Math361 Homework 08
April 24, 2014
1. Claim: If m∗ (A) = 0 for some A ⊂ R, then m∗ (A ∪ B) = m∗ (B) for any subset B in R.
Proof. Since B ⊂ A ∪ B, and then by countable sub additivity, we have m∗ (B) ≤ m∗ (A ∪ B) ≤
m∗ (B) + m∗ (A) = m∗ (B)
2. Claim: Any subset A ⊂ R consisting of one single point, i.e., A = {a} for some a ∈ R, has zero
outer measure, and so any countable subset of R has zero outer measure, and so the subset of all
rational numbers Q has outer measure zero.
Proof. Let A = {a}. Then A ⊂ (a − , a + ) for all > 0. Since |(a − , a + )| = 2, we can say
m∗ (A) ≤ 2 for all > 0, meaning that m∗ (A) = 0
Let B be a countable subset of R. Then we can find enumerate the elements of B into a sequence
bn . We can make another sequence of singleton
P ∗sets Bn = {bn } such∗that B = ∪Bn . By countable
additivity, we have m∗ (B) = m∗ (∪Bn ) ≤
m (Bn ) = 0 and so m (B) = 0. Since the rationals
are countable, we can say m∗ (Q) = 0.
3. Claim: m∗ ([0, 1] − Q) = 1.
Proof. Consider that [0, 1] − Q ⊂ [0, 1]. Thus, m∗ ([0, 1] − Q) ≤ 1.
Consider that the union of [0, 1]−Q and [0, 1]∩Q is [0, 1]. Thus we have by countable sub additivity
1 = m∗ ([0, 1]) ≤ m∗ ([0, 1] − Q) + m∗ ([0, 1] ∩ Q)
Since [0, 1] ∩ Q ⊂ Q, it must be that m∗ ([0, 1] ∩ Q) ≤ 0 =⇒ m∗ ([0, 1] ∩ Q) = 0. So we have
1 ≤ m∗ ([0, 1] − Q)
This proves both sides of the inequality, and so equality holds.
4. Claim: For any fixed real number t ∈ R and subset A ⊂ R, if we denote its t-translate by t + A =
{t + a|a ∈ A} ⊂ R, then m∗ (A) = m∗ (t + A)
Proof. Let {In } be a cover of A. Then we can translate each In over by t such that (a, b) maps to
(a + t, b + t). Notice that |(a, b)| = b − a = |(a + t, b + t)| or |In | = |In + t|. Call this new collection
of intervals {In + t}
Fix x ∈ (a, b). Then x + t ∈ (a + t, b + t). Thus, since {In } covers A, {In + t} covers A + t.
P
Since A + t ⊂ ∪{In + t}, we have m∗ (A + t) ≤ m∗ (∪{In + t}) ≤
|In + t|, that last move by
subaddivitity.
P
But since |In | = |In + t|, we have m∗ (A + t) ≤
|In |.
1
By definition of the infinum, for all > 0, we can find a cover of A called {In } such that
m∗ (A) + .
P
That is, since m∗ (A + t) ≤
|In | holds for all covers of A, we can say
P
|In | <
m∗ (A + t) < m∗ (A) + Thus, we let → 0 and get
m∗ (A + t) ≤ m∗ (A)
This holds for all A ⊂ R and t ∈ R
Let A0 = t + A and t0 = −t. Then by the conclusion above, we have m∗ (t0 + A0 ) ≤ m∗ (A0 ). But
t0 + A0 = −t + t + A = A and A0 = t + A. Thus, we have m∗ (A) ≤ m∗ (t + A)
This proves equality.
5. Claim: If we fix t ≥ 0 and A ⊂ R, and we define the t-dilation of A to be the set tA = {ta|a ∈
A} ⊂ R, then m∗ (tA) = tm∗ (A)
Proof. Fix an interval
b). Then |I| = b − a and |tI| = t(b − a), or that t|I| = |tI|. This
P I = (a,P
further implies that
t|In | =
|tIn |, if In is just an interval.
Let {In } cover A. Then we can dilate each interval by t. Let {tIn } be the dilated collection of
intervals.
Fix x ∈ (a, b).P Then tx ∈ (ta, tb). So we have tA ⊂ ∪{tIn }. This tells us that m∗ (tA) ≤
m∗ (∪{tIn }) ≤
|tIn |, the last move by subaddivitity.
P
P
P
But we knew that
t|In | = |tIn | and so m∗ (tA) ≤ t|In |.
P
By definition of the infinum, for allP > 0, we can find a cover of A called {In } such that
|In | <
m∗ (A) + . This also tells us that
t|In | < tm∗ (A) + t
Thus, we can conclude
m∗ (tA) < tm∗ (A) + t
and let go to zero to get
m∗ (tA) ≤ tm∗ (A)
which holds for all A ⊂ R and t ≥ 0.
Let t = 0. Then tA = {0} and we are done, since tm∗ (A) = 0 = m∗ (0).
Let t 6= 0. Then define A0 = tA and t0 = t−1 . Thus, we have m∗ (t0 A0 ) ≤ t0 m∗ (A0 ). But t0 A0 = A
and A0 = tA and so we have m∗ (A) ≤ t−1 m∗ (tA) or tm∗ (A) ≤ m∗ (tA)
This proves equality.
2