221 Analysis 2, 2008–09
Suggested solutions to exercise sheet 1
1. Show that if A is a ring of subsets of a set X, then A ∩ B ∈ A whenever
A ∈ A and B ∈ A.
Solution. We have A ∩ B = (A ∪ B) \ (B \ A) ∪ (A \ B) (to see this, draw
a Venn diagram). Since A is closed under \ and ∪, we have A ∩ B ∈ A.
Alternatively, observe that A∩B = A\(A\B); this shows that closure under
the set difference is enough to ensure that A is closed under intersections.
2. (a) Suppose that X and I are sets, and that for every i ∈ I, we are given a
ring Ai of subsets of X. Prove that
\
A=
Ai
i∈I
is a ring of subsets of X.
(b) If S ⊆ P(X), let r(S) be given by
\
r(S) = {A : A is a ring of subsets of X and S ⊆ A}.
Prove that
(i) r(S) is a ring containing S, and
(ii) if B is any ring of subsets with S ⊆ B, then r(S) ⊆ B.
[This shows that r(S) is the smallest ring containing S; we say that r(S) the
ring generated by S.]
(c) Show that if X = R and S = {(a, b] : a, b ∈ R} then r(S) is the interval
ring I. [You may assume that I is a ring of subsets of R; this is proven in
an exercise below.]
Solution. (a) Since each Ai is a ring, we have ∅ ∈ Ai for all i ∈ I, so
∅ ∈ A.
If A, B ∈ A then A, B ∈ Ai for all i ∈ I. Since each Ai is a ring, we have
A \ B ∈ Ai and A ∪ B ∈ Ai for all i ∈ I. So A \ B ∈ A and A ∪ B ∈ A.
Hence A is a ring.
(b) (i) r(S) is an intersection of rings, so by (a) it is a ring; since each A in
the intersection contains S, so does r(A).
(ii) If B is a ring and S ⊆ B then B participates in the intersection defining
r(S), so r(S) ⊆ B.
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(c) Since S ⊆ I and I is a ring, by property (ii) we have r(S) ⊆ I.
On the other hand, if S ⊆ A where A is any ring of subsets of R, then since A
is closed under finite unions and I is the set of finite unions of elements of S,
we must have I ⊆ A. So by the definition of r(S), it follows that I ⊆ r(S);
the reverse inclusion was established in the last paragraph, so r(S) = I.
3. Let X be a set, let A ⊆ P(X) with ∅ ∈ A, and let m : A → [0, ∞] be a
measure on A. Show that if A, B and A \ B are in A with B ⊆ A,
(a) m(B) ≤ m(A) [we say that m is increasing], and
(b) if m(B) < ∞, then m(A \ B) = m(A) − m(B).
What happens in (b) if m(B) = ∞?
Solution. Since B ⊆ A, we have A = B ∪ (A \ B), a disjoint union. Using
the (countable) additivity of m, we see that
m(A) = m(B ∪ (A \ B)) = m(B) + m(A \ B).
Since m(A \ B) ≥ 0, this shows that m(A) ≥ m(B).
If m(B) < ∞ then we can subtract m(B) from both sides to obtain (b).
If m(B) = ∞ then m(A) = m(B) + m(A \ B) ≥ m(B), so m(A) = ∞ too.
So the expression m(B) − m(A) is undefined in this case.
Also, when m(B) = ∞, the measure of A \ B can take any value in [0, ∞]; for
example, if m is Lebesgue measure on the line then m(R \ R) = m(∅) = 0 and
m(R \ (−∞, 0)) = m[0, ∞) = ∞.
4. (a) Show that for any a, b, c, d ∈ R,
(a, b] ∩ (c, d] = (max(a, c), min(b, d)],
(a, b] ∩ (−∞, d] = (a, min(b, d)] and
(a, b] ∩ (c, ∞) = (max(a, c), b].
[If b ≤ a then (a, b] is the empty set.]
(b) Suppose that B = (a1 , b1 ] ∪ · · · ∪ (an , bn ] where a1 < b1 < a2 < b2 < · · · <
bn−1 < an < bn . Write B ∁, the complement of B in R, as a union of n + 1
intervals.
(c) Show that the interval ring
I = {(a1 , b1 ] ∪ · · · ∪ (an , bn ] : n ≥ 1, ai , bi ∈ R, 1 ≤ i ≤ n}
is a ring of subsets of R. [Hint: A \ B = A ∩ B ∁.]
