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Lecture 5: The Borel Sets of R in Statistics 451 (Fall 2013), Lecture notes of Algebra

A part of the lecture notes for statistics 451 (fall 2013) at an unspecified university. In lecture #5, professor michael kozdron introduces the concept of borel sets of r, which is a σ-algebra generated by the open sets. The definition of open and closed sets, the difference between the collection of open sets o and the borel σ-algebra b, and provides a proof of theorem 5.3, which states that the borel σ-algebra is generated by intervals of the form (−∞, a], where a is a rational number. The document also includes exercises to help students understand the concepts.

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Statistics 451 (Fall 2013) September 13, 2013
Prof. Michael Kozdron
Lecture #5: The Borel Sets of R
We will now begin investigating the second of the two claims made at the end of Lecture #3,
namely that there exists a σ-algebra B1of subsets of [0,1] on which it is possible to define
a uniform probability.
Our goal for today will be to define the Borel sets of R. The actual construction of the
uniform probability will be deferred for several lectures.
Recall that a set ERis said to be open if for every xEthere exists some >0
(depending on x)suchthattheinterval(x, x +)iscontainedinE. Also recall that
intervals of the form (a, b)for−∞ <a<b<are open sets. A set FRis said to be
closed if Fcis open. Note that both Rand are simultaneously both open and closed sets.
If we consider the collection Oof all open sets of R, then it follows immediately that Ois not
aσ-algebra of subsets of R.(Thatis,ifAOso that Ais an open set, then, by definition,
Acis closed and so Ac/O.) However, we know that σ(O), the σ-algebra generated by O,
exists and satisfies Oσ(O)2R. This leads to the following definition.
Definition. The Borel σ-algebra of R,writtenB,istheσ-algebra generated by the open
sets. That is, if Odenotes the collection of all open subsets of R,thenB=σ(O).
Since Bis a σ-algebra, we see that it necessarily contains all open sets, all closed sets,
all unions of open sets, all unions of closed sets, all intersections of closed sets, and all
intersections of open sets.
Exercise 5.1. The purpose of this exercise is to is to remind you of some facts about open
and closed sets. Suppose that {E1,E
2,...}is an arbitrary collection of open subsets of R,
and suppose that {F1,F
2,...}is an arbitrary collection of closed subsets of R.Provethat
(a)
j=1 Ejis necessarily open,
(b)
j=1 Ejneed not be open,
(c)
j=1 Fjis necessarily closed, and
(d)
j=1 Fjneed not be closed.
We of t e n c a l l a c o u n t a b l e intersection of op en s e t s a Gδset (from the German Gebeit for
open and Durchschnitt for intersection) and a countable union of closed sets an Fσset (from
the French ferm´e for closed and somme for union).
The following theorem characterizes open subsets of Rand will occasionally be of use.
5–1
pf3

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Statistics 451 (Fall 2013) September 13, 2013 Prof. Michael Kozdron

Lecture #5: The Borel Sets of R

We will now begin investigating the second of the two claims made at the end of Lecture #3, namely that there exists a σ-algebra B 1 of subsets of [0, 1] on which it is possible to define a uniform probability.

Our goal for today will be to define the Borel sets of R. The actual construction of the uniform probability will be deferred for several lectures.

Recall that a set E ⊆ R is said to be open if for every x ∈ E there exists some ￿ > 0 (depending on x) such that the interval (x − ￿, x + ￿) is contained in E. Also recall that intervals of the form (a, b) for −∞ < a < b < ∞ are open sets. A set F ⊆ R is said to be closed if F c^ is open. Note that both R and ∅ are simultaneously both open and closed sets.

If we consider the collection O of all open sets of R, then it follows immediately that O is not a σ-algebra of subsets of R. (That is, if A ∈ O so that A is an open set, then, by definition, Ac^ is closed and so Ac^ ∈/ O.) However, we know that σ(O), the σ-algebra generated by O, exists and satisfies O ￿ σ(O) ⊆ 2 R^. This leads to the following definition.

