Math 318 HW #5 Solutions
1. Show that the sets Ex defined as part of showing that m∗ is not countably additive (cf.
Example 7.7) form equivalence classes. Use this to conclude that for any x, y ∈ [0, 1], either
Ex = Ey or Ex ∩ Ey = ∅.
Proof. We want to show that the relation x ∼ y when y ∈ Ex is an equivalence relation. The
relation ∼ is certainly reflexive, as x ∈ Ex . Also, if x ∼ y, then by definition y ∈ Ex , meaning
that y − x = r for some r ∈ Q. But then
x − y = −r ∈ Q,
so x ∈ Ey , and hence y ∼ x.
For transitivity, suppose x ∼ y and y ∼ z, meaning that y ∈ Ex and z ∈ Ey . In turn, this
means that y − x = r1 and z − y = r2 for some r1 , r2 ∈ Q. But then
z − x = (z − y) + (y − z) = r2 + r1 ∈ Q,
so z ∈ Ex , which means x ∼ z.
Therefore, we’ve shown that ∼ is an equivalence relation. Hence, for any x, y ∈ [0, 1], either
x ∼ y or x 6∼ y.
If x ∼ y, then for any z ∈ Ex , x ∼ z, so transitivity implies y ∼ z, which means z ∈ Ey , so
we have that Ex ⊆ Ey . The reverse containment is given by a similar argument, so Ex = Ey .
On the other hand, if x 6∼ y, then no element of Ex can be an element of Ey (and vice versa),
so we see that Ex ∩ Ey = ∅.
2. Exercise 9.8.
(a) Show that the two expressions for µ in Example 6.5 are equivalent.
S
Proof. Suppose, first, that A ⊂ ni=1 (ai , bi ). Let f be a step function which is equal to
1 on [ai , bi ] for each i, and zero everywhere else. Then f ≥ χA and
Z
1
f=
0
(bi − ai ).
i=1
Therefore, we see that
( n
) Z
n
X
[
(bi − ai ) : A ⊂
(ai , bi ) ⊆
i=1
n
X
1
f : f a step function and f ≥ χA on [0, 1] .
0
i=1
Hence,
Z
inf
1
( n
)
n
[
X
f : f a step function and f ≥ χA on [0, 1] ≤ inf
(bi − ai ) : A ⊂
(ai , bi ) .
0
i=1
1
i=1
On the other hand, suppose f is a step function on [0, 1] such that f ≥ χA . Let
[a1 , b1 ], . . . , [an , bn ] be the intervals on which f ≥ 1. Then
A⊆
n
[
[ai , bi ] ⊆
i=1
n
[
(ai − 1/m, bi + 1/m)
i=1
for any positive integer m. Therefore,
xm :=
n h
X
i=1
n
n
i X
2n X
+
(bi − ai )
(bi + 1/m) − (ai − 1/m) =
[bi − ai + 2/m] =
m
i=1
i=1
Pn
S
P
forms a sequence in { ni=1 (bi − ai ) : A ⊂ ni=1 (ai , bi )} which converges to
i=1 (bi −
R1
ai ) ≤ 0 f . Therefore,
( n
)
Z 1
n
X
[
inf
(bi − ai ) : A ⊂
f.
(ai , bi ) ≤ lim xm ≤
i=1
m→∞
i=1
0
Since the choice of f was
left is less than or
nR arbitrary, we see that the infimum on the o
1
equal to all elements of 0 f : f a step function and f ≥ χA on [0, 1] , and so we know
that
( n
)
Z 1
n
X
[
inf
f : f a step function and f ≥ χA on [0, 1] .
(bi − ai ) : A ⊂
(ai , bi ) ≤ inf
i=1
0
i=1
Having shown inequalities both ways, we can conclude that
( n
)
Z 1
n
X
[
inf
f : f a step function and f ≥ χA on [0, 1] = inf
(bi − ai ) : A ⊂
(ai , bi ) ,
0
i=1
i=1
as desired.
(b) Show that “Jordan content” is consistent with lengths of intervals.
Proof. For any interval I in [0, 1] with endpoints a and b (i.e. (a, b), [a, b), (a, b], or [a, b],
and even including the possibility that a = b), we know that I ⊂ (a − 1/m, b + 1/m) for
all positive integers m. Hence,
µ(I) ≤ lim (b − a + 2/m) = b − a.
m→∞
On the other hand, (a, b) ⊆ I, so every collection of open intervals containing I must
contain b − a and, hence, have length ≥ b − a, so µ(I) ≥ b − a.
Thus, since the inequality goes both ways, we see that µ(I) = b − a, and so the Jordan
content is indeed consistent with lengths of intervals.
(c) Show that µ(Q ∩ [0, 1]) = 1 and µ([0, 1]\Q) = 1.
2
S
Proof. Suppose Q ∩ [0, 1] ⊆ ni=1 (ai , bi ) with the intervals (ai , bi ) mutually disjoint and
arranged so that bi ≤ ai+1 for all i. Since every interval contains a rational number (by
the density of Q in R), it cannot be the case that bi < ai+1 ; hence, we have bi = ai+1
for all i. Therefore,
n
[
[0, 1]\ (ai , bi )
i=1
Pn
is a finite collection of points, and so i=1 (bi − ai ) = 1. Since our choice of finite cover
of Q ∩ [0, 1] by open intervals was arbitrary, we conclude that µ(Q ∩ [0, 1]) = 1.
