The reader is assumed to be familiar with the Riemann integral, but if not, the above is more general and
gives the principal results for the Riemann integral of continuous functions very easily. Therefore, here is a
slight digression to show this.

In the special case that f :

[a,b]

→ ℝ, and γ

(t)

= t, the Rieman Stieltjes sums reduce to ordinary
Riemann sums

∑n
f (τi)(ti − ti−1)
i=1

and the above argument reduces to showing the existence of the Riemann integral ∫_{a}^{b}f

(t)

dt for a
continuous function. Note that it is obvious that if f

(t)

≥ 0 for all t, then

∫ b
f (t)dt ≥ 0
a

and so

∫ b ∫ b ∫ b
(|f |− f )dx ≥ 0 ⇒ |f |dx ≥ fdx
∫a ∫a a∫
b b b
a (|f |+ f )dx ≥ 0 ⇒ a |f |dx ≥ − a fdx

which together imply that

∫ |∫ |
b || b ||
a |f|dx ≥ || a fdx|| (4.6)

(4.6)

the triangle inequality.

As to vector valued functions and Riemann integration, if f :

[a,b]

→ ℝ^{p} is continuous, what is meant
by ∫_{a}^{b}f

(s)

ds? We mean the obvious thing:

(∫ ) ∫
b b
a f (s)ds ≡ a fi(s)ds
i

Now consider the following sequence of steps using the Cauchy Schwarz inequality, Proposition 1.10.1, from
the first to the second line.

∫
b
a |fi(s)|ds ≡ Ji

( )
∑p ∫ b 2 ∑p ∫ b ∫ b∑p
|fi(s)|ds = Ji |fi(s)|ds = Ji|fi(s)|ds
i=1 a i=1 ( a ) ( a i=1 )
∫ b ∑p 1∕2 ∑p 2 1∕2
≤ J2i |fi(s)| ds
a i=1 i=1
(∑p )1 ∕2∫ b( ∑p )1∕2
= J2i |fi(s)|2 ds
i=1 a i=1
( p ( ∫ b )2 )1∕2∫ b
= (∑ |fi(s)|ds ) |f (s)|ds
i=1 a a

Divide by the first term on the right. This gives

( ∑p (∫ b )2) 1∕2 ||∫ b || ∫ b
( |fi(s)|ds ) ≡ || f (s)ds||≤ |f (s)|ds
i=1 a |a | a

which holds even if ∑_{i=1}^{p}

( )
∫b|f (s)|ds
a i

^{2} = 0. This proves the following convenient version of the triangle
inequality, this time for vector valued functions.

Proposition 4.1.5Let f :

[a,b]

→ ℝ^{p}be continuous. Then

| |
||∫ b || ∫ b
|| a f (s)ds|| ≤ a |f (s)|ds

Also one obtains easily the fundamental theorem of calculus.

Theorem 4.1.6Let f :

[a,b]

→ ℝ be continuous and real valued. Suppose F^{′}

(x)

= f

(x)

for all x ∈

(a,b)

where F is a continuous function on

[a,b]

. Then

∫ b
f (x)dx = F (b)− F (a)
a

Proof:Let ε > 0. There is a δ > 0 such that if

∥P∥

< δ, then for P =

{x ,⋅⋅⋅,x }
0 n

,

| |
||∫ b ∑n ||
|| f (x)dx − f (τi) (xi − xi−1)|| < ε
a i=1

Now

||∫ b || ||∫ b ∑n ||
|| f (x)dx − (F (b)− F (a))||= || f (x)dx− (F (xi)− F (xi−1))||
| a | | a i=1 |

By intermediate value theorem from calculus, there is τ_{i}∈

(xi− 1,xi)

such that

F (xi)− F (xi− 1) = F′(τi)(xi − xi−1),F′ = f

Thus

| | | |
||∫ b ∑n || ||∫ b ||
= || f (x)dx − f (τi)(xi − xi−1)|| = || f (x)dx− (F (b)− F (a))|| < ε
a i=1 a

because

∥P∥

< δ. Since ε > 0, it follows that ∫_{a}^{b}f

(x)

dx −

(F (b)− F (a))

= 0. ■

In case of vector valued continuous functions with f

(s)

= F^{′}

(s)

so F_{i}^{′}

(s)

= f_{i}

(s)

,

∫
b ( ∫ b ∫b )T
a f (s)ds ≡ a f1(s)ds ⋅⋅⋅ a fp(s)ds
( )T
= F1 (b)− F1 (a) ⋅⋅⋅ Fp (b) − Fp(a)
= F (b) − F (a)

The other form of the fundamental theorem of calculus is also obtained.

Theorem 4.1.7Letf :

[a,b]

→ ℝ^{p}be continuous and let F

(t)

≡∫_{a}^{t}f

(s)

ds. Then F^{′}

(t)

= f

(t)

for all t ∈

(a,b)

.

Proof:Let t ∈

(a,b)

and let

|h|

be small enough that everything of interest is in

[a,b]

. First suppose
h > 0. Then

|| || || ∫ t+h ∫ t+h ||
||F(t+-h)−-F-(t)− f (t)|| = ||1 f (s)ds− 1 f (t)ds||
h |h t h t |

∫ ∫
1- t+h 1- t+h
≤ h t |f (s)− f (t)|ds ≤ h t εds = ε

provided that ε is small enough due to continuity of f at t. A similar inequality is obtained if h < 0 except
in the argument, you will have t + h < t so you have to switch the order of integration in going to the
second line and replace 1∕h with 1∕