2. PROBABILITY SPACES
2.1 Probability functions
Whenever probabilities cannot reasonably be assigned to all
subsets of the sample space, it is important to distinguish
between events and other subsets of the sample space.
The probability function P:F[0,1] assigns probabilities
to certain subsets (called events) of the sample space .
Not all subsets of  need to be events. If A is not in the
domain F of the probability function P, then P(A) is
undefined and A is not an event.
Experiment: A die is tossed twice.
Possible outcomes: (1,1),(1,2),…,(1,6), (2,1),(2,2),…,(2,6),
(3,1),(3,2),…,(3,6), (4,1),(4,2),…,(4,6),
(5,1),(5,2),…,(5,6), (6,1),(6,2),…,(6,6)
Sample space: ={(1,1),(1,2),…,(6,5),(6,6)}
Simple events: {(1,1)},{(1,2)},…,{(6,5)},{(6,6)}
Simple event probabilities: P({(1,1)})=…=P({(6,6})= 361
In this example, probabilities can easily be assigned to all
subsets of the sample space just by summing simple event
probabilities, hence all subsets are events. For example,
P({(4,5),(4,6),(5,6)})=P({(4,5)})+P({(4,6)})+P({(5,6)})
= 361 + 361 + 361 = 363 = 121 .
2.1
2.2 Uncountably infinite sample spaces
An uncountably infinite sample space  (e.g., an interval)
has not only simple subsets (e.g., subintervals) but also more
complicated ones, many of which are so odd that it is
absolutely impossible to assign probabilities to them in a
reasonable manner (see Appendix A).
Example of a probability function P that is defined on all
semiclosed subintervals of =(0,1]:
P((a,b])=b-a
The probability of an interval (a,b] is given by its length.
The domain of P is F={(a,b]:0ab1}.
Clearly, we should not call any function P a probability
function unless it meets some obvious requirements. In our
example, there seems to be no problem in that respect.
Indeed,
and
P()=P((0,1])=1-0=1, P()=P((a,a])=a-a=0,
P((0,b])+P((0,b]c)=(b-0)+P((b,1])=b+(1-b)=1.
Moreover, if a1a2b1b2, then
P((a1,b1](a2,b2])=P((a1,b2])
=b2-a1
=(b1-a1)+(b2-a2)-(b1-a2)
=P((a1,b1])+P((a2,b2])-P((a2,b1])
=P((a1,b1])+P((a2,b2])-P((a1,b1](a2,b2])).
2.2
2.3 Extending the domain of a probability function
Let P be the probability function defined by P((a,b])=b-a for
all semiclosed subintervals of =(0,1]. The domain of P is
given by F={(a,b]:0ab1}.
If AF and BF, then ABF, but Ac, Bc, and AB need
not necessarily be elements of F.
A=(0.1,0.3], B=(0.5,0.7])  AB=F,
Ac=(0,0.1](0.3,1]F,
Bc=(0,0.5](0.7,1]F,
AB=(0.1,0.3](0.5,0.7]F
A=(0.1,0.6], B=(0.5,1])  AB=(0.5,0.6]F,
Ac=(0,0.1](0.6,1]F,
Bc=(0,0.5]F,
AB=(0.1,1]F
Despite the fact that some complements and unions of
elements of F are not in F, we could easily assign
probabilities to them by putting
P(Ac)=1-P(A)
and
P(AB)=P(A)P(B)-P(AB),
respectively. It seems therefore reasonable to increase the
domain of P by those subsets of  to which we can assign
probabilities in a consistent manner.
2.3
2.4 -algebras
Henceforth we will only consider probability functions
whose domains are closed under the operations of taking
complements, countable intersections and countable unions.
A set F of subsets of a nonempty set  is called a -algebra
(or -field) if it satisfies
,F,
AF  AcF,
and A1,A2,…F  A1A2…F  A1A2…F.
 Exercise 2.a: Let ={1,2,3}. Examine whether or not
and
F1={,{1},{2},{3},{1,2},{1,3},{2,3},},
F2={,},
F3={,{1},{2,3},},
F4={,{1},{2},{2,3},{1,3},} are -algebras.
F1, F2, and F3 are -algebras, but F4 is not a -algebra,
because {1}{2}={1,2}F4 and {1,3}{2,3}={3} F4.
The smallest -algebra of subsets of the real line =(-,)
that contains all intervals is called the Borel -algebra
and its elements are called Borel sets.
2.4
2.5 Probability distributions
A function P:F[0,1] that is defined on a -algebra F is
called a probability distribution (or probability measure)
if it satisfies
(i) P()=0, P()=1,
(ii) P(Ac)=1-P(A) if AF,
and (iii) P(A1A2...)=P(A1)+P(A2)+…
if A1,A2,… are pairwise disjoint elements of F.
The triple (,F,P) is called a probability space.
 Exercise 2.b: Let ={1,2,3} and F={,{1},{2,3},}.
Examine whether or not the functions given by
and
P1()=0, P1({1})=0, P1({2,3})=1, P1()=1,
P2()=0, P2({1})= 87 , P2({2,3})=- 17 , P2()=1,
P3()=0, P3({1})= 15 , P3({2,3})= 52 , P3()=1,
P4()=0, P4({1})= 23 , P4({2,3})= 13 , P4()=1,
P5()=0, P5({1})=0, P5({2,3})=0, P5()=1,
respectively, are probability distributions.
Obviously, only P1 and P4 are probability distributions.
2.5
2.6 Further exercises
 Exercise 2.c: Let P be the probability function defined
by P((a,b])=b-a for all semiclosed subintervals of =(0,1].
Define both P((a,b)) and P([a,a]) in a consistent manner.
1
1
1
P((a,b))=P((a,b- 101 ](b- 101 ,b- 100
](b- 100
,b- 1000
]…)
1
1
1
=P((a,b- 101 ])+P((b- 101 ,b- 100
])+P((b- 100
,b- 1000
])+…
1
1
1
=((b- 101 )-a)+((b- 100
)-(b- 101 ))+((b- 1000
)-(b- 100
))+…
1
1
1
=b-a- 101 +( 101 - 100
)+( 100
- 1000
)+…
=b-a
P([a,a])=1-P([a,a]c)
=1-P((0,a)(a,1])
=1-(P((0,a))+P((a,1]))
=1-((a-0)+(1-a))
=0
 Exercise 2.d: Let ={0,1,2,3}. Examine whether or not
F={,{0},{3},{1,2},{0,1,2},{1,2,3},{0,3},}
is a -algebra.
 Exercise 2.e: Find the smallest -algebra of subsets of
={4,6,8,10,12,…} that contains {4,6} and {4,6,8}.
2.6