Problem 1
Part a
If E has infinite measure, then just take A = X.
Let E ⊂ X and let ǫ > 0. By the definition of the induced outer S
measure
∞
µ∗ , There exists a countable collection Aj of sets in A such that E ⊂ j=1 Aj
and
∞
X
µ0 (Aj ) ≤ µ∗ (E) + ǫ.
j=1
Let A =
S∞
j=1
Aj . Then A is in Aσ , and

µ(A) = µ∗ 
≤
∞
X
∞
[
j=1

Aj 
µ∗ (Aj )
j=1
≤ µ∗ (E) + ǫ.
Part b
Let E ⊂ X be a µ∗ -measurable set with finite measure. For every integer j,
selectTBj to be a set in Aσ with E ⊂ Bj such that µ∗ (Bj ) ≤ µ∗ (E) + ǫ. Define
∞
B = j=1 Bj . Then Bj ∈ Aσδ and E ⊂ B. Finally, B \ E is a subset of Bj \ E
for every j. Because E is µ∗ -measurable, it follows that µ∗ (Bj \ E) is at most
µ∗ (Bj ) − µ∗ (E), and therefore E has µ∗ -measure no more than 1j . Since this
holds for all j, it follows that µ∗ (B \ E) is equal to zero.
Conversely, Suppose E ⊂ X and B ∈ Aσδ with E ⊂ B and µ∗ (B \ E) = 0.
Let F be an arbitrary subset of X. We need to show that µ∗ (F ) ≥ µ∗ (F ∩
E) + µ∗ (F ∩ E c ). Note that we can write F ∩ B as (F ∩ (B \ E)) ∪ F ∩ E.
So by subadditivity, µ∗ (F ∩ B) ≤ µ∗ (F ∩ E) + µ∗ (F ∩ (B \ E)), and µ∗ (F ∩
(B \ E)) is zero by monotonicity and the fact that µ∗ (B \ E) is zero. Therefore,
µ∗ (F ∩ E) ≥ µ∗ (F ∩ B), and clearly µ∗ (F ∩ E c ) ≥ µ∗ (F ∩ B c ) by monotonicity.
So µ∗ (F ∩ E) + µ∗ (F ∩ E c ) ≥ µ∗ (F ∩ B) + µ∗ (F ∩ B c ) = µ∗ (F ), where the
last equality follows from the fact that B is a measurable set. Note that this
direction of the proof did not make use of the assumption that µ∗ (E) was finite.
Part c
By the second part of part b, we only need to show that, for any measurable
set E, there exists a set B ∈ Aσδ such that µ∗ (B \ E) = 0. Suppose µ0 is
a σ-finite measure, and cover X by a countable collection of sets X1 , X2 , . . . ,
in A. Then each µ∗ (E ∩ Xj ) is finite for each j, because E ∩ Xj is covered
by Xj ∈ A. Therefore, we can apply the result of part b to each E ∩ Xj to
arrive at sets Bj in Aσδ with µ∗ (Bj \ (E ∩ Xj )) = 0 and E ∩ Xj ⊂ Bj . Define
1
S∞
B = j=1 Bj . Each element of B \ E is an element of Bj \ (E ∩ Xj ) for some
S
j, so B \ E ⊂ ∞
j=1 Bj \ (E ∩ Xj ), and therefore B \ E has outer measure zero
by countable subadditivity.
Problem 2
Suppose that µ is a σ-finite measure. Let E ∈ M and consider a set of the
form E ∪ F such that F ⊂ N for some µ-measurable set N with µ(N ) = 0.
Then E ∪ N is a set in M = Mσδ such that µ(E) = µ∗ (E) ≤ µ∗ (E ∪ F ) ≤
µ∗ (E ∪ N ) = µ(E ∪ N ) = µ(E), so by part c of the previous problem, it follows
that E ∪ F is µ∗ -measurable, and therefore that any set in M is measurable
with respect to µ∗ .
