Real Analysis HW 3 Solutions
Problem 29:
(i) Show that rational equivalence defines an equivalence relation on any set.
(ii) Explicitly find a choice set for the rational equivalence relation on Q.
(iii) Define two numbers to be irrationally equivalent provided their difference is irrational
or zero. Is this an equivalence relation on R? Is this an equivalence relation on Q?
Solution:
(i) To see that the definition of ‘rational equivalence’ on any set ⊆ R defines an equivalence
relation ∼, we must check that ∼ is reflexive, symmetric, and transitive. Reflexivity
and symmetry are immediate since for any x, y ∈ A, x−x = 0 ∈ Q and if x−y = q ∈ Q
then y − x = −q ∈ Q. Finally if x, y, z ∈ A with x ∼ y and y ∼ z, then x − y = q1 ∈ Q
and y − z = q2 ∈ Q, and so x − z = q1 + q2 ∈ Q.
(ii) Obviously every element of Q is rationally equivalent to every other element of Q.
Therefore any singleton element {q} ⊆ Q is a choice set.
(iii) First we note that ‘irrational equivalence’ is not an equivalence relation on R, since it
fails transitivity, owing to
√ the fact that√the sum of two irrational numbers need not
be irrational (i.e. 1/2 + 2 and 1/2 − 2). However since there are no two distinct
elements of Q which are ‘irrationally equivalent’, we have a trivial equivalence relation
on Q which partitions Q into the set of singletons of Q.
Problem 32. Does Lemma 16 remain true if Λ is allowed to be finite or to be uncountably
infinite? Does it remain true if Λ is allowed to be unbounded.
Solution. Recall Lemma 16 from the text:
Lemma 16 Let E be a bounded measurable set of real numbers. Suppose there
is a bounded, countably infinite set of real numbers Λ for which the collection of
translates of E, (λ + E)λ∈Λ , is disjoint. Then m(E) = 0.
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Lemma 16 does not remain true if Λ is allowed to be finite. Let E = [0, 1) and Λ = {0, 1}.
Then E + 1 and E are disjoint, but E is not measure 0.
If Λ is uncountable, then we may always pick a countable subset which satisfies the conditions
of the theorem. Hence the Lemma holds if Λ is uncountable.
The Lemma is not true if Λ is unbounded. Consider E = [0, 1) and Λ = Z. Then E + λ are
all disjoint for all λ ∈ Λ.
Problem 33. Let E be a non-measurable set of finite outer measure. Show that there is a
Gδ set G that contains E for which
m∗ (E) = m∗ (G), while m∗ (G ∼ E) > 0
T
∞
Proof. Define G in the usual way, G = ∞
k=1 Ok where {Ok }k=1 are open sets satisfying
m∗ (Ok ) < m∗ (E) + k1 . Using monotonicity and that fact that m∗ (G) ≤ m∗ (E) + 1/k for
every k ≥ 1, we readily conclude that m∗ (G) = m∗ (E). However we cannot have that
m∗ (G ∼ E) = 0, since that would imply that E = G ∩ (G ∼ E)C is measurable. Therefore
m∗ (G ∼ E) > 0.
Problem 38. Let the function f : [a, b] → R be Lipschitz, that is, there is a constant c ≥ 0
such that for all [u, v] ∈ [a, b], |f (u) − f (v)| ≤ c|u − v|. Show that f maps a set of measure
zero onto an set of measure zero. Show that f maps an Fσ set onto an Fσ set. Conclude
that f maps a measurable set to a measurable set.
