Because
I
n
= (
a
n
, b
n
)
⊂
A
we have
(2)

f
(
b
n
)

f
(
a
n
)

=

Z
b
n
a
n
f
0
(
x
)
dx

=
Z
b
n
a
n
f
0
(
x
)
dx
=
Z
b
n
a
n

f
0
(
x
)

dx
;
similarly, because
J
k
⊂
B
,
(3)

f
(
d
k
)

f
(
c
k
)

=

Z
d
k
c
k
f
0
(
x
)
dx

=

Z
d
k
c
k
f
0
(
x
)
dx
=
Z
d
k
c
k

f
0
(
x
)

dx.
Let
² >
0. Since (by disjointness)
∞
X
n
=1
m
(
I
n
) =
m
(
A
)
≤
m
([
a, b
])
<
∞
,
there is
N
such that
∑
∞
n
=
N
+1
m
(
I
n
)
< ²/
(2
C
) where
C
is a bound for

f
0

;

f
0
(
x
)
 ≤
C
for all
x
∈
[
a, b
]. Similarly, there
is
K
such that
∑
∞
k
=
K
+1
m
(
J
k
)
< ²/
(2
C
). Now consider the intervals
I
1
, I
2
, . . . , I
N
, J
1
, J
2
, . . . , J
M
. These are pairwise
disjoint open subintervals of [
a, b
]. We can order them by their left endpoints and relabel them as
K
1
, K
2
, . . . , K
N
+
M
,
where
K
1
is the leftmost of these
N
+
M
intervals,
K
2
the one following
K
1
, etc. Let us also relabel the endpoints, so
K
j
= (
σ
j
, τ
j
) for
j
= 1
, . . . , N
+
M
. Adding if needed, the endpoints
a, b
, the points
a
≤
σ
1
< τ
1
< σ
2
<
· · ·
< σ
N
+
M
<
τ
N
+
M
≤
b
forma partition of [
a, b
] and by definition of
V
b
a
(
f
) we have
V
b
a
(
f
)
≥

f
(
σ
1
)

f
(
a
)

+

f
(
τ
1
)

f
(
σ
1
)

+

f
(
σ
2
)

f
(
τ
1
)

+

f
(
τ
2
)

f
(
σ
2
)

+
· · ·
+

f
(
τ
N
+
M

f
(
σ
N
+
M
)

+

f
(
b
)

f
(
τ
N
+
M
)

≥

f
(
τ
1
)

f
(
σ
1
)

+

f
(
τ
2
)

f
(
σ
2
)

+
· · ·
+

f
(
τ
N
+
M

f
(
σ
N
+
M
)

.
The last sum is the sum over the intervals
K
1
, . . . , K
N
+
M
, which are just the intervals
I
1
, . . . , I
N
, J
1
, . . . , J
M
in some
order. That is, each
f
(
τ
j
)

f
(
σ
j
) is either one of

f
(
b
n
)

f
(
a
n
)

or

f
(
d
k
)

f
(
c
k
)

; that is, we proved
V
b
a
(
f
)
≥
N
X
n
=1

f
(
b
n
)

f
(
a
n
)

+
M
X
k
=1

f
(
d
n
)

f
(
c
k
)

.
In view of (2), (3), we proved
(4)
V
b
a
(
f
)
≥
N
X
n
=1
Z
b
n
a
n

f
0
(
x
)

dx
+
M
X
k
=1
Z
d
k
c
k

f
0
(
x
)

dx
=
Z
S
N
n
=1
I
n
∪
S
M
k
=1
J
k

f
0

,
the last equality holding because all the intervals ar disjoint. Now
Z
A

f
0

=
∞
X
n
=1
Z
I
n

f
0

=
Z
S
N
n
=1
I
n

f
0

+
Z
S
∞
n
=
N
+1
I
n

f
0

Since

f
0
 ≤
C
,
Z
S
∞
n
=
N
+1
I
n

f
0
 ≤
Cm
(
∞
[
n
=
N
+1
I
n
) =
C
∞
X
n
=
N
+1
m
(
I
n
)
<
²
2
.
Thus
Z
A

f
0

<
Z
S
N
n
=1
I
n

f
0

+
²
2
Similarly,
Z
B

f
0

<
Z
S
M
k
=1
J
k

f
0

+
²
2
,
7
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so that
Z
S
N
n
=1
I
n
∪
S
M
k
=1
J
k

f
0

=
Z
S
N
n
=1
I
n

f
0

+
Z
S
M
k
=1
J
k

f
0

>
Z
A
∪
B

f
0
 
².
By (4),
Z
A
∪
B

f
0

< V
b
a
(
f
) +
².
But
f
0
= 0 in [
a, b
]
\
(
A
∪
B
), thus
Z
b
a

f
0
(
x
)

dx
=
Z
A
∪
B

f
0

< V
b
a
(
f
) +
².
Since
² >
0 is arbitrary, this proves
Z
b
a

f
0
(
x
)

dx
≤
V
b
a
(
f
)
.
8
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 Spring '11
 Speinklo
 Empty set, Open set, Topological space, Dominated convergence theorem, Lebesgue integration, Nonmeasurable set

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