(c) Given the definition of
μ
*
(
A
) as an infimum, proving (1) is equivalent to proving: If
{
I
n
}
is an arbitrary countable
collection of open intervals covering
A
, then
(2)
μ
*
(
A
1
) +
μ
*
(
A
2
)
≤
∞
X
n
=1
˜
‘
(
I
n
)
.
We can keep following Royden for a while longer. To prove (2) Royden introduces (and so do we)
I
1
n
=
I
n
∩
(
a,
∞
),
I
2
n
=
I
n
∩
(
∞
, a
]; then
A
1
⊂
S
∞
n
=1
I
1
n
,
A
2
⊂
S
∞
n
=1
I
2
n
. Because we have an outer measure, we have
μ
*
(
A
1
)
≤
∞
X
n
=1
μ
*
(
I
1
n
)
,
μ
*
(
A
2
)
≤
∞
X
n
=1
μ
*
(
I
2
n
)
thus
(3)
μ
*
(
A
1
) +
μ
*
(
A
2
)
≤
∞
X
n
=1
(
μ
*
(
I
1
n
) +
μ
*
(
I
2
n
)
)
.
Royden completes the proof of (2) by using a result he proved before, namely that
m
*
(
I
) =
‘
(
I
) for all intervals
I
, plus
the obvious fact that
‘
(
I
1
n
)+
‘
(
I
2
n
) =
‘
(
I
n
). If we had the similar results for
μ
*
,
˜
‘
instead of
m
*
, ‘
, respectively, we could
wrap up things just as Royden. It turns out that
μ
*
(
I
) =
˜
‘
(
I
) for all intervals
I
is true if
f
is continuous (which is why
I thought continuity was essential) but fails to be true if
f
has jumps. Noticing that the intervals
I
n
, I
1
n
are open, while
I
2
n
is of the form (
·
,
·
], we only will have to figure out the relation between
μ
*
and
˜
‘
for intervals of this type. So we
leave things on hold at (3) and take a detour to explore these relations. The ideas used in doing all this are all already
in Royden; I am merely modifying them to apply in the case we are considering.
Claim 1
Let
I
be an interval of endpoints
a, b
,
a < b
. Then
(4)
f
(
b

)

f
(
a
+)
≤
μ
*
(
I
)
≤
f
(
b
+)

f
(
a

)
.
(If
a
=
∞
, by definition
f
(
∞
) =
f
(
∞
); if
b
=
∞
,
f
(
∞
) =
f
(
∞
+).)
To see that
μ
*
(
I
)
≤
f
(
b
+)

f
(
a

) we notice that one can find a sequence
{
J
n
}
of open intervals,
J
n
= (
c
n
, d
n
) where
c
n
≤
a < b
≤
d
n
and
{
c
n
}
goes to
a
,
d
n
to
b
as
n
→ ∞
. In fact, if
a /
∈
I
, so
I
is open on the side of
a
, just set
c
n
=
a
for all
n
. If
a
∈
I
, then
a
∈
R
(
a
is not
∞
); set
c
n
=
a

1
/n
. Similarly, if
b /
∈
I
, set
d
n
=
b
for all
n
; otherwise set
d
n
=
b
+ 1
/n
. Now
I
⊂
J
n
(which can be interpreted as
I
⊂
J
n
∪ ∅ ∪ ∅ ∪ · · ·
) so that by the definition of
μ
*
we see that
μ
*
(
I
)
≤
˜
‘
(
J
n
) =
f
(
d
n
)

f
(
c
n
)
for all
n
∈
N
. Letting
n
→ ∞
we get
μ
*
(
I
)
≤
f
(
b
+)

f
(
a

)
,
the second inequality in (4).
To prove the first inequality in (4), we may assume that
μ
*
(
I
)
<
∞
, otherwise the
inequality is obvious. Given then
² >
0, there is a countable collection
{
J
n
}
of open intervals such that
I
⊂
∞
[
n
=1
J
n
and
such that if
J
n
= (
c
n
, d
n
) for
n
∈
N
,
(5)
∞
X
n
=1
˜
‘
(
J
n
) =
∞
X
n
=1
(
f
(
d
n
)

f
(
c
n
))
≤
μ
*
(
I
) +
².