we are done;
k
= 1. If not,
d
n
1
∈
I
and there is
n
, 1
≤
n
≤
N
such that
c
n
< d
n
1
< d
n
. The way we selected
n
1
, this
forces any such
n
to satisfy
c
n
≥
a
+
η
so
c
n
1
< c
n
< d
n
1
< d
n
. Of all such
n
’s, let
n
2
be one for which
d
n
is maximal
so that now we have 1
≤
n
2
≤
N
,
c
n
1
< c
n
2
< d
n
1
< d
n
2
and if
n
is such that 1
≤
n
≤
N
,
c
n
1
< c
n
< d
n
1
< d
n
, then
d
n
≤
d
n
2
. Of course, there is no reason to assume that
n
2
> n
1
. If
d
n
2
> b

η
, we are done;
k
= 2. If not . . . , I hope
you get the drift. In this way we get a subset
{
n
1
, . . . , n
k
}
of
{
1
, . . . ,
N
}
such that
[
a
+
η, b

η
]
⊂
k
[
j
=1
(
c
n
j
, d
n
j
)
and
c
n
1
< a
+
η
,
d
n
k
> b

η
c
n
1
< c
n
2
< d
n
1
< c
n
3
<
· · ·
< c
n
k
< d
n
k

1
< d
n
k
.
Because
f
is increasing,
f
(
a
+
η
)
≥
f
(
c
n
1
),
f
(
b

η
)
≤
f
(
d
n
k
) and thus
f
(
b

η
)

f
(
a
+
η
)
≤
f
(
d
n
k
)

f
(
c
n
k
)
=
f
(
d
n
k
)

f
(
d
n
k

1
) +
f
(
d
n
k

1
)

f
(
d
n
k

2
) +
· · ·
+
f
(
d
n
2
)

f
(
d
n
1
) +
f
(
d
n
1
)

f
(
c
n
1
)
=
k
X
j
=2
f
(
d
n
j
)

f
(
d
n
j

1
) +
f
(
d
n
1
)

f
(
c
n
1
)
.
The best way to avoid being intimidated by a long string of calculations is by doing them oneself.
This is not
hard!!
Don’t let it get you down. Anyway, we use once more the fact that
f
is increasing, thus
f
(
d
n
j
)

f
(
d
n
j

1
)
≤
f
(
d
n
j
)

f
(
c
n
j

1
) and we see that
f
(
b

η
)

f
(
a
+
η
)
≤
k
X
j
=1
f
(
d
n
j
)

f
(
c
n
j

1
) =
k
X
j
=1
˜
‘
(
J
n
k
)
≤
∞
X
n
=1
˜
‘
(
J
n
)
.
Combining with (5), we proved that
f
(
b

η
)

f
(
a
+
η
)
≤
μ
*
(
I
) +
²
for all
η >
0. Letting
η
↓
0,
f
(
b

)

f
(
a
+)
≤
μ
*
(
I
) +
².
Since
² >
0 was arbitrary, we proved the first inequality in (4).
Contrary to what happens with Royden, we have
to consider also the case of an unbounded
I
, because it is possible here for
I
to be unbounded and yet
μ
*
(
I
)
<
∞
.
This will happen if
f
(
x
) has a finite limit for
x
→ ∞
or
x
→ ∞
. Assume first that
a
=
∞
,
b
finite. We can let
J
k
= (

k, b
) where

k < b
. Then
J
k
⊂
I
and because
μ
*
is an outer measure,
μ
*
(
J
k
)
≤
μ
*
(
I
); by what we proved
f
(
b

)

f
(

k
+)
≤
f
(
J
k
)
≤
μ
*
(
I
)
and we can let
k
→ ∞
to get
μ
*
(
I
)
≥
f
(
b

)

f
(
∞
).
As similar argument works if
a
∈
R
and
b
=
∞
, or if
a
=
∞
, b
=
∞
.
Claim 1 is established.
Claim 2.
If
I
= (
a, b
) is an open interval, then
(6)
μ
*
(
I
) =
f
(
b

)

f
(
a
+)
.
In fact, because of (4), it suffices to prove that
μ
*
(
I
)
≤
f
(
b

)

f
(
a
+). I’ll assume that
∞
< a < b <
∞
; the case
in which one or both of
a, b
is not finite is actually simpler and if you understood what I do in the case I consider,
you should have no trouble dealing with the infinite endpoint(s) case. We recall that the set of discontinuities of an
increasing function is countable. The complement of a countable set being dense, we can find sequences