μ
*
(
I
1
n
) +
μ
*
(
I
2
n
) =
f
(
b
n

)

f
(
a
+) +
f
(
a
+)

f
(
a
n
+) =
f
(
b
n

)

f
(
a
n
+)
as asserted. Now
f
(
b
n

)

f
(
a
n
+)
≤
f
(
b
n
)

f
(
a
n
)
˜
‘
(
I
n
) so that (3) and (8) imply
μ
*
(
A
1
) +
μ
*
(
A
2
)
≤
∞
X
n
=1
˜
‘
(
I
n
)
,
completing the proof of (2).
This concludes the treatment of part (b), except for a some comments. Borel sets are measurable. Not every countable
or even finite set has measure zero in this model.
For example, if
f
has a jump at
c
, then from
μ
*
((
c

², c
]) =
f
(
c
+)

f
(
c

²
+) it is quite easy to deduce that
μ
*
(
{
c
}
) =
f
(
c
+)

f
(
c

)
.
We only evaluated the exact outer measure of open intervals and of half open intervals open on the left. But it is easy
to see that if
I
= [
a, b
], then
μ
*
(
I
) =
f
(
b
+)

f
(
a

) and if
I
= [
a, b
), then
μ
*
(
I
) =
f
(
b

)

f
(
a

).
(c) This is not much different from part (b). You might notice first that
μ
*
of an interval works out to the same value
as in part (b). Once that is established, all goes in (essentially) the same way.
2. Let
{
A
n
}
be a sequence of measurable subsets of
R
. Assume they are all contained in some measurable set
E
with
m
(
E
)
<
∞
. Prove:
m
ˆ
∞
[
n
=1
∞
\
m
=
n
A
m
!
≤
lim inf
n
→∞
m
(
A
n
)
≤
lim sup
n
→∞
m
(
A
n
)
≤
m
ˆ
∞
\
n
=1
∞
[
m
=
n
A
m
!
.
Conclude that if
S
∞
n
=1
T
∞
m
=
n
A
n
=
T
∞
n
=1
S
∞
m
=
n
A
n
=: lim
n
A
n
, then
m
‡
lim
n
A
n
·
= lim
n
→∞
m
(
A
n
)
.
By considering the case of
A
n
=
A
when
n
is even,
A
n
=
B
when
n
is odd, where
A, B
are measurable,
A
∩
B
=
∅
,
m
(
A
)
>
0,
m
(
B
)
>
0, show that it is possible to have
m
ˆ
∞
[
n
=1
∞
\
m
=
n
A
n
!
<
lim inf
n
→∞
m
(
A
n
)
≤
lim sup
n
→∞
m
(
A
n
)
< m
ˆ
∞
\
n
=1
∞
[
m
=
n
A
n
!
.
Solution.
Let
E
be a measurable subset of
R
with
m
(
E
)
<
∞
and let
A
n
be a measurable subset of
E
for all
n
∈
N
.
For the first inequality to be proved, set
B
n
=
T
∞
m
=
n
A
m
. Then
B
1
⊂
B
2
⊂ · · ·
(and it goes without saying, so I won’t
say it, that the
B
n
’s are measurable) and
m
ˆ
∞
[
n
=1
∞
\
m
=
n
A
n
!
=
m
ˆ
∞
[
n
=1
B
n
!
= lim
n
→∞
m
(
B
n
)
.
On the other hand,
B
n
⊂
A
m
for all
m
≥
n
, thus
m
(
B
n
)
≤
m
(
A
m
) for all
m
≥
n
, hence
m
(
B
n
)
≤
inf
m
≥
n
m
(
A
m
). We
know from our studies of liminf and limsup that the sequence
{
inf
m
≥
n
m
(
A
m
)
}
is increasing and thus has a limit, and
we know what this limit is. We can return to our calculations above and complete them as follows:
m
ˆ
∞
[
n
=1
∞
\
m
=
n
A
m
!
=
m
ˆ
∞
[
n
=1
B
n
!
= lim
n
→∞
m
(
B
n
)
≤
lim
n
→∞
inf
m
≥
n
m
(
A
m
) = lim inf
n
→∞
m
(
A
n
)
.
This proves the first inequality.
5
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Note.
The symbol
m
is unfortunately overloaded; I hope it is clear from the context when it is a subindex, when it
represents Lebesgue measure. The basic unwritten rule not followed by Royden is NOT to use a letter that is commonly
used for a variable to represent a fixed object. That is why one introduced symbols like
R
,
N
, etc., to avoid preempting
the use of
R, N
, etc. I’ll also use this opportunity to say that there was an unfortunate typo in the statement of the
inequalities; two subindices of
n
that should have been
m
.
Since I had stated the inequalities already in class, and
because the typos are obvious, I do expect people to have corrected them.
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 Spring '11
 Speinklo
 CN, Lebesgue measure, open intervals

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