Re: question: Lebesgue measure
From: Stephen J. Herschkorn (sjherschko_at_netscape.net)
Date: 12/30/04
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Date: Thu, 30 Dec 2004 17:48:36 -0500
tetrahedron wrote:
>I was wondering how many nonmeasurable sets exist (e.g. as subsets of
>the reals). Since AC is needed to prove their existence, is it
>possible to make any statement about their number at all?
>
>Second, I don't understand why the Cantor set is uncountable. It's not
>even dense. Why am I wrong?
>
>
There are beth_2 (= cardinality of P(R)) nonmeasuarable sets:
Clearly, there at most beth_2 such sets. Let X be a nonmeasuarable
set with cardinality c (= beth_1); the typically constructed example
will do. Let A = {Y in P(X): Y is measurable}. Then the map Y |->
X\Y is an injection from A to P(X) \ A. This implies that |P(X) \
A| = |P(X)|.
The Cantor set C = {sum(k=1..infty,b_k 3^-k) : b_k = 0 or 2 for each
k.} Set up a bijection to C --> [0,1] by replacing b_k 3^-k with
(b_k / 2) 2^-k.
-- Stephen J. Herschkorn sjherschko@netscape.net
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