Conversely, suppose that S is a µ∗ -measurable set. Then we can apply the
previous problem, part c to S c , which is also a µ∗ -measurable set. That is,
There exists a set B ∈ Mσδ = M such that µ∗ (B \ S c ) is equal to zero. Now,
consider the set B c ⊂ S. Note that S \ B c = S ∩ B is exactly the same set
as B \ S c . So B \ S c has outer measure zero and therefore S is a union of B c ,
which is µ-measurable, and S ∩ B, which has outer measure zero. Note that
any set of outer measure zero is a subset of a set of µ-measure zero by part a of
the previous problem- it must be contained in sets in M of arbitrarily small µ
measure and hence in a set of µ-measure zero.
Problem 3
Part a
We want to take advantage of proposition 1.7 in Folland. We need to show that
sets of the form (a, b]∩Q form an elementary family of subsets of Q. Clearly, the
empty set is of this form. Suppose I take E = (a1 , b1 ] ∩ Q and F = (a2 , b2 ] ∩ Q;
E ∩ F is the (possibly trivial) set (max(a1 , a2 ), min(b1 , b2 )] ∩ Q, which is of the
desired form. Finally, consider ((a, b] ∩ Q)c , where the complement is taken in
Q. If a and b are finite, this set is ((−∞, a] ∩ Q) ∪ ((b, ∞] ∩ Q) , which is a
union of 2 sets in the elementary family. A similar, but simpler, argument can
be applied if one of a, b = ∞. So the sets of the form (a, b] form an elementary
family; therefore, sets of finite unions of such sets form an algebra.
Part b
It’s enough to show that, for any x ∈ Q, the singleton set {x} is in the σalgebra generated by A. This is enough because any subset of Q is countable
and therefore a countable union of singletons.
T∞
Note that any singleton set {x} can be expressed as n=1 (x − n1 , x] ∩ Q.
This is a countable intersection of elements of A, and therefore in the σ-algebra
generated by A.
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Part c
First, we must show that µ0 is a premeasure on A. Clearly µ0 (∅) = 0. Furthermore, countable additivity holds: any disjoint union of sets, not all of which
are empty, has measure ∞, and at least one such set in the union must have
measure ∞.
One extension µ1 of this premeasure to the power set of Q can be obtained
by defining µ1 (S) to be ∞ for any nonempty set S. This is clearly a measure
on the power set of Q for the same reasons µ0 is.
A different extension µ2 is given by the counting measure: each nonempty
set in A has infinitely many elements, so the counting measure on Q extends
the premeasure µ0 .
Of course, there are other extensions too; for example, you could take any
finite nonzero multiple of the counting measure.
Problem 4
We can assume that the Lebesgue measure of E is finite: if it is infinite, just
consider E ∩[a, a+1] for some a for which this intersection has positive Lebesgue
measure.
For any ǫ > 0, there is a covering of E by intervals Ij such that
P
m(I
)
<
m(E) + ǫ. Note that we can assume the intervals are essentially
j
j
disjoint: If there exist Ij , Ik with k > j such that m(Ij ∩ Ik ) ≥ 0, then we can
just replace Ik by Ik \ Ij and still cover E.
Now, we will apply a version of the pigeonhole principle: Given any nonnegative numbers a1 , a2 , . . . an and b1 , b2 , . . . bn , we have that
a1 + a2 + . . . + an
aj
≤ max .
b1 + b2 + . . . + bn
bj
By taking a limit, this inequality can be seen to hold for infinite convergent
sums as well, with the max replacedPby a sup. Setting aj = m(Ij ∩ E) and
bj = m(Ij ), we get that the fraction
Pj
j
aj
bj
a
is equal to
m(E)
m(E)+ǫ .
So there exists a
m(E)
m(E)
j such that bjj ≥ m(E)+ǫ
. This is sufficient to prove the result because m(E)+ǫ
can be made arbitrarily close to 1 by taking ǫ to be sufficiently small.
It may be worth mentioning that you can actually do this for an interval I
centered at almost every x ∈ E: This follows from a general principle called the
Lebesgue differentiation theorem.
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