221 Analysis 2, 2008–09
Suggested solutions to exercise sheet 2
1. Show that if m1 and m2 are measures on a ring A and λ1 , λ2 ∈ [0, ∞] then
m(A) = λ1 m1 (A) + λ2 m2 (A) defines a measure m on A.
Solution. Since m1 (A), m2 (A) ∈ [0, ∞] and λ1 , λ2 ∈ [0, ∞] we have m(A) ∈
[0, ∞] for every A ∈ A. So m : A → [0, ∞].
For every λ ∈ [0, ∞], we have λ · 0 = 0. So m(∅) = λ1 m(∅) + λ2 m(∅) =
λ1 · 0 + λ2 · 0 = 0 + 0 = 0.
Suppose that A1 , A2 , A3 , . . . is a sequence of disjoint sets in A and
P∞that A =
S
∞
A
∈
A.
Since
m
and
m
are
measures,
we
have
m
(A)
=
1
2
1
i=1 m1 (Ai )
i=1 i
P∞
and m2 (A) = i=1 m2 (Ai ), so
m(A) = λ1
∞
X
i=1
m1 (Ai )+λ2
∞
X
∗
m2 (Ai ) =
i=1
∞
X
i=1
λ1 m1 (Ai )+λ2 m2 (Ai ) =
∞
X
m(Ai ).
i=1
[You should check carefully that the equality marked ∗ is valid in all cases,
regardless of whether λ1 , λ2 or any of the series takes the value ∞.] So
m(∅) = 0 and m is countably additive; so m is a measure.
2. (a) Show that N × N is countable by finding a bijection N → N × N. [Hint:
draw a picture and think about the sets Sk = {(i, k + 1 − i) : i = 1, 2, . . . , k}
for k = 1, 2, 3, . . . ]
(b) If X is a set of the form X = {xij : i ∈ N, j ∈ N}, show that X can be
written as X = {y1, y2 , y3 , . . . }.
Solution. (a) The sets S1 , S2 , S3 , . . . form a partition of N × N (this is
most easily seen by drawing a picture). Consider the function f : N → N × N
defined recursively by
(
(i + 1, j − 1)
if f (n) = (i, j) and j > 1,
f (1) = (1, 1),
f (n + 1) =
(1, i + 1)
if f (n) = (i, 1).
If f (n) = f (m) for some n ≤ m then for some k we have f (n) ∈ Sk , so for
some i we have (i, k + 1 − i) = f (n) = f (m) = (i + (m − n), k + 1 − (m − n)),
so m = n and f is one-to-one. As n runs through N, f (n) exhausts each of
the sets S1 , S2 , . . . in turn, so f is onto. [More formally, it is not too hard
to show that f (|S1 | + · · · + |Sk−1| + i) = (i, k + 1 − i), so every element of
S
k≥1 Sk = N × N is in the image of f .] So f is a bijection N → N × N; so
N × N is countable.
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(b) Let us write x(i,j) = xij for i, j ∈ N. Let f be a bijection N → N × N and
let yn = xf (n) for n ∈ N. We have
X = {x(i,j) : (i, j) ∈ N × N} = {xf (n) : n ∈ N} = {y1 , y2 , y3, . . . }.
3. Suppose that A is a ring of sets, m is a measure on A and that BS
1 , B2 , . . . , Bn
are m-measurable sets with Bi ∩ Bj = ∅ for i 6= j. If Cn = nj=1 Bj then
show that for any Y ⊆ Cn ,
∗
m (Y ) =
n
X
m∗ (Y ∩ Bj ).
j=1
Solution. For n = 1 we have m∗ (Y ) = m∗ (Y ∩ B1 ) + m∗ (Y \ B1 ) =
m∗ (Y ∩ B1 ) + m∗ (Y \ C1 ) = m∗ (Y ∩ B1 ) + m∗ (∅) = m∗ (Y ∩ B1 ).
Suppose inductively that the statement holds for n − 1 sets. Since Y ⊆ Cn
and B1 , . . . , Bn are disjoint, we have Y \ Bn ⊆ Cn \ Bn = Cn−1 , so
m∗ (Y ) = m∗ (Y ∩ Bn ) + m∗ (Y \ Bn )
= m∗ (Y ∩ Bn ) +
n−1
X
(since Bn is measurable)
m∗ ((Y \ Bn ) ∩ Bj ) (by our inductive assumption)
j=1
= m∗ (Y ∩ Bn ) +
n−1
X
m∗ (Y ∩ Bj )
(since B1 , . . . , Bn are disjoint)
j=1
=
n
X
m∗ (Y ∩ Bj ).
j=1
By induction, the statement holds for all n ≥ 1.
4. Show that if A is a σ-algebra of subsets of a set X, then
∞
\
Ai ∈ A whenever A1 , A2 , A3 , . . . is a sequence of sets in A.
i=1
Solution. Let A1 , A2 , A3 , . . . be a sequence of sets in A. For each i ≥ 1 we
have A∁i ∈ A since A is closed under
complements. Since A is closed under
S
∁
A
countable unions, we have also ∞
i=1 i ∈ A. Taking the complement of this
set and applying De Morgan’s laws gives
[
∞
∞
\
∁ ∁
Ai
=
Ai ∈ A.
i=1
i=1
5. Let A be a σ-algebra of subsets of R containing the interval ring I. Prove
that:
(a) {x} is in A for any x ∈ R;
(b) every subinterval of R is in A.
2
Solution. (a) For any b ∈ R and n ≥ 1 we have (b− n1 , b] ∈ I, so (b− n1 , b] ∈
A. By the previous exercise,
{b} =
∞
\
(b − n1 , b] ∈ A.
n=1
(b) There are several cases to consider. Let a, b ∈ R.
(i) Finite intervals: (a, b] ∈ I, so (a, b] ∈ A. Now [a, b] = (a, b] ∪ {a} ∈ A
since (a, b] ∈ A and {a} ∈ A, and (a, b) = (a, b]\ {b} ∈ A since (a, b] ∈ A and
{b} ∈ A, and A is closed under set differences. So [a, b) = {a} ∪ (a, b) ∈ A.
S
(ii) Half-lines: (a, ∞) = ∞
n=1 (a, n] ∈ A since (a, n] ∈ A for each n ≥ 1 and
A is closed under countable unions. So [a, ∞) = {a} ∪ (a, ∞) ∈ A; taking
complements of these sets gives (−∞, a], (−∞, a) ∈ A.
(iii) ∅ ∈ A so R = ∅∁ ∈ A.
6. Explain what is wrong with the following argument that every subset of R
is in M, the collection of Lebesgue measurable subsets of R:
We know that M is a σ-algebra containing I, so from the previous exercise,
{x} ∈ M for any x ∈ R. For any set B ⊆ R,
[
B=
{x},
x∈B
so B is a union of sets in M. Since M forms a σ-algebra, a union of sets
in M must be in M; so B ∈ M.
For which sets B ⊆ R is this argument valid?
Solution. Although M does form a σ-algebra, this implies that it is closed
under countable unions, not arbitrary unions. The argument given will only
show that B ∈ M if B is countable.
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