Real Analysis
Chapter 1 Solutions
Jonathan Conder
3. (a) Let M be an infinite σ-algebra of subsets of some set X. There exists a countably infinite subcollection C ⊆ M,
and we may choose C to be closed under taking complements (adding in missing complements if necessary).
For each x ∈ X, define Dx := ∩{C ∈ C | x ∈ C}, so that Dx ∈ M. Let x, y ∈ X and suppose y ∈ Dx . Then
y ∈ C for all C ∈ C with x ∈ C. Moreover, if y ∈ C for some C ∈ C, then y ∈
/ C c ∈ C, so x ∈
/ C c and hence
x ∈ C. This implies that, if C ∈ C, then x ∈ C iff y ∈ C. In particular Dx = Dy . If x, y ∈ X and there exists
z ∈ Dx ∩ Dy , then Dx = Dz = Dy , so the collection D := {Dx | x ∈ X} is pairwise disjoint. If C ∈ C, then
C = ∪{Dx | x ∈ C}, since x ∈ Dx ⊆ C for all x ∈ C. Therefore, the image of the map ∪ : 2D → 2X contains C,
so there exists a surjection from some subset of 2D onto C. If D were finite, then 2D would be as well, which is
a contradiction because C is infinite. Therefore D is an infinite, pairwise disjoint subcollection of M.
(b) Let M be an infinite σ-algebra of subsets of some set X. By the previous exercise, there exists an infinite, pairwise
disjoint subcollection D of M. Since Q is countable, there exists an injection q : Q → D. Define r : 2Q → M by
r(A) := ∪a∈A q(a). Then r is well-defined because each A ⊆ Q is countable, and injective because D is pairwise
disjoint. There exists an injection from R → 2Q , for example the map x 7→ (−∞, x] ∩ Q (which is injective by
the least upper bound property of R). Composing these injections gives an injective map from R → M, which
shows that card(M) ≥ c.
4. Let A be an algebra which is closed under countable increasing unions. To show that A is a σ-algebra, it suffices
to show that it is closed under arbitrary countable unions. To this end, let {En }n∈N be a countable collection of
members of A. For each n ∈ N define Fn := ∪nk=1 Ek . Since A is an algebra it is clear that (Fn )n∈N is an increasing
sequence of members of A, so ∪n∈N En = ∪n∈N Fn ∈ A as required. Conversely, it is plain that every σ-algebra is
closed under countable increasing unions.
5. Let E be a collection of subsets of some set X, and let M be the σ-algebra generated by E. Define
N := ∪{A | A is the σ-algebra generated by some countable subcollection of E}.
It is clear that ∅, X ∈ N, E ⊆ N and N is closed under taking complements. Let {En }n∈N be a countable collection
of members of N. For each n ∈ N there exists a countable subcollection En of E such that En is in the σ-algebra
generated by En . Since union of countably many countable sets is countable, F := ∪n∈N En is a countable subcollection
of E. Let A be the σ-algebra generated by F, so that A ⊆ N. For each n ∈ N the σ-algebra generated by En is a subset
of A, because A is a σ-algebra containing En . This implies that En ∈ A for all n ∈ N, and hence ∪n∈N En ∈ A ⊆ N.
Therefore N is closed under countable unions, so it is a σ-algebra containing E. This implies that M ⊆ N.
Conversely, let E ∈ N. Then E belongs to the σ-algebra A generated by some countable subcollection of E. Since M
is a σ-algebra containing this subcollection, A ⊆ M and hence E ∈ M. This shows that N ⊆ M and hence N = M.
6. Note that µ(∅) = µ(∅ ∪ ∅) = µ(∅) = 0. Given a sequence (An )n∈N of disjoint sets in M, there exists a sequence
(En )n∈N of sets in M and a sequence (Fn )n∈N of subsets of measure zero sets from M such that An = En ∪ Fn for all
n ∈ N. Note that (En )n∈N is pairwise disjoint, and ∪n∈N Fn is a subset of a measure zero set. Therefore
X
X
µ(∪n∈N An ) = µ((∪n∈N En ) ∪ (∪n∈N Fn )) = µ(∪n∈N En ) =
µ(En ) =
µ(An ).
n∈N
n∈N
This shows that µ is a measure. Let A ⊆ X, and suppose there exists B ∈ M such that A ⊆ B and µ(B) = 0. Then
B = E ∪ F for some E ∈ M and F a subset of a measure zero set N ∈ M. It follows that A ⊆ E ∪ N, where E ∪ N ∈ M
and µ(E ∪ N ) ≤ µ(E) + µ(N ) = µ(E) + 0 ≤ µ(B) = 0. Therefore A = ∅ ∪ A ∈ M, which shows that µ is complete.
