7/27/2019 Borel Semi
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More Borel -algebras
We defined B(R) to be the -algebra of subsets of R generated by
thecollection of all open intervals
(a, b) : < a < b < },
Similarly we can define the Borel -algebra B(R2) as the -algebra
generatedby the family of sets
(a, b) (c, d) : a < b and c < d
.
The collection B(R2) contains every subset of the plane that we
meet in day today life ( but not every subset belongs to it: there
are weird sets that dont.)
Reminder about notation: ifA and B are two sets then AB, their
cartesianproduct, is the set of all ordered pairs (x, y) with x A
and y B. There is apotential clash of notation here- ifx and y are
real numbers then the ordered pair(x, y) R2 must not be confused
with the open interval {z R : x < z < y}.In the above
definition of B(R2) the set (a, b) (c, d) is the rectangle
{(x, y) R2 : a < x < b and c < y < d}.
Exercise 0.1 Identify the following subset of R2,
A =
qQ
(, q) (, q),
and deduce that the triangle {(x, y) R2 : x > 1, y > 1, x
+ y < 0} belongsto the Borel -algebra. Generalize your argument
to show that the interior ofany convex polygon belongs to the Borel
-algebra.
A =
qQ
(, q) (, q) = {(x, y) R2 : x + y < 0}.
Each rectangle (, q) (, q) belongs to B(R2) and Q is countable
soA B(R2) since -algebras are closed under countable unions. Taking
theintersection of A with the two Borel sets {(x, y) R2 : x > 1}
and {(x, y) R2 : y > 1} we see that the triangle also belongs to
B(R2).
To generalize this we note that the interior of convex polygons
are obtained asthe intersection of finitely many half-plane of the
form {(x, y) R2 : x + y 0
[a, b] (, ) : a b}
The first two, collections generate B(R2); the third does not.
To show that(C1) = (C2) it suffices to verify that each member of
C1 belongs to (C2) and
every member of C2 belongs to (C1). For then, using the fact
that (Ci) is thesmallest -algebra containing Ci, we may deduce both
that (C1) (C2) andthat (C2) (C1).
Now any open rectangle(a, b)(c, d) is the countable union of
closed rectangles[a, b] [c, d] with a, b, c, d rational numbers
satisfying a < a < b < b andc < c < d < d. Also
any closed rectangle [a, b] [c, d] is the countableintersection of
open rectangles (a, b) (c, d) with a, b, c, d rational
numberssatisfying a < a b < b and c < c d < d. This
establishes that thecollection of closed rectangles generates the
same-algebra as the collection ofopen rectangles.
In a similar way we assert that any open disc is a countable
union of openrectangles contained in it, and that any open
rectangle is a countable union of
open discs contained within it: show that any open rectangle A
can be writtenas
n1
qAQ
B(q,1
n).
where B(q, 1n
) = {z R2 : z q < 1n
}. Thus the collection of open discs,and that of open rectangles
generate the same -algebra.
Let F =
A (, ) : A B(R)
. This is a -algebra (CHECK IT!) andit contains every member of
the third collection, but it does not contain everymember of B(R2)
( for example it does not contain (0, 1) (0, 1)).
Figure 1 illustrates the construction of a subset A of R2 known
as Sier-pinskis carpet. To construct it we start with the
square
A0 = {(x, y) R2 : 0 x, y < 1}
and cut out (the interior of ) the central square. This leaves
us with the set
A1 = A0 \ {(x, y) R2 : 1/3 x, y < 2/3}.
The next step involves cutting out 8 smaller squares leaving a
set A2. We repeatthis process of cutting out smaller squares
indefinitely: the carpet A is then
2
