6.1 Integrals of continuous functions
85
Theorem 6.1.12.
Let
f
∈ R
[
a,b
]
,m
≤
f
≤
M
. If
ϕ
is continuous on
[
m,M
]
and
h
(
x
) :=
ϕ
(
f
(
x
))
for
x
∈
[
a,b
]
, then
h
∈ R
[
a,b
]
.
Proof.
Fix
ε >
0. Since
ϕ
is uniformly continuous on [
m,M
], ﬁnd 0
< δ < ε
such that

s

t

< δ
=
⇒ 
ϕ
(
s
)

ϕ
(
t
)

< ε,
s,t
∈
[
m,m
]
.
Since
f
∈ R
, ﬁnd
P
=
{
x
0
,...,x
n
}
such that
Osc
(
f,P
)
< δ
2
.
M
i
,m
i
are extrema of
f
,
M
0
i
,m
0
i
are for
h
. Subdivide the set of indices
{
1
,...,n
}
into
two classes:
i
∈
A
⇐⇒
M
i

m
i
< δ,
i
∈
B
⇐⇒
M
i

m
i
≥
δ.
For
i
∈
A
, have
M
0
i

m
0
i
≤
ε
by choice of
δ
. For
i
∈
B
, have
M
0
i

m
0
i
≤
2sup
m
≤
t
≤
M

ϕ
(
t
)

.
By prev bound of
δ
2
,
δ
X
i
∈
B
(
x
i

x
i

1
)
≤
X
i
∈
B
(
M
i

m
i
)(
x
i

x
i

1
)
< δ
2
X
i
∈
B
(
x
i

x
i

1
)
< δ.
Then
Osc
(
h,P
) =
X
i
∈
A
(
M
0
i

m
0
i
)(
x
i

x
i

1
) +
X
i
∈
B
(
M
0
i

m
0
i
)(
x
i

x
i

1
)
≤
ε
(
b

a
) + 2
δ
sup

ϕ
(
t
)

< ε
(
b

a
+ 2sup