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Real Analysis
Chapter 1 Solutions
Jonathan Conder
Let λ : M → [0, ∞] be another measure which extends µ, and let A ∈ M. Then A = E ∪ F for some E ∈ M and F a
subset of a measure zero set N ∈ M. It follows that
µ(E) = λ(E) ≤ λ(A) ≤ λ(E) + λ(F ) ≤ λ(E) + λ(N ) = µ(E) + µ(N ) = µ(E).
Therefore λ(A) = µ(E) = µ(A), so λ = µ. This shows that µ is the unique measure on M which extends µ.
7. Set µ :=
Pn
j=1 aj µj ,
and note that µ(∅) = 0. If {Ei }i∈N is a pairwise disjoint collection of members of M, then
µ(∪i∈N Ei ) =
n
X
j=1
aj µj (∪i∈N Ei ) =
n
X
aj
j=1
X
µj (Ei ) =
n
XX
aj µj (Ei ) =
i∈N j=1
i∈N
X
µ(Ei )
i∈N
(note that we are allowed to change the order of summation because all the terms are non-negative; alternatively you
can consider the convergent/divergent cases separately). This shows that µ is a measure on (X, M).
8. Let (X, M, µ) be a measure space and {En }n∈N a countable collection of measurable sets. Then (∩∞
k=n Ek )n∈N is an
∞
increasing sequence of measurable sets with µ(∩k=n Ek ) ≤ µ(En ) for all n ∈ N, so
∞
∞
µ(lim inf En ) = µ(∪n∈N (∩∞
k=n Ek )) = lim µ(∩k=n Ek ) = lim inf µ(∩k=n Ek ) ≤ lim inf µ(En ).
n→∞
n→∞
n→∞
n∈N
∞
Moreover (∪∞
k=n Ek )n∈N is a decreasing sequence of measurable sets with µ(∪k=n Ek ) ≥ µ(En ), so
∞
∞
µ(lim sup En ) = µ(∩n∈N (∪∞
k=n Ek )) = lim µ(∪k=n Ek ) = lim sup µ(∪k=n Ek ) ≥ lim sup µ(En )
n→∞
n∈N
n→∞
n→∞
provided that the first term of the sequence has finite measure, i.e. µ(∪∞
k=1 Ek ) < ∞.
9. Let (X, M, µ) be a measure space and E, F ∈ M. Then E = (E ∩ F ) t (E \ F ) and (E \ F ) t F = E ∪ F, so
µ(E) + µ(F ) = µ(E ∩ F ) + µ(E \ F ) + µ(F ) = µ(E ∩ F ) + µ(E ∪ F ).
10. Clearly µE (∅) = µ(∅ ∩ E) = µ(∅) = 0. If {En }n∈N is a pairwise disjoint collection of members of M, then
X
X
µE (∪n∈N En ) = µ((∪n∈N En ) ∩ E) = µ(∪n∈N (En ∩ E)) =
µ(En ∩ E) =
µE (En ).
n∈N
n∈N
Therefore µE is a measure.
11. Suppose that µ is continuous from below. Let {En }n∈N be a pairwise disjoint collection of measurable sets, and for
each n ∈ N define Fn := ∪ni=1 En . Since µ is continuous from below and finitely additive,
µ(∪i∈N Ei ) = µ(∪n∈N Fn ) = lim µ(Fn ) = lim
n→∞
n→∞
n
X
i=1
µ(Ei ) =
∞
X
µ(Ei ).
i=1
This shows that µ is a measure. Conversely, if µ is a measure it is continuous from below by Theorem 1.8.
Now suppose that µ(X) < ∞ and µ is continuous from above. If (En )n∈N is an increasing sequence of measurable
sets, then (Enc )n∈N is a decreasing sequence of measurable sets, which implies that
µ(∪n∈N En ) = µ(X) − µ((∪n∈N En )c ) = µ(X) − µ(∩n∈N Enc ) = µ(X) − lim µ(Enc ) = lim µ(En ).
n→∞
n→∞
This shows that µ is continuous from below, so it is a measure by the previous paragraph. Conversely, if µ is a (finite)
measure it is continuous from above by Theorem 1.8.
