Let
J
n
= (
c
n
+1
, c
n
),
K
n
= (
d
n
, d
n
+1
) for
n
∈
N
;
H
= (
c
1
, d
1
). Let also
² >
0 be given. Because
f
is continuous at
c
n
,
there is
δ
n
>
0 such that

x

c
n

< δ
n
implies

f
(
x
)

f
(
c
n
)

< ²/
2
n
+2
. In particular,
f
(
c
n
+
δ
n
/
2)

f
(
c
n

δ
n
/
2) =
f
(
c
n
+
δ
n
/
2)

f
(
c
n
) +
f
(
c
n
)

f
(
c
n

δ
n
/
2)
<
²
2
n
+1
.
Similarly, there is
η
n
>
0 such that

x

d
n

< η
n
implies

f
(
x
)

f
(
d
n
)

< ²/
2
n
+2
. In particular,
f
(
d
n
+
η
n
/
2)

f
(
d
n

η
n
/
2) =
f
(
d
n
+
η
n
/
2)

f
(
d
n
) +
f
(
d
n
)

f
(
d
n

η
n
/
2)
<
²
2
n
+1
.
Let
L
n
(
c
n

δ
n
/
2
, c
n
+
δ
n
/
2),
M
n
(
d
n

η
n
/
2
, d
n
+
η
n
/
2) for
n
∈
N
. Since this may look a bit complicated, I will insist
that the idea is quite simple and natural; draw a picture! The family of open intervals
{
J
n
} ∪ {
K
n
} ∪ {
L
n
} ∪ {
M
n
} ∪ {
H
}
is a countable family covering (
a, b
) and if we add all the
˜
‘
values of these intervals we notice
∞
X
n
=1
˜
‘
(
J
n
)
=
lim
n
→∞
n
X
k
=1
(
f
(
c
k
)

f
(
c
k
+1
)) =
f
(
c
1
)

lim
n
→∞
f
(
c
n
+1
) =
f
(
c
1
)

f
(
a
+)
.
∞
X
n
=1
˜
‘
(
K
n
)
=
lim
n
→∞
n
X
k
=1
(
f
(
d
k
+1
)

f
(
d
k
)) = lim
n
→∞
f
(
d
n
+1
)

f
(
d
1
) =
f
(
b

)

f
(
d
1
)
.
˜
‘
(
H
)
=
f
(
d
1
)

f
(
c
1
)
.
∞
X
n
=1
˜
‘
(
L
n
)
=
∞
X
n
=1
f
c
n
+
δ
n
2
¶

f
c
n

δ
n
2
¶¶
<
∞
X
n
=1
e
2
n
+1
=
²
2
.
∞
X
n
=1
˜
‘
(
M
n
)
=
∞
X
n
=1
‡
f
‡
d
n
+
η
n
2
·

f
‡
d
n

η
n
2
··
<
∞
X
n
=1
e
2
n
+1
=
²
2
.
Because we have a countable cover of
I
, the sum of the expressions on the left of the first of all these equali
ties/inequalitites dominates
μ
*
(
I
); adding up we thus get
μ
*
(
I
)
< f
(
b

)

f
(
a
+) +
².
Since
² >
0 is arbitrary,
μ
*
(
I
)
≤
f
(
b

)

f
(
a
+) follows,
establishing Claim 2.
Claim 3.
If
I
= (
a, b
],
∞ ≤
a < b <
∞
, then
(7)
μ
*
(
I
) =
f
(
b
+)

f
(
a
+)
Assume first that
a >
∞
. We can proceed exactly as we did in proving the first inequality in (4), with two difference.
The countable cover
{
J
n
}
is defined in the same way in terms of
²
, but then we use the fact that [
a
+
η, b
] is compact
to extract a finite subcover of [
a
+
η, b
]. That is the first difference; that
b

η
does not appear; it is replaced by
b
. The
second difference is that where we used
f
(
b

η
)
≤
f
(
d
n
k
), now we use that
f
(
b
+)
≤
f
(
d
n
k
). Everything else is the
same, and we end with
μ
*
(
I
)
≥
f
(
b
+)

f
(
a
+). For the converse of this inequality, select any
c
∈
(
a, b
) at which
f
is
continuous. For a small
δ
consider the open intervals (
a, c
)
,
(
c

δ, c
+
d
)
,
(
c, b
+
δ
). These intervals cover
I
; because
μ
*
is an outer measure and by Claim 2:
μ
*
(
I
)
≤
μ
*
((
a, c
))+
μ
*
((
c

δ, c
+
δ
))+
μ
*
((
c, b
+
δ
)) =
f
(
c

)

f
(
a
+)+
f
((
c
+
δ
)

)

f
((
c

δ
)+)+
f
((
b
+
d
)

)

f
(
c
+)
.
Using the continuity of
f
at
c
, letting
δ
→
0, we get
μ
*
(
I
)
≤
f
(
b
+)

f
(
a
+),
establishing Claim 3
.
With this we can return to where we were on hold, namely at (3). With the notation as defined there, we have
(8)
μ
*
(
I
1
n
) +
μ
*
(
I
2
n
) =
μ
*
(
I
n
) =
f
(
b
n

)

f
(
a
n
+)
4