On the other hand, the irrationals are also dense, so the same argument shows that
any cover of [0, 1]\Q can also only miss finitely many points, and so µ([0, 1]\Q) = 1 as
well.
(d) Prove or disprove each of the defining properties of measure for Jordan content.
Proof. The first three are all obviously true of the Jordan content. Since (c) implies that
the Jordan content cannot be a measure (else it would violate Proposition 8.3), we know
that the Jordan content must not satisfy countable additivity. In fact, it doesn’t even
satisfy finite additivity, since
µ (Q ∩ [0, 1]) ∪ ([0, 1]\Q) = µ([0, 1]) = 1 < 2 = 1 + 1 = µ(Q ∩ [0, 1]) + µ([0, 1]\Q).
3. Exercise 9.27. If m∗ (A) + m∗ (B) = m∗ (A) + m∗ (B), prove that A is measurable and B is
measurable.
Proof. If A is not measurable, then m∗ (A) 6= m∗ (A). From Lemma 8.7 we know that m∗ (A) ≤
m∗ (A), so it must be the case that m∗ (A) < m∗ (A). Therefore,
m∗ (A) + m∗ (B) < m∗ (A) + m∗ (B)
(1)
since m∗ (B) ≤ m∗ (B), again by Lemma 8.7. Therefore, if A is not measurable, then the
equality in (1) does not hold. Equivalently, the contrapositive says that if (1) does hold, then
A must be measurable.
The same argument shows that (1) implies that B is measurable, so we conclude that (1)
implies that both A and B are measurable.
4. Let S be the set of all intersections of Q with arbitrary closed, open, and half-open subintervals
of [0, 1], including degenerate closed subintervals consisting of a single point, and finite and
countable unions thereof. Define µ : S → R by
µ(Acd ) = d − c,
where Acd is the intersection of Q with any of the intervals (c, d), [c, d], [c, d), or (c, d]. Show
that µ is not countably additive and, hence, not a measure.
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Proof. It is straightforward to see that µ is not countably additive. By definition, for any
x ∈ Q ∩ [0, 1],
{x} = Q ∩ [x, x] = Axx ,
so
µ({x}) = µ(Axx ) = x − x = 0.
Let {r1 , r2 , . . .} be an enumeration of Q ∩ [0, 1]. Then
Q=
∞
[
{ri } =
i=1
∞
[
Ar i r i .
i=1
Now, on one hand
µ(Q ∩ [0, 1]) = µ(A01 ) = 1 − 0 = 1,
but on the other hand
∞
X
µ({ri }) = 0.
i=1
Therefore, we have a situation where A01 =
S∞
µ(A01 ) >
i=1 Ari ri ,
∞
X
but
µ(Ari ri ),
i=1
so µ is not countably additive.
5. (a) Let S be a collection of subsets of [0, 1] that is closed under countable unions. Suppose
µ : S → R is a set function satisfying the first three conditions for being a measure. In
addition, suppose µ is
S
P
i. finitely additive (meaning µ(A) = ni=1 µ(Ai ) for A = ni=1 Ai with Ai ∈ S and
Ai ∩ Aj = ∅ whenever i 6= j); and
P
S∞
ii. countably sub-additive (meaning µ(A) ≤ ∞
i=1 µ(Ai ) for A =
i=1 Ai with Ai ∈ S
and Ai ∩ Aj = ∅ whenever i 6= j).
Prove that µ is a measure on S.
Proof. Since we already know µ satisfies the first three conditions for being a measure, we
only need to verify that µ is countably additive. Suppose µ is not countably additive.
Then
S∞ there exists a countable collection of disjoint sets {A1 , A2 , . . .} such that A =
i=1 Ai and
∞
X
µ(A) <
µ(Ai ).
i=1
Therefore, µ(A) is not an upper bound on the sequence of partial sums
Since this is an increasing sequence, there exists some N such that
µ(A) <
N
X
i=1
4
µ(Ai ).
Pn
i=1 µ(Ai ).
But then, using the finite additivity of µ, we know that
!
N
N
X
[
µ(A) <
µ(Ai ) = µ
Ai ≤ µ
i=1
i=1
∞
[
!
Ai
,
(2)
i=1
SN
where
S∞ the last inequality is due to the monotonicity of µ and the fact that i=1 Ai ⊂
i=1 Ai .
Now, notice that (2) says that µ(A) < µ(A), which is a contradiction. From this contradiction, then, we can conclude that our assumption that µ was not countably additive
was untenable, so we’ve proved that µ is countably additive.
(b) Explain why Exercise 9.29 is an easy corollary of part (a).
Answer. The outer measure m∗ satisfies the first three conditions of being a measure,
and Lemma 8.6 says that m∗ is countably subadditive. If m∗ were finitely additive, then
(a) would imply that m∗ is countably additive, which we’ve seen is not true. Therefore,
we know that m∗ is not finitely additive, which is the result requested in Exercise 9.29.
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