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Real Analysis
Chapter 1 Solutions
Jonathan Conder
14. Suppose for a contradiction that there exists C ∈ (0, ∞) such that every measurable subset F ⊆ E satisfies µ(F ) ≤ C
or µ(F ) = ∞. Set M := sup{µ(F ) | F ⊆ E is measurable and µ(F ) < ∞}, and note that 0 ≤ M ≤ C. For each
n ∈ N there exists a measurable subset En ⊆ E such that M − n−1 ≤ µ(En ) < ∞. Set Fn := ∪ni=1 En for each n ∈ N
P
and define F := ∪n∈N Fn . Note that M − n−1 ≤ µ(En ) ≤ µ(F ) and also µ(Fn ) ≤ ni=1 µ(En ) < ∞ for all n ∈ N, so
M ≤ µ(F ) = limn→∞ µ(Fn ) ≤ M. This shows that µ(F ) = M, so µ(E \ F ) = ∞. Since µ is semifinite, there exists a
measurable subset A ⊆ E \ F such that 0 < µ(A) < ∞. This contradicts the definition of M, because A ∪ F ⊆ E but
µ(F ) < µ(A) + µ(F ) = µ(A ∪ F ) < ∞. Therefore, for any C ∈ (0, ∞) there exists a measurable subset F ⊆ E such
that C < µ(F ) < ∞.
17. Let A, B ⊆ X be disjoint µ∗ -measurable sets, and let E ⊆ X. Then (A ∪ B) ∩ Ac = B and hence
µ∗ (E ∩ (A ∪ B)) = µ∗ (E ∩ (A ∪ B) ∩ A) + µ∗ (E ∩ (A ∪ B) ∩ Ac ) = µ∗ (E ∩ A) + µ∗ (E ∩ B).
It follows by induction that
Pn
k=1 µ
∗ (E
∩ Ak ) = µ∗ (E ∩ (∪nk=1 Ak )) ≤ µ∗ (E ∩ (∪∞
k=1 Ak )) for all n ∈ N. Therefore
∞
X
µ∗ (E ∩ Ak ) ≤ µ∗ (E ∩ (∪∞
k=1 Ak )),
k=1
which implies that µ∗ (E ∩ (∪∞
k=1 Ak )) =
P∞
k=1 µ
∗ (E
∩ Ak ) because µ∗ is subadditive.
18. (a) Let E ⊆ X and ε ∈ (0, ∞). If µ∗ (E) = ∞ then X ∈ A ⊆ Aσ , E ⊆ X and µ∗ (X) ≤ µ∗ (E) + ε. Otherwise
(
)
X
∗
µ0 (An ) An ∈ A for all n ∈ N and E ⊆ ∪n∈N An ,
µ (E) = inf
n∈N
so there exists a sequence (An )n∈N in A such that E ⊆ ∪n∈N An and
P
then A ∈ Aσ , E ⊆ A and µ∗ (A) ≤ n∈N µ0 (An ) ≤ µ∗ (E) + ε.
P
n∈N µ0 (An )
≤ µ∗ (E) + ε. If A := ∪n∈N An ,
(b) Let E ⊆ X such that µ∗ (E) < ∞. Suppose that E is µ∗ -measurable. For each n ∈ N there exists Bn ∈ Aσ such
that E ⊆ Bn and µ∗ (Bn ) ≤ µ∗ (E) + n−1 . Define B := ∩n∈N Bn , so that B ∈ Aσδ and E ⊆ B. Since µ∗ (E) < ∞,
it follows that µ∗ (B \ E) ≤ µ∗ (Bn ∩ E c ) = µ∗ (Bn ) − µ∗ (Bn ∩ E) = µ∗ (Bn ) − µ∗ (E) ≤ n−1 for all n ∈ N. This
implies that µ∗ (B \ E) = 0.
Conversely, suppose there exists B ∈ Aσδ such that E ⊆ B and µ∗ (B \ E) = 0. If F ⊆ X, then
µ∗ (F ∩ E) + µ∗ (F ∩ E c ) ≤ µ∗ (F ∩ B) + µ∗ (F ∩ B c ) + µ∗ (F ∩ (B \ E)) ≤ µ∗ (F ∩ B) + µ∗ (F ∩ B c ) = µ∗ (F )
because B ∈ Aσδ is µ∗ -measurable. Clearly µ∗ (F ) ≤ µ∗ (F ∩ E) + µ∗ (F ∩ E c ), so E is µ∗ -measurable.
19. If E ⊆ X is µ∗ -measurable, then µ∗ (X) = µ∗ (X ∩ E) + µ∗ (X ∩ E c ) = µ∗ (E) + µ∗ (E c ) by definition, so µ∗ (E) = µ∗ (E).
