Mathematics 105, Spring 2004 — Problem Set V Solutions
1
V.A
Show that any countable subset of
R
n
is measurable, and has measure zero.
Solution.
First of all, a set containing a single point
y
is measurable since it is closed, and
has measure zero since it is a closed rectangle of volume zero. (If you prefer to consider
only rectangles having strictly positive volumes to be rectangles, then you can instead note
that
{
y
}
is covered by the single rectangle [
y
1

ε, y
1
+
ε
]
× · · · ×
[
y
n

ε, y
n
+
ε
], which has
volume 2
n
ε
n
, and this tends to zero as
ε
→
0.)
Now any countable set
E
is a countable union of sets each of which contains a single
point, hence is a countable union of sets
E
j
of measure zero, and hence

E
 ≤
∑
j

E
j

=
∑
j
0 = 0.
V.B
Let
A
1
⊃
A
2
⊃
A
3
⊃ · · ·
be an infinite sequence of nested measurable subsets of
R
n
.
Let
A
be their intersection. Show that if

A
1

<
∞
then

A

= lim
n
→∞

A
n

.
Solution.
First of all note that it suffices to prove this in the special case where
A
=
∅
.
Indeed, in the general case define
˜
A
j
=
A
j
\
A
. Then the sets
˜
A
j
are still measurable (since
the measurable sets form a
σ
algebra), and their intersection is empty. From the special
case we therefore conclude that

˜
A
j
 →
0 as
j
→ ∞
. Now
A
j
is the disjoint union
˜
A
j
∪
A
,
so

A
j

=

A

+

˜
A
j
 → 
A

.
Thus we may suppose henceforth that
∩
j
A
j
=
∅
.
Define auxiliary sets
B
j
by
B
j
=
A
j
\
A
j
+1
for all
j
≥
1. Then
A
1
=
∪
∞
j
=1
B
j
, and the sets
B
j
are pairwise disjoint because
the sets
A
j
are nested. Since each
A
j
is measurable, so is
R
n
\
A
j
, and hence so is
B
j
=
A
j
∩
(
R
n
\
A
j
+1
).
Since the sets
B
j
are measurable and pairwise disjoint,
 ∪
∞
j
=1
B
j

=
∑
∞
j
=1

B
j

. This
union equals
A
1
, which has finite measure.
We conclude that the infinite series
∑
j

B
j

converges.
Thus lim
N
→∞
∑
j
≥
N

B
j
 →
0 as
N
→ ∞
.
But
A
N
=
∪
j
≥
N
B
j
, so

A
N
 ≤
∑
j
≥
N

B
j
 →
0.
Comment.
Note how the
countable
additivity of Lebesgue measure is the key to evaluating
the limit. This connection between countable and additivity and various limiting operations
is an essential feature of the theory, and is what makes Lebesgue measure and integration
more successful than the older theory of Jordan content.
Comment.
The conclusion doesn’t hold in general without the hypothesis that
A
1
has finite
measure. Consider for instance the case where
A
n
= [
n,
∞
). Then

A
n

=
∞
for all
n
, yet
∪
n
A
n
=
∅
.
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 Spring '04
 MichaelChrist
 Math, Topology, Sets, Empty set, Topological space, Lebesgue measure

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