Possible Proof of the Lusin-Purves Theorem

From: Acid Pooh (poopdeville_at_gmail.com)
Date: 10/27/04 

Date: 27 Oct 2004 04:28:18 -0700

The Lusin-Purves theorem states that X,Y are standard Borel spaces and
B is a Borel set in XxY such that each cross-section B_x is countable,
then the projection of B onto X is a Borel set. I've seen several
proofs of this fact, but I think I might have come up with a novel
(though elementary) one. This sounds kind of crankish, especially
since I don't seem to use the countability of each section in any
significant way. Constructive criticism/counter-examples (but please,
be gentle. I haven't slept in a few days)

Pf/ Since X and Y are Standard, XxY is Standard. We may assume that
B is Borel since we can enlarge the topology on XxY so that B is
closed and open. Thus B is Polish. The Cantor-Bendixson theorem
implies that B can be written as B = O union P, where O is countable,
P perfect, and O intersect P empty. Let a be any point in p and a_n
any sequence in P converging to a. Since the projection mapping pi_X
is continuous, if a_n -> a, pi_X(a_n) -> pi_X(a). Since every point
in P is a limit point, pi_X(P) is closed, hence Borel.

Consider the family of sets which satisfy the condition B_x intersect
P is empty. Since each of these sets is disjoint from P and
countable, it follows that it is contained in O (in fact, the union of
such B_x's is O). Since O is countable, there are only countably many
x's in X such that B_x intersect P is empty. Thus the projection
pi_X({(x,y) in B: B_x intersect P is empty}) is Borel.

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Any help salvaging this would be appreciated.
'cid 'ooh