Conversely, let E ⊆ X and suppose that µ∗ (E) = µ∗ (E). By the previous exercise, for each n ∈ N there exist
An , Bn ∈ Aσ such that E ⊆ An , E c ⊆ Bn , µ∗ (An ) ≤ µ∗ (E) + n−1 and µ∗ (Bn ) ≤ µ∗ (E c ) + n−1 . If A := ∩n∈N An , then
µ∗ (A \ E) ≤ µ∗ (An ∩ Bn ) = µ∗ (An ) − µ∗ (An \ Bn )
= µ∗ (An ) − µ∗ (Bnc )
= µ∗ (An ) − (µ∗ (X) − µ∗ (Bn ))
≤ µ∗ (E) + n−1 − µ∗ (X) + µ∗ (E c ) + n−1
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Real Analysis
Chapter 1 Solutions
Jonathan Conder
= µ∗ (E) − µ∗ (E) + 2n−1
= 2n−1
for all n ∈ N, and hence µ∗ (A \ E) = 0. Moreover E ⊆ A and A ∈ Aσδ , so by the previous exercise E is µ∗ -measurable.
20. (a) If (An )n∈N is a sequence in M∗ such that E ⊆ ∪n∈N An , then
X
X
µ(An ).
µ∗ (E) ≤ µ∗ (∪n∈N An ) ≤
µ∗ (An ) =
n∈N
n∈N
Therefore µ∗ (E) ≤ µ+ (E), by definition. If there exists A ∈ M∗ such that E ⊆ A and µ∗ (A) = µ∗ (E), then
µ+ (E) ≤ µ(A) +
∞
X
µ(∅) = µ∗ (A) = µ∗ (E),
n=2
µ∗ (E)
µ+ (E).
µ∗ (E)
so
=
Conversely, suppose that
= µ+ (E). By exercise 18 part (a), for each n ∈ N there
exists An ∈ M∗σ = M∗ such that E ⊆ An and µ+ (An ) ≤ µ+ (E) + n−1 . Define A := ∩n∈N An , so that A ∈ M∗
and E ⊆ A. Moreover µ+ (A) ≤ µ+ (An ) ≤ µ+ (E) + n−1 for all n ∈ N, so µ+ (A) ≤ µ+ (E). It follows that
µ∗ (A) = µ+ (A) = µ+ (E) = µ∗ (E), because A ∈ M∗ and E ⊆ A.
(b) Let µ0 be a premeasure on an algebra A ⊆ M∗ such that µ∗ is induced by µ0 , and let E ⊆ X. By exercise 18
part (a), for each n ∈ N there exists An ∈ Aσ ⊆ M∗ such that E ⊆ An and µ∗ (An ) ≤ µ∗ (E) + n−1 . Define
A := ∩n∈N An , so that A ∈ M∗ and E ⊆ A. Moreover µ∗ (A) ≤ µ∗ (An ) ≤ µ∗ (E) + n−1 for all n ∈ N, so
µ∗ (A) ≤ µ∗ (E). It follows that µ∗ (A) = µ∗ (E), because E ⊆ A. By the previous exercise, this implies that
µ∗ (E) = µ+ (E). Therefore µ∗ = µ+ .
(c) Define µ∗ : 2X → [0, ∞] by µ∗ (∅) = 0, µ∗ ({0}) = 2, µ∗ ({1}) = 2 and µ∗ (X) = 3. Clearly µ∗ (A) ≤ µ∗ (B) for all
A ⊆ B ⊆ X. Moreover µ∗ is subadditive, because µ∗ (X) < µ∗ ({0}) + µ∗ ({1}). Therefore µ∗ is an outer measure.
Note that {0} is not µ∗ -measurable, because µ∗ (X) = 3 < 2 + 2 = µ∗ (X ∩ {0}) + µ∗ (X ∩ {0}c ). This implies that
P
µ+ ({0}) = inf{ n∈N µ∗ (An ) | An ⊆ X for all n ∈ N and An = X for some n ∈ N} = µ∗ (X) = 3 6= 2 = µ∗ ({0}).
21. Let E ⊆ X be locally µ∗ -measurable, and let A ⊆ X. If µ∗ (A) = ∞, then clearly µ∗ (A ∩ E) + µ∗ (A ∩ E c ) ≤ µ∗ (A).
Otherwise, by exercise 18 part (a), for each n ∈ N there exists a µ∗ -measurable set An ⊆ X such that A ⊆ An and
µ∗ (An ) ≤ µ∗ (A) + n−1 . In particular, if n ∈ N then µ∗ (An ) < ∞, and hence An ∩ E and An ∩ E c = An ∩ (An ∩ E)c
are µ∗ -measurable (because E is locally µ∗ -measurable). It follows that
µ∗ (A ∩ E) + µ∗ (A ∩ E c ) ≤ µ∗ (An ∩ E) + µ∗ (An ∩ E c ) = µ(An ∩ E) + µ(An ∩ E c ) = µ(An ) = µ∗ (An ) ≤ µ∗ (A) + n−1
for all n ∈ N, so µ∗ (A ∩ E) + µ∗ (A ∩ E c ) ≤ µ∗ (A). Clearly µ∗ (A) ≤ µ∗ (A ∩ E) + µ∗ (A ∩ E c ) in either case, so E is
µ∗ -measurable. This shows that µ is saturated.