Proof: Let > 0 and let E ⊂ [a, b] be a set of measure zero, then
P∞we know that there
∞
exists a collection of open intervals {Ik }k=1 that cover E such that k=1 `(Ik ) ≤ /c, where
c is the Lipschitz constant for f . Now let ak = inf x∈Ik f (x) and bk = supx∈Ik f (x). Note
that f (Ik ) ⊆ [ak , bk ]. By the Lipschitz property of f , we know that for any u, v ∈ [a, b],
|f (u) − f (v)| ≤ c · |u − v|. Taking the sup over u, v ∈ Ik on both sides of the inequality, we
find
bk − ak = sup |f (u) − f (v)| ≤ c · sup |u − v| = c · `(Ik ).
u,v∈Ik
u,v∈Ik
It follows that m(f (Ik )) ≤ bk − ak ≤ c · `(Ik ), and since f (E) ⊆
m(f (E)) ≤
∞
X
m(f (Ik )) ≤
k=1
∞
X
S∞
k=1
f (Ik ) this implies
c · `(Ik ) < .
k=1
Therefore f (E) is measure 0.
To show that f maps Fσ sets onto Fσ sets, let F be a Fσ set. By definition
we can write
S∞
∞
F as the countable union of a collection of closed sets {Fk }k=1 , F = Sk=1 Fk . Since f is
continuous, we know that f (Fk ) is a closed set and therefore the union ∞
k=1 f (Fk ) is an Fσ
set. It follows that
!
∞
∞
[
[
f (F ) = f
Fk =
f (Fk )
k=1
k=1
2
and therefore f maps Fσ sets to Fσ sets.
Finally suppose E ⊂ [a, b] is a measurable set of positive measure (we already proved the
case for 0 measure), then we know there exists an Fσ set F ⊆ E such that m(E ∼ F ) = 0.
We use the set F to partition E into an Fσ set and a set of measure 0, E = F ∪ (E ∼ F ).
Therefore
f (E) = f (F ) ∪ f (E ∼ F ).
However, we know f (F ) is an Fσ set and f (E ∼ F ) is a set of measure zero, therefore f (E)
is measurable.
Problem 45. (Not assigned) Show that a strictly increasing function that is defined on
an interval has a continuous inverse.
Solution: Clearly since f : (a, b) → Range(f ) is strictly monotone it is one-to-one and so
it has a well defined inverse function f −1 : Range(f ) → (a, b). Now fix x ∈ Range(f ), and
define the function
fx : α 7→ f (f −1 (x) + α).
Note fx (0) = x and that since fx (α) is strictly increasing in α, it is invertible with inverse
fx−1 (y) = f −1 (y) − f −1 (x). Let > 0, and define
δ=
min
{α : |α|=}
|x − fx (α)|.
Note that for any y ∈ Range(f ) such that |x − y| < δ, there exist α0 , with y = fx (α0 ).
Since |x − fx (α)| is strictly increasing in |α| we must have |α0 | < . This implies that
α0 = fx−1 (y) = f −1 (y) − f −1 (x) and so
|f −1 (y) − f −1 (x)| = |α0 | < .
Problem 46. Let f be a continuous function and B be a Borel set. Show that f −1 (B) is a
Borel set.
Proof. Let B denote the Borel sets. We define the set
E = {E ⊂ R|f −1 (E) ∈ B}
to be the collection of sets E for which f −1 (E) is Borel. Note that, since f is continuous
E contains all the open sets. Whats left is to show that E is a σ-algebra. To do this, we
consider a set E ∈ E. We note that f −1 (E C ) = f −1 (E)C ∈ B, therefore E C ∈ E. Similarly
for any collection of open sets {Ek }∞
k=1 ⊂ E we have
!
∞
∞
[
[
−1
f
Ek =
f −1 (Ek ) ∈ B
k=1
k=1
3
S
and therefore ∞
k=1 Ek ∈ E. It follows that E is a σ-algebra containing the open sets, and so
by the definition of the Borel sets, B ⊆ E. Therefore by the definition of E if B ∈ B ⊆ E,
then f −1 (B) ∈ B.
Problem 47: Use the preceding two problems to show that a continuous strictly increasing
function that is defined on an interval maps Borel sets to Borel sets.
Solution: Since f is continuous and strictly monotone, it has a continuous inverse g = f −1
(see problem 45). Since g is continuous, we have by problem 46 that if B is Borel, g −1 (B) =
f (B) is Borel.
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