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

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    Figure 1: Showing the sets from the first four steps in the construction of Sierpinskiscarpet.

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    defined to be the subset of points belonging to the original square A0 which are

    never removed at any step of the construction Believe us-there really are pointsthat are left! (In fact A contains uncountably many points.).

    Exercise 0.3 Explain why the carpet A belongs to the collection of subsetsB(R2). Calculate the area of the sets An where An is the set constructed bystep n of the algorithim. What do you think the area of A is?

    The original square belongs to the Borel -algebra and so does each squarewhich is removed at each stage of the construction. Thus the Borel -algebra,by virtue of being closed under finite set operations contains each An. Now wecan write

    A =

    n=1

    An

    from which we deduce that A also belongs to B(R2

    ) using the closure undercountable intersections property. Observe that in constructing An+1 from Anwe remove 1/9 of the area of An. Since the area of A0 is 1 we deduce thatthe area of An = (8/9)

    n. Because A is a subset of An for each n, the onlyreasonable value to assign to be the area of A is 0.

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