22. (a) Clearly µ|M = µ, so by exercise 6 from section 1.3 it suffices to show that M∗ = M. To this end, let A ∈ M.
Then A = E ∪ F for some E ∈ M and F ⊆ X such that F ⊆ B for some B ∈ M with µ(B) = 0. Clearly
A ⊆ E ∪ B ∈ M = Mσδ and µ∗ ((E ∪ B) \ A) ≤ µ∗ (B) = 0. By exercise 18 part (c), it follows that A ∈ M∗ .
Conversely, let A ∈ M∗ . By exercise 18 part (c) there exists B ∈ Mσδ = M such that A ⊆ B and µ∗ (B \ A) = 0.
Since B \ A ∈ M∗ , exercise 18 part (b) implies that B \ A ⊆ E and µ∗ (E \ (B \ A)) = 0 for some E ∈ M. Note
that B \ E ∈ M and A ∩ E ∈ M∗ . Moreover µ∗ (A ∩ E) ≤ µ∗ (E \ (B \ A)) = 0, so there exists F ∈ M such that
A ∩ E ⊆ F and µ∗ (F \ (A ∩ E)) = 0, by exercise 18 part (b). It follows that
µ(F ) = µ∗ (F ) ≤ µ∗ (F \ (A ∩ E)) + µ∗ (A ∩ E) = 0.
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Real Analysis
Chapter 1 Solutions
Jonathan Conder
Since A = (B \ E) ∪ (A ∩ E), it follows that A ∈ M. Therefore M∗ = M.
e and µ
e ∈ M,
b be the completion of µ. Given A ∈ M∗ with µ(A) < ∞, the converse of the previous exercise
(b) Let E
implies that A ∈ M (using exercise 18 part (b) instead of part (c)). Hence A = E ∪ F for some E ∈ M and
F ⊆ X such that F ⊆ B for some B ∈ M with µ(B) = 0. In particular µ
b(A) = µ(E) = µ∗ (E) ≤ µ∗ (A) < ∞.
e is locally µ
e ∩ A) < ∞ the previous exercise implies that
e ∩ A ∈ M, as E
b-measurable. Since µ∗ (E
Therefore E
e ∩ A ∈ M∗ (again using exercise 18 part (b) instead of part (c)). This shows that E
e is locally µ∗ -measurable,
E
e ∈ M∗ by exercise 21.
so E
e ∈ M∗ and A ∈ M be such that µ
Conversely, let E
b(A) < ∞. Then A = E ∪ F for some E ∈ M and F ⊆ X such
that F ⊆ B for some B ∈ M with µ(B) = 0. It follows that
e ∩ A) ≤ µ∗ (A) ≤ µ∗ (E) + µ∗ (F ) = µ(E) + 0 = µ
µ∗ (E
b(A) < ∞.
e ∩ A ∈ M∗ and hence
This implies that A ∈ M∗ (by the previous exercise using part (b) instead of part (c)), so E
e
e ∩ A ∈ M (again by the previous exercise). This shows that E
e is locally µ
e ∈ M.
E
b-measurable, so E
e \ M, then µ
e = M∗ . By exercise 6 from section 1.3, µ
e
e
Therefore M
b and µ agree on M. If E ∈ M
b(E) = ∞ by
e
∗
definition. Moreover, if E ∈ M = M and µ(E) < ∞, then E ∈ M by the previous exercise (using part (b)
e \ M.
e
instead of part (c)). This implies that µ
b and µ also agree on M
23. (a) Let E := {(a, b] ∩ Q | a, b ∈ R}. Clearly ∅ = (0, 0] ∩ Q ∈ E. If (a1 , b1 ] ∩ Q ∈ E and (a2 , b2 ] ∩ Q ∈ E then their
intersection is (max{a1 , a2 }, min{b1 , b2 }] ∩ Q ∈ E. Moreover, the complement of (a, b] ∩ Q ∈ E is ((−∞, a] ∩ Q) ∪
((b, ∞] ∩ Q), which is a disjoint union of elements of E provided that a ≤ b. If a > b then the complement of
(a, b] ∩ Q is just (−∞, ∞] ∩ Q ∈ E. This shows that E is an elementary family of subsets of Q, so the collection
of finite disjoint unions of members of E is an algebra. If (a1 , b1 ] ∩ Q ∈ E and (a2 , b2 ] ∩ Q ∈ E are not disjoint,
then their union is (min{a1 , a2 }, max{b1 , b2 }] ∩ Q ∈ E. Therefore A is the collection of finite disjoint unions of
members of E, so A is an algebra.
