Math 318 HW #6 Solutions
1. (a) Exercise 16.7. Prove Corollary
1 ⊃ A2 ⊃ A3 ⊃ · · · are meaT∞ 10.10, which says that if∗ AT
∗
surable subsets of E, then i=1 Ai is measurable and m ( ∞
i=1 Ai ) = limi→∞ m (Ai ).
Proof. Let Bi = E\Ai for each i. Then
B1 ⊂ B2 ⊂ B3 ⊂ · · ·
and so, by Corollary 10.9,
S∞
i=1 Bi
m∗
is measurable and
!
∞
[
Bi = lim m∗ (Bi ).
i→∞
i=1
Therefore,
∞
\
i=1
Ai =
∞
\
∞
[
(E\Bi ) = E\
i=1
is measurable by Corollary 8.4 and
!
∞
\
m∗
Ai = 1 − m∗
i=1
Bi
i=1
∞
[
!
Bi
= 1 − lim m∗ (Bi ).
i=1
i→∞
But then m∗ (Bi ) = 1 − m∗ (Ai ) for each i, so we have
!
∞
\
∗
m
Ai = 1 − lim (1 − m∗ (Ai )) = lim m∗ (Ai ),
i=1
i→∞
i→∞
as desired.
(b) Exercise 16.36. Show that Corollary 10.10 is false for unbounded sets Ai , i = 1, 2, . . ..
Where does the proof of Corollary 10.10 break down in this case?
Answer. For each i = 1, 2, . . ., define Ai = [i, +∞). Then each Ai is a Borel set and,
hence, measurable, with m∗ (Ai ) = +∞. Therefore,
lim m∗ (Ai ) = +∞,
i→∞
but
T∞
i=1 Ai
= ∅, which obviously has measure zero.
2. Exercise 16.23. Construct a perfect nowhere dense set D ⊂ E such that m(D) = 1/2. (Hint:
follow the construction of Example 12.5, but throw away less than the middle third at each
step.)
Answer. Let D0 = [0, 1] and form D1 by throwing away an open interval of length 1/4 from
the middle of D0 .
Now D1 consists of two closed intervals; to form D2 , remove an open interval of length 1/16
from each. Then D2 consists of four closed intervals; to form D3 , remove an open interval of
length 1/64 from the middle of each.
1
In general, Dn consists of 2n closed intervals, and we form Dn+1 by removing an open interval
of length 1/4n+1 from each. Then define
D=
∞
\
Dn .
n=1
By construction,
m([0, 1]\D) =
∞
X
n
2 ·
n=0
1
4n+1
=
∞
X
1
n=0
1
1/4
=
= 1/2,
n
4 2
1 − 1/2
·
so m(D) = 1/2 as well.
Now, the proof that D is perfect and nowhere dense is essentially the same as the proof that
the Cantor set is perfect and nowhere dense.
3. (a) Find a trivial proof of Theorem 12.9 which illustrates why this theorem is not very useful
as stated.
Proof. Let B = [0, m(A)]. Then m(B) = m(A) − 0 = m(A). But this set B may have
nothing to do with A; it could very well be the case that A ∩ B = ∅.
(b) Prove the following, actually useful, version of Theorem 12.9: Given a measurable set
A ⊆ [0, 1], there exists a Borel set B ⊆ [0, 1] such that the symmetric difference A4B
has measure zero.
(Recall that the symmetric difference of two sets is defined as A4B = (A\B) ∪ (B\A).)
Proof. Since A is measurable, we know that, for each n, there exists an open set Bn ⊃ A
such that m(Bn ) < m(A) + 1/n. We can construct these sets so that B1 ⊃ B2 ⊃ · · · (if
not, then let B10 = B1 , B20 = B2 ∩ B1 , let B30 = B3 ∩ B20 , etc., and then B10 ⊃ B20 ⊃ · · ·
as desired).
Define
∞
\
B=
Bn .
n=1
B is the countable intersection of open sets (which are countable unions of open intervals
and, thus, Borel sets), so B is a Borel set. Also, B ⊃ A, so
A4B = (A\B) ∪ (B\A) = B\A.
By Problem 1(a) above,
m(B) = lim m(Bn ) ≤ lim (m(A) + 1/n) = m(A).
n→∞
n→∞
Since A ⊂ B, we have m(A) ≤ m(B) by monotonicity, so we conclude that m(B) =
m(A). Therefore,
m(A4B) = m(B\A) = m(B) − m(A) = 0,
as desired.
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