2
Solution. (a) We have
x ∈ (a, b] ∩ (c, d] ⇐⇒
⇐⇒
⇐⇒
⇐⇒
⇐⇒
x ∈ (a, b] & x ∈ (c, d]
(x > a & x ≤ b) & (x > c & x ≤ d)
(x > a & x > c) & (x ≤ b & x ≤ d)
x > max(a, c) & x ≤ min(b, d)
x ∈ (max(a, c), min(b, d)].
So
(a, b] ∩ (c, d] = (max(a, c), min(b, d)].
We could prove the other two identities in a similar fashion; but observe that
the argument above works just as well if any of a, b, c, d are ±∞. So
(a, b] ∩ (−∞, d] = (max(a, −∞), min(b, d)] = (a, min(b, d)] and
(a, b] ∩ (c, ∞) = (max(a, c), min(b, ∞)] = (max(a, c), b].
(b) A point x is in the complement of B iff x ∈ (bi , ai+1 ] where 1 ≤ i < n, or
x ≤ a1 , or x > bn . So
B ∁ = (−∞, a1 ] ∪ (b1 , a2 ] ∪ (b2 , a3 ] ∪ · · · ∪ (bn−1 , an ] ∪ (bn , ∞).
(c) The empty set ∅ = (0, 0] is in I, and it is clear from the definition of I
that it is closed under finite unions.
Let A, B ∈ A; then we may write
A = A1 ∪ · · · ∪ Am
and B = (a1 , b1 ] ∪ . . . (an , bn ]
where each Ai is of the form (a, b], and B is written as in (b) (since every
element of I can be written in this way). Let D0 = (−∞, a1 ], Dj = (bj , aj+1]
for 1 ≤ j < n and Dn = (bn , ∞) so that B ∁ = D0 ∪ D1 ∪ · · · ∪ Dn . Then
∁
A\B =A∩B =A∩
n
[
Dj =
j=0
=
n [
m
[
j=0
Ai ∩ Dj =
i=1
n
[
(A ∩ Dj )
j=0
n
m
[[
(Ai ∩ Dj ).
j=0 i=1
By part (a), we have Ai ∩ Dj ∈ I for each 1 ≤ i ≤ m and 0 ≤ j ≤ n. Since I
is closed under finite unions, the union A \ B of these (n + 1)m sets is also
in I.
5. Let A = A1 ∪ · · · ∪ An where Ai = (ai , bi ] and a1 < b1 < a2 < · · · < bn−1 <
an < bn , and suppose there are real numbers bj , cj such that A can also be
written as
∞
[
A=
Bj where Bj = (cj , dj ].
j=1
Prove that there is a partition N = S1 ∪S2 ∪· · ·∪Sn such that for i = 1, 2, . . . , n
we have
[
Ai =
Bj .
j∈Si
[We write N = {1, 2, 3, . . . }; a partition is simply a disjoint union.]
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Solution. For i = 1, 2, . . . , n − 1, let Si = {j ∈ N : Bj 6= ∅, Bj ⊆ Ai }, and
let Sn = {j ∈ N : Bj ⊆ An }.
If k 6= ℓ and j ∈ Sk ∩ Sℓ then Bj ⊆ (ak , bk ] ∩ (aℓ , bℓ ] = ∅, so Bj = ∅ which
forces k = ℓ = n, a contradiction. So Sk ∩ Sℓ = ∅ if k 6= ℓ.
If j ∈ N \ (S1 ∪ · · · ∪ Sn ) then ∅ =
6 Bj ⊆ A buy Bj is not entirely contained
in any Si . So there exist k < ℓ and x ∈ Ak ∩ Bj and y ∈ Aℓ ∩ Bj . [Why?]
Since Bj is an interval containing x and y and x < y, we have [x, y] ⊆ Bj .
But x ≤ bk < aℓ < y, so aℓ ∈ [x, y] ⊆ Bj ⊆ A which is false, since aℓ 6∈ A. So
S1 ∪ · · · ∪ Sn = N.
S
Let Ci = j∈Si Bj ; we claim that Ci = Ai . By construction, Bj ⊆ Ai for
j ∈ SSi , so Ci ⊆ Ai . Hence Ci ∩ Ai = Ci and Ck ∩ Ai = ∅ ifSk 6= i. Since
A = j∈N Bj and S1 , . . . , Sn is a partition of N, we have A = nk=1 Ci , so
Ai = A ∩ Ai =
n
[
k=1
Ck ∩ Ai =
n
[
(Ck ∩ Ai ) = Ci .
k=1
4