Download Exercise 4.16 Show that the class F of subsets A of R such that A or

Exercise 4.16 Show that the class F of subsets A of R such that A or Ac is discrete
(finite or countably infinite) is a σ-field. Define on F a set function P by P(A) = 0 if
A is discrete and P(A) = 1 if Ac is discrete. Show that P so defined is a probability
measure.
Proof. Firstly, recall that the set of real numbers R is uncountable, so if A ⊂ R is
discrete, then Ac is uncountable and vice versa. Secondly, the union of countably many
countable sets is countable.
Take F as defined above. We want to show that F is a σ-algebra, so we need to
show that
(i) ∅ ∈ F
(ii) if A1 , A2 , · · · ∈ F then
S∞
∈F
i=1 Ai
(iii) if A ∈ F then Ac ∈ F
By definition, the empty set is finite so condition (i) holds trivially. For the second
condition, recall that the union of countably many countable sets is countable. Hence, if
A1 , A2 , . . . are all countable so is their union, so ∪∞
i=1 Ai ∈ F. If at least one of the sets
is uncountable, then the union is uncountable so its complement is discrete. In either
c
case, ∪∞
i=1 Ai ∈ F. For the third condition, if A ∈ F then either A is discrete or A is
discrete. If A is discrete, then Ac is uncountable. But the complement of Ac is A, which
is discrete so Ac ∈ F. The other ways follows by a similar argument. This proves that
F is a σ-algebra.
Next, let P(A) = 0 if A is discrete and P(A) = 1 is Ac is discrete, where A ∈ F.
To show that P is a probability measure on (Ω, F) we need to show that
(i) P(∅) = 0 and P(Ω) = 1
(ii) if A1 , A2 , . . . is a collection of pairwise disjoint members of F, so that Ai ∩ Aj = ∅
for all pairs i, j satisfying i 6= j, then
P
∞
[
∞
X
Ai =
P (Ai ).
i=1
i=1
For the first condition, the empty set is finite and Ω = R is uncountable, so P(∅) = 0
and P(Ω) = 1. For the second condition, note that if A1 , A2 , . . . are all countable than
both sides are equal to zero so the equality holds. Assume that Ak is uncountable for
some k (i.e. Ack is finite). Since the sets are pairwise disjoint it follows that Ai ⊆ Ack for
all i 6= k. But every subset of a countable set is also countable, so Ai is countable for
all i 6= k. Hence,
1=P
∞
[
Ai
i=1
= P (Ak ) +
| {z }
=1
| {z }
uncountable
Hence, P is a probability measure on (Ω,F).
1
X
i6=k
P (Ai ) = 1
| {z }
=0