SOLUTIONS OF HW4
MINGFENG ZHAO
February 24, 2011
1. (Page 40, 30) If E ∈ L and m(E) > 0, then for any 0 < α < 1, there is an open interval I such
that m(E ∩ I) > αm(I).
Proof. Case I: Let E ∈ L and ∞ > m(E) > 0, if there exists some 0 < α0 < 1 such that for any open
interval I such that
(1)
m(E
\
I) ≤ α0 m(I).
By the interior regularity of m, 1.18 Theorem in Page 36, we know that
m(E) = {m(U ) : E ⊂ U , and U is open}.
So for any > 0, there exists some open set U such that E ⊂ U , and
m(E) + ≥ m(U ).
Since U is open in R, then we can write U as a countable disjoint union of open intervals, that is,
U =
∞
[
Ik ,
k=1
where {Ik }∞
k=1 are disjoint open intervals in R. Hence we have
m(E)
\ = m E
U
= m E
\
∞
[
!!
Ik
k=1
= m
∞
[
\ E
Ik
k=1
1
!
2
MINGFENG ZHAO
=
∞
X
\ m E
Ik
Since {Ik }∞
k=1 are disjoint
k=1
≤
∞
X
α0 m(Ik )
By (1)
k=1
=
α0
∞
X
m(Ik )
k=1
=
α0 m
∞
[
!
Since {Ik }∞
k=1 are disjoint
Ik
k=1
So α0 ≥
m(E)
m(E)+ ,
=
0 m(U )
≤
α0 (m(E) + )
when take → 0, we get α0 ≥ 1, which contradicts with 0 < α0 < 1.
Case II: If m(E) = ∞.
For each n ≥ 1, let En = E ∩ [−n, n], by the continuity of the measure from below, we know there
exists N ≥ 1 such that whenever n ≥ N , we have m(En ) ≥ m(EN ) ≥ 1. By the result of Case I, we
have for any α > 0, there exists an open interval I such that
m(EN ∩ I) > αm(I).
Since EN ⊂ En for all n ≥ N , and I is a bounded interval, we can find a large n1 ≥ N such that
I ⊂ En1 . Hence
m(E ∩ I)
= m(En1 ∩ I)
≥ m(EN ∩ I)
>
αm(I).
SOLUTIONS OF HW4
3
2. (Page 40, 31) If E ∈ L and m(E) > 0, the set E − E = {x − y : x, y ∈ E} contains an interval
centered at 0.
Proof. Take 1 > α ≥
3
4,
by the result of the Problem 30 in Page 40, we know that there exists an
open interval I = (a, b) such that m(E ∩ I) > αm(I).
Claim: − 12 m(I), 12 m(I) ⊂ E − E.
If not, then there exists some x ∈ − 12 m(I), 12 m(I) , but x ∈
/ E − E. Obviously, we know x 6= 0. So
we can get I − x = (a − x, b − x), and if x < 0, we have a < a − x < b < b − x, so I − x ∩ I = (a − x, b),
that is, m(I − x ∩ I) = b − a + x = m(I) − |x| > 12 m(I). If x > 0, we have a − x < a < b − x < b, so
I − x ∩ I = (a, b − x), that is, m(I − x ∩ I) = b − x − a = m(I) − |x| > 12 m(I). So in both cases, we
have m(I − x ∩ I) = m(I) − |x| > 12 m(I). And also we know that m(I − x) = m(I).
Since x ∈
/ E − E, then E − x ∩ E = φ. So
m(I) ≥ m(I ∩ (E ∪ (E − x)))
= m(I ∩ E) + m(I ∩ (E − x))
>
αm(I) + m(I ∩ (E − x))
Which implies that m(I ∩ (E − x)) < (1 − α)m(I). Since m(E ∩ I) > αm(I), then m(E ∩ I − x) =
m((E − x) ∩ (I − x)) > αm(I − x). So we get
m(I)
= m(I − x)
≥ m((I − x) ∩ (I ∪ (E − x))
= m((I − x) ∩ I) + m((I − x) ∩ (E − x)) − m(I ∩ (I − x) ∩ (E − x))
>
1
m(I) + αm(I) − m(I ∩ (I − x) ∩ (E − x))
2
Hence m(I ∩ (I − x) ∩ (E − x)) > (α − 21 )m(I). Therefore, we get α −
that α < 34 , contradiction.
1
2
< 1 − α, which implies
4
MINGFENG ZHAO
Department of Mathematics, University of Connecticut, 196 Auditorium Road, Unit 3009, Storrs, CT
06269-3009
E-mail address: mingfeng.zhao@uconn.edu