(b) Let M be the σ-algebra generated by A. Since Q is countable and M is closed under countable unions, it suffices
to show that {x} ∈ M for all x ∈ Q. Given x ∈ Q and n ∈ N, it is clear that (x − n1 , x] ∩ Q ∈ M. Therefore
{x} = ∩n∈N ((x − n1 , x] ∩ Q) ∈ M, as required.
(c) By definition µ0 (∅) = 0. Let (En )n∈N be a sequence of disjoint members of A whose union lies in A. If En = ∅
P
for all n ∈ N, then ∪n∈N En = ∅ and hence µ0 (∪n∈N En ) = 0 = n∈N µ0 (En ). Otherwise ∪n∈N En 6= ∅, and
P
Em 6= ∅ for some m ∈ N, so µ0 (∪n∈N ) = ∞ and n∈N µ0 (En ) ≥ µ0 (Em ) = ∞. Therefore µ0 is a premeasure on
A. Define µ1 , µ2 : 2Q → [0, ∞] by


∞, E 6= ∅
∞, E contains 2−n m for some m ∈ Z and n ∈ N
µ1 (E) =
and µ1 (E) =
0, otherwise.
0,
E=∅
Clearly µ1 |A = µ0 = µ2 |A , because every non-empty interval contains a rational of the form 2−n m for some
m ∈ Z and n ∈ N. For the same reason that µ0 is a premeasure, µ1 is a measure. A very similar argument
implies that µ2 is a measure. But µ1 6= µ2 because µ1 ({3−1 }) = ∞ whereas µ2 ({3−1 }) = 0.
24. (a) Since µ(A) + µ(Ac ∩ B) = µ(A ∪ B) = µ(B) + µ(B c ∩ A), by symmetry it suffices to show that µ(Ac ∩ B) = 0.
To this end, note that E ⊆ A ∪ B c = (Ac ∩ B)c (if a member of E is not in B c , it is in B ∩ E = A ∩ E ⊆ A). So
µ(X) = µ∗ (X) = µ∗ (E) ≤ µ∗ ((Ac ∩ B)c ) = µ((Ac ∩ B)c ) = µ(X) − µ(Ac ∩ B),
and hence µ(Ac ∩ B) = 0 as required.
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Real Analysis
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Jonathan Conder
(b) It is clear that ME is a σ-algebra on E and that ν(∅) = 0. Let {En }n∈N be a pairwise disjoint subset of ME . For
each n ∈ N there exists An ∈ M such that En = An ∩ E. Set A := ∪i∈N ∪∞
j=i+1 (Ai ∩ Aj ) and define Bn := An \ A
for each n ∈ N. It is easily checked that {Bn }n∈N is a pairwise disjoint subset of M and that En = Bn ∩ E for
all n ∈ N. Therefore
ν(∪n∈N En ) = ν(∪n∈N (Bn ∩ E)) = ν((∪n∈N Bn ) ∩ E) = µ(∪n∈N Bn ) =
X
µ(Bn ) =
n∈N
X
ν(En ),
n∈N
which shows that ν is a measure on ME .
25. Let (Cn )n∈N be a sequence of compact intervals covering R, and fix n ∈ N. There exists a Gδ set Vn ∈ Mµ and a null
set Nn ∈ Mµ such that E ∩ Cn = Vn \ Nn . Let (Vnk )k∈N be a sequence of open sets such that Vn = ∩k∈N Vnk . For each
k ∈ N define an open set Unk := Vnk ∪ Cnc , and set U := ∩n,k∈N Unk . Then U is a Gδ set and U \ E ⊆ ∪n∈N Nn . Indeed,
if x ∈ U \ E there exists n ∈ N such that x ∈ Cn , which implies that x ∈ Vnk for all k ∈ N and hence x ∈ Vn but
x∈
/ E ∩ Cn = Vn \ Nn . Moreover E ⊆ U, because E ⊆ Unk for all n, k ∈ N. It follows that E = U \ (U \ E), where
U \ E is a null set.
If n ∈ N, then E ∩ Cn = Hn ∪ Nn for some Fσ set Hn ∈ Mµ and some null set Nn ∈ Mµ . Clearly ∪n∈N Hn and ∪n∈N Nn
are respectively Fσ and null sets. Moreover E = ∪n∈N (E ∩ Cn ) = (∪n∈N Hn ) ∪ (∪n∈N Nn ).