Definition. The Borel σ-algebra of R, written B, is the σ-algebra generated by the open sets. That is, if O denotes the collection of all open subsets of R, then B = σ(O).

Since B is a σ-algebra, we see that it necessarily contains all open sets, all closed sets, all unions of open sets, all unions of closed sets, all intersections of closed sets, and all intersections of open sets.

Exercise 5.1. The purpose of this exercise is to is to remind you of some facts about open and closed sets. Suppose that {E 1 , E 2 ,.. .} is an arbitrary collection of open subsets of R, and suppose that {F 1 , F 2 ,.. .} is an arbitrary collection of closed subsets of R. Prove that

(a)

j=1 E^ j^ is necessarily open, (b)

j=1 E^ j^ need not be open, (c)

j=1 F^ j^ is necessarily closed, and (d)

j=1 F^ j^ need not be closed.

We often call a countable intersection of open sets a G (^) δ set (from the German Gebeit for open and Durchschnitt for intersection) and a countable union of closed sets an F (^) σ set (from the French ferm´e for closed and somme for union).

The following theorem characterizes open subsets of R and will occasionally be of use.

Theorem 5.2. If E ⊆ R is an open set, then there exist at most countably many disjoint open intervals I (^) j , j = 1, 2 ,.. ., such that

E =

￿^ ∞

j=

I (^) j.

Proof. The trick is to define an equivalence relation on E as follows. If a, b ∈ E, we say that a and b are equivalent, written a ∼ b, if the entire open interval (a, b) is contained in E. This equivalence relationship partitions E into a disjoint union of classes. We do not know a priori that there are at most countably many such classes. Therefore, label these classes I (^) j , j ∈ J, where J is an arbitrary index set. Note that I (^) j is, in fact, an interval for the following reason. If a (^) j , b (^) j ∈ I (^) j , then a (^) j ∼ b (^) j so that the entire open interval (a (^) j , bj ) is contained in I (^) j. As a (^) j , b (^) j were arbitrary we see that I (^) j is, in fact, an interval. The next step is to show that I (^) j is open. Let x ∈ I (^) j be arbitrary. Since x ∈ E and E is open, we know that there exists an ￿ > 0 such that (x − ￿, x + ￿) ⊆ E. However, we clearly have a ∼ x for every a ∈ (x − ￿, x + ￿) which implies that this ￿-neighbourhood of x is contained in I (^) j. Thus, I (^) j is open as required. The final set is to show that there are at most countably many I (^) j. This follows from the fact that each I (^) j must contain at least one rational number. Since there are countably many rationals, there can be at most countably many I (^) j.

It is worth noting that we have not yet proved that B ￿= 2 R^ ; in other words, we have not yet proved that there exist non-Borel sets. It turns out that the set H constructed in Lecture # is non-Borel, although we will not prove this at present. In fact, there is no simple procedure to determine whether a given set A ⊆ R is Borel or not. However, one way to understand B is that it is generated by intervals of the form (−∞, a] as the next theorem shows.

Theorem 5.3. The Borel σ-algebra B is generated by intervals of the form (−∞, a] where a ∈ Q is a rational number.

Proof. Let O 0 denote the collection of all open intervals. Since every open set in R is an at most countable union of open intervals, we must have σ(O 0 ) = B. Let D denote the collection of all intervals of the form (−∞, a], a ∈ Q. Let (a, b) ∈ O 0 for some b > a with a, b ∈ Q. Let

a (^) n = a +

n so that a (^) n ↓ a as n → ∞, and let

b (^) n = b −

n so that b (^) n ↑ b as n → ∞. Thus,

(a, b) =

￿^ ∞

n=

(an , bn ] =

￿^ ∞

n=

{(−∞, b (^) n ] ∩ (−∞, an ] c^ }

which implies that (a, b) ∈ σ(D). That is, O 0 ⊆ σ(D) so that σ(O 0 ) ⊆ σ(D). However, every element of D is a closed set which implies that

σ(D) ⊆ B.