝑓, w) − 𝑅(𝑓𝑛 , w)| + |𝑅(𝑓𝑛 , w) − 𝐼𝑛 | + |𝐼𝑛 − 𝐼|

𝜖 𝜖
< + + 𝜖2 = 𝜖.
4 4
This proves our claim.
9 Exercises.

(1) Prove that if 𝑓 : [𝑎, 𝑏] −→ R is a Riemann integrable function, and

𝑚 ≤ 𝑓 (𝑥) ≤ 𝑀, ∀𝑥 ∈ [𝑎, 𝑏]

then ∫︁ 𝑏
𝑚(𝑏 − 𝑎) ≤ 𝑓 (𝑥)𝑑𝑥 ≤ 𝑀 (𝑏 − 𝑎).
𝑎

(2) Prove that if 𝑓, 𝑔 ∈ ℛ([𝑎, 𝑏]), then 𝑓.𝑔 ∈ ℛ([𝑎, 𝑏]).

(3) Prove that if 𝑓, 𝑔 ∈ ℛ([𝑎, 𝑏]), and ℎ(𝑥) = max{𝑓 (𝑥), 𝑔(𝑥)}, then ℎ ∈ ℛ([𝑎, 𝑏]).
Deduce that if 𝑓 is Riemann integrable, then |𝑓 | is Riemann integrable.

(4) Prove that if 𝑓 ∈ ℛ([𝑎, 𝑏]), then

⃒∫︁ 𝑏 ⃒ ∫︁ 𝑏
⃒ ⃒
⃒ 𝑓 (𝑥) 𝑑𝑥 ⃒≤ |𝑓 (𝑥)| 𝑑𝑥.
⃒ ⃒
𝑎 𝑎

(5) Let 𝑓 ∈ ℛ([𝑎, 𝑏]) and let 𝑚 = inf 𝑥∈[𝑎,𝑏] 𝑓 (𝑥), 𝑀 = sup𝑥∈[𝑎,𝑏] 𝑓 (𝑥). Prove that if
𝑔 : [𝑚, 𝑀 ] −→ R is continuous, then 𝑔 ∘ 𝑓 is Riemann integrable.

4
1.1 The Fundamental Theorem of Calculus
Let’s now try to explain the phrase “integration is the inverse operation of dif-
ferentiation”.
The explanation is given by the famous theorem of Newton and Leibnitz which we
shall prove at the end of this section. Proofs are easy.
But firstly, we need to learn more about the value of the integral, especially for
continuous functions
10 Theorem. [The mean value theorem for the Riemann integral]
Let 𝑓 ∈ ℛ([𝑎, 𝑏])

(a) If 𝑚 ≤ 𝑓 (𝑥) ≤ 𝑀 for every 𝑥 ∈ [𝑎, 𝑏], then

∫︁ 𝑏
𝑚(𝑏 − 𝑎) ≤ 𝑓 (𝑥) 𝑑𝑥 ≤ 𝑀 (𝑏 − 𝑎).
𝑎

(b) If 𝑓 is continuous, then there exists a point 𝑐 ∈ [𝑎, 𝑏] such that

∫︁ 𝑏
𝑓 (𝑥) 𝑑𝑥 = 𝑓 (𝑐)(𝑏 − 𝑎).
𝑎

We next define the indefinite integral

11 Definition. Let 𝑓 ∈ ℛ([𝑎, 𝑏]) and let 𝑐 ∈ [𝑎, 𝑏], then the function 𝐹𝑐 : [𝑎, 𝑏] −→ R
defined by ∫︁ 𝑥
𝐹𝑐 (𝑥) = 𝑓 (𝑡) 𝑑𝑡
𝑐
is called an indefinite integral of 𝑓 .
12 Proposition. Any two indefinite integrals of the same function 𝑓 differ by a
constant.
13 Theorem. [The Fundamental Theorem of Calculus]
Let 𝑓 ∈ ℛ([𝑎, 𝑏]), and let 𝐹 be an indefinite integral of 𝑓 , then

(a) 𝐹 is continuous.
∫︀ 𝑏
(b) 𝑎 𝑓 (𝑡) 𝑑𝑡 = 𝐹 (𝑏) − 𝐹 (𝑎).

(c) If 𝑓 is continuous at 𝑥0 ∈]𝑎, 𝑏[, then 𝐹 is differentiable at 𝑥0 and 𝐹 ′ (𝑥0 ) = 𝑓 (𝑥0 ).

Proof. —

(a) Let 𝑥, 𝑦 ∈ [𝑎, 𝑏] and suppose 𝑦 < 𝑥, then

⃒∫︁ 𝑥 ∫︁ 𝑦
⃒ ∫︁ 𝑥
⃒ ⃒
|𝐹 (𝑥) − 𝐹 (𝑦)| = ⃒
⃒ 𝑓 (𝑡)𝑑𝑡 − 𝑓 (𝑡)𝑑𝑡⃒⃒ ≤ |𝑓 (𝑡)| 𝑑𝑡.
𝑐 𝑐 𝑦

As 𝑓 is∫︀bounded, then there is a constant 𝑀 such that |𝑓 (𝑡)| ≤ 𝑀 for all 𝑡 ∈ [𝑎, 𝑏],
𝑥
hence 𝑦 |𝑓 (𝑡)| 𝑑𝑡 ≤ 𝑀 (𝑥 − 𝑦), so |𝐹 (𝑥) − 𝐹 (𝑦)| ≤ 𝑀 (𝑥 − 𝑦), which proves
continuity.
∫︀ 𝑏 ∫︀ 𝑎 ∫︀ 𝑏
(b) 𝐹 (𝑏) − 𝐹 (𝑎) = 𝑐 𝑓 (𝑡)𝑑𝑡 − 𝑐 𝑓 (𝑡)𝑑𝑡 = 𝑎 𝑓 (𝑡)𝑑𝑡.

5
 (c) By the mean value theorem, for any ℎ, we have
∫︁ 𝑥0 +ℎ
𝐹 (𝑥0 + ℎ) − 𝐹 (𝑥0 ) = 𝑓 (𝑡) 𝑑𝑡 = ℎ𝑓 (𝑐ℎ ),
𝑥0

for some 𝑐ℎ between 𝑥0 and 𝑥0 + ℎ.

So,
𝐹 (𝑥0 + ℎ) − 𝐹 (𝑥0 )
lim = lim 𝑓 (𝑐ℎ ),
ℎ→0 ℎ ℎ→0

By the continuity of 𝑓 at 𝑥0 ; as ℎ tends to 0 𝑐ℎ must tend to 𝑥0 , and so 𝑓 (𝑐ℎ )

tends to 𝑓 (𝑥0 ).

14 Exercises.

(1) Let 𝑔 ∈ 𝐶([𝑎, 𝑏]) and let 𝑎(𝑥), 𝑏(𝑥) be two continuously differentiable functions
from an interval [𝑐, 𝑑] into [𝑎, 𝑏]. Define 𝐹 : [𝑐, 𝑑] −→ R by
∫︁ 𝑏(𝑥)
𝐹 (𝑥) = 𝑔(𝑡) 𝑑𝑡.
𝑎(𝑥)

Prove that 𝐹 is differentiable on ]𝑐, 𝑑[ and that

𝐹 ′ (𝑥) = 𝑏′ (𝑥) 𝑔(𝑏(𝑥)) − 𝑎′ (𝑥) 𝑔(𝑎(𝑥)).