26. Let E ∈ Mµ and suppose that µ(E) < ∞. Given ε ∈ (0, ∞), there exists an open set U ∈ Mµ such that E ⊆ U and
µ(U ) < µ(E) + 2ε . Let (Un )n∈N be a sequence of disjoint open intervals such that ∪n∈N Un = U. Then
X
µ(Un ) = µ(U ) < ∞,
n∈N
so there exists N ∈ N such that
P∞
n=N +1 µ(Un )
< 2ε . Define A := ∪N
n=1 Un . It follows that
µ(E4A) ≤ µ(E \ A) + µ(A \ E) ≤ µ(E \ U ) + µ(U \ A) + µ(U \ E) = 0 +
∞
X
µ(Un ) + µ(U ) − µ(E) < ε.
n=N +1
28. Let a, b ∈ R. Since µF is continuous from above,
µF ([a, b]) = µF (∩n∈N (a − n−1 , b]) = lim µF ((a − n−1 , b]) = lim (F (b) − F (a − n−1 )) = F (b) − F (a−).
n→∞
n→∞
It follows that µF ({a}) = µF ([a, a]) = F (a) − F (a−), in which case µF ([a, b)) = µF ([a, b]) − µF ({b}) = F (b−) − F (a−)
and µF ((a, b)) = µF ([a, b)) − µF ({a}) = F (b−) − F (a).
29. (a) Suppose that E ⊆ N but m(E) > 0. Define R := Q ∩ [0, 1), and for each r ∈ R set Er := E + r. Clearly each Er
is measurable with m(Er ) = m(E), and ∪r∈R Er ⊆ [0, 2). Let r, s ∈ R and suppose that Er intersects Es . Then
there exists t ∈ Er ∩ Es , so that t − r, t − s ∈ E ⊆ N. Since t − r = (t − s) + (s − r) and s − r ∈ Q, the definition
of N implies that t − s = t − r. Therefore r = s, which shows that {Er }r∈R is pairwise disjoint. Hence
∞=
X
r∈R
m(E) =
X
m(Er ) = m(∪r∈R Er ) ≤ m([0, 2)) = 2,
r∈R
which is a contradiction. Therefore m(E) = 0.
6
Real Analysis
Chapter 1 Solutions
Jonathan Conder
(b) Suppose that m(E) > 0, but every subset of E is measurable. Since E = ∪n∈Z (E ∩ [n, n + 1)), there exists
n ∈ Z such that m(E ∩ [n, n + 1)) > 0. Define F := (E ∩ [n, n + 1)) − n, so that F ⊆ [0, 1), m(F ) > 0 and every
subset of F is measurable. Also define R := Q ∩ [−1, 1], and for each r ∈ R set Nr := N + r. It is clear that
[0, 1) ⊆ ∪r∈R Nr , and hence F = ∪r∈R (F ∩ Nr ). If r ∈ R then F ∩ Nr ⊆ F and (F ∩ Nr ) − r ⊆ N, which implies
that both subsets are measurable (by containment in F and translational invariance) and have measure zero (by
P
P
the previous exercise and translational invariance). Therefore m(F ) ≤ r∈R m(F ∩ Nr ) = r∈R 0 = 0, which
is a contradiction so not every subset of E is measurable.
30. Let E ∈ L with m(E) > 0, and suppose there exists α ∈ (0, 1) such that m(E ∩ I) ≤ αm(I) for all open intervals I.
Without loss of generality m(E) < ∞ (since m is semifinite, we may replace E by a subset of finite positive measure).
Define ε := m(E)(1 − α), so that ε > 0. Since E ∈ L and m is outer regular, there exists an open set U ⊆ R such
that E ⊆ U and m(U ) < m(E) + ε. As U is open, there exists a pairwise disjoint collection {Ii }i∈N of open intervals
such that U = ∪i∈N Ii , and these intervals are bounded because m(U ) < ∞. If if i ∈ N then
m(Ii ) = m(Ii \ E) + m(E ∩ Ii ) ≤ m(Ii \ E) + αm(Ii )
and hence (1 − α)m(Ii ) ≤ m(Ii \ E). Since ∪i∈N (Ii \ E) = U \ E has measure m(U ) − m(E), it follows that
(1 − α)m(U ) = (1 − α)
X
i∈N
m(Ii ) =
X
(1 − α)m(Ii ) ≤
i∈N
X
m(Ii \ E) = m(∪i∈N (Ii \ E)) < ε = (1 − α)m(E)
i∈N
and hence m(U ) < m(E) ≤ m(U ), which is impossible. Thus, for each α ∈ (0, 1) there exists an open interval I such
that m(E ∩ I) > αm(I). The same clearly holds for α ∈ (−∞, 0].
