a, b
], there exists a partition
P
=
{
x
0
, x
2
. . . , x
n
}
of [
a, b
] such that
U
(
f, P
)

L
(
f, P
)
< δ
2
.
We claim that for this partition we also have
U
(
g
◦
f, P
)

L
(
g
◦
f, P
) =
n
i
=1
[
M
i
(
g
◦
f
)

m
i
(
g
◦
f
)] Δ
x
i
<
.
To show this, we separate the set of indices of the partition
P
into two
disjoint sets.
A
=
{
i
:
M
i
(
f
)

m
i
(
f
)
< δ
}
and
B
=
{
i
:
M
i
(
f
)

m
i
(
f
)
≥
δ
}
.
Then if
i
∈
A
and
x, y
∈
[
x
i

1
, x
i
], we have

f
(
x
)

f
(
y
)
 ≤
M
i
(
f
)

m
i
(
f
)
< δ,
so that

g
◦
f
(
x
)

g
◦
f
(
y
)

<
.
But then
M
i
(
g
◦
f
)

m
i
(
g
◦
f
)
≤
.
It follows that
i
∈
A
[
M
i
(
g
◦
f
)

m
i
(
g
◦
f
)] Δ
x
i
≤
i
∈
A
Δ
x
i
≤
(
b

a
)
.
4
On the other hand, if
i
∈
B
, then [
M
i
(
f
)

m
i
(
f
)]
/δ
≥
1, so that
i
∈
B
Δ
x
i
≤
1
δ
i
∈
B
[
M
i
(
f
)

m
i
(
f
)] Δ
x
i
≤
1
δ
[
U
(
f, P
)

L
(
f, P
)]
< δ <
.
Thus since
M
i
(
g
◦
f
)

m
i
(
g
◦
f
)
≤
2
K
for all
i
, we have
i
∈
B
[
M
i
(
g
◦
f
)

m
i
(
g
◦
f
)] Δ
x
i
≤
2
K
i
∈
B
Δ
x
i
<
2
K
.
Now when we combine all the indices we obtain
U
(
g
◦
f, P
)

L
(
g
◦
f, P
)
=
∑
i
∈
A
[
M
i
(
g
◦
f
)

m
i
(
g
◦
f
)] Δ
x
i
+
∑
i
∈
B
[
M
i
(
g
◦
f
)

m
i
(
g
◦
f
)] Δ
x
i
≤
(
b

a
) + 2
K
=
(
b

a
+ 2
K
)
<
.
5
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 Math, Continuous function, Metric space, dt, dx