31. Let E ∈ L with m(E) > 0, and set α := 34 . By the previous exercise there exists a open interval I ⊆ R with endpoints
a, b ∈ R such that m(E ∩ I) > αm(I). Suppose there exists x ∈ (− 12 m(I), 12 m(I)) such that x ∈
/ E − E. Then
−x ∈
/ E − E, and clearly 0 ∈ E − E, so we may assume x > 0. Moreover E ⊆ (E c + x) ∩ (E c − x). Note that
(a, a + 2x] ⊆ I, because 2x < m(I) = b − a. Define n := max{k ∈ N | a + 2kx < b}, so that
m(E ∩ (a, a + 2nx]) =
≤
n
X
k=1
n
X
(m(E ∩ (a + 2(k − 1)x, a + (2k − 1)x]) + m(E ∩ (a + (2k − 1)x, a + 2kx]))
(m((E c − x) ∩ (a + 2(k − 1)x, a + (2k − 1)x]) + m((E c + x) ∩ (a + (2k − 1)x, a + 2kx]))
k=1
≤
≤
n
X
k=1
n
X
(m(E c ∩ (a + (2k − 1)x, a + 2kx] − x) + m(E c ∩ (a + 2(k − 1)x, a + (2k − 1)x] + x))
(m(E c ∩ (a + (2k − 1)x, a + 2kx]) + m(E c ∩ (a + 2(k − 1)x, a + (2k − 1)x]))
k=1
≤ m(E c ∩ (a, a + 2nx]).
This implies that m(E ∩ (a, a + 2nx]) ≤ 12 (m(E ∩ (a, a + 2nx]) + 21 m(E c ∩ (a, a + 2nx])) = 21 m((a, a + 2nx]) = nx.
Note that 4nx ≥ m(I), since otherwise k := 2n satisfies a + 2kx < b (but k > n). It follows that
m(E ∩ I) ≤ m(E ∩ (a, a + 2nx]) + m((a + 2nx, b)) ≤ m(I) − nx ≤ m(I) −
3m(I)
m(I)
=
= αm(I).
4
4
This is a contradiction, so there does not exist x ∈ (− 12 m(I), 12 m(I)) such that x ∈
/ E − E.
7
Real Analysis
Chapter 1 Solutions
Jonathan Conder
33. Choose a surjection q : N → Q ∩ [0, 1], and for each n ∈ N define In := (q(n) − 3−n , q(n) + 3−n ). For each n ∈ N,
∞
define Dn = In \ (∪∞
k=n+1 Ik ). If m, n ∈ N and m < n, then Dm ∩ Dn = ∅ because Dn ⊆ In ⊆ ∪k=m+1 Ik . Moreover
2·3
−n
= m(In ) ≤ m(Dn ) +
m(∪∞
k=n+1 Ik )
∞
X
≤ m(Dn ) +
k=n+1
m(Ik ) = m(Dn ) +
∞
X
2 · 3−k = m(Dn ) + 3−n ,
k=n+1
and hence m(Dn ) ≥ 3−n , for all n ∈ N. For each n ∈ N there exists a Borel set An ⊆ Dn such that 0 < m(An ) < 3−n ,
which can be found by intersecting Dn with an interval of the form [3−n−1 k, 3−n−1 (k + 1)] for some k ∈ Z (not all
of the intersections can have zero measure). Define A := [0, 1] ∩ (∪n∈N An ), and let I be a subinterval of [0, 1] with
midpoint c. There exists n ∈ N such that 4 · 3−n < m(I), and there are infinitely many rationals in (c − 3−n , c + 3−n ),
so there exists k ∈ N with k ≥ n and q(k) ∈ (c − 3−n , c + 3−n ). It follows that Ik ⊆ (c − 2 · 3−n , c + 2 · 3n ) ⊆ I.
Therefore Ak ⊆ A ∩ I (since Ak ⊆ Ik ⊆ [0, 1]), so m(A ∩ I) ≥ m(Ak ) > 0. Moreover
m(A ∩ I) = m((∪n∈N An ) ∩ I) = m(Ak ) + m((∪n∈N\{k} An ) ∩ I) < m(Dk ) + m(I \ Dk ) = m(I),
because m(Ak ) < m(Dk ) and ∪n∈N\{k} An ⊆ ∪n∈N\{k} Dn ⊆ Dkc .
8