```Date: Dec 30, 2012 3:17 AM
Author: fom
Subject: How WM is cheating - fat Cantor set measure

To see that the measure associated withfat Cantor sets can be attributed tofintism in the final analysis, onemust look at the infinite productof measure spaces.Just as with the Tychonoff topologyon infinite product topologies,infinite products of measure spaceshave a local finiteness constraint.That is, since the measures ofmeasurable sets in the product willbe obtained by infinitary products,m_oo(E)=Pi_n(m_n(E_n)) for n=0 to n=oothe measure on the domain of eachcomponent measure space must beunity.  That is, if(X_i,S_i,m_i)is a component measure space, thenm_i(X_i)=1The local finiteness condition assertsthat measurable sets in the producthave measurable components satisfying-(m_n(E_n)=1)for only finitely many of its components.a measureable set in the product.  Thus,for measurable sets in the product,0<=m_oo(E)=Pi_n(m_n(E_n))<=1 for n=0 to n=ooFor present purposes, letX_0={0,1}S_0={{0,1}, {1}, {0}, null}m_0({0,1})=1m_0({1})=1/2m_0({0})=1/2m_0(null)=0and, for i>0 let(X_i,S_i,m_i)=(X_0,S_0,m_0)and call the infinite product(X_oo,S_oo,m_oo)Now, each point of the Cartesianproduct X_ooX=(x_0,x_1,...)is a countably infinite sequence of 1'sand 0's.  Each finite initial segmentcorresponds with a measurable set --that is, a cylinder of points agreeingon the first n coordinates.  So, for aninitial segment of length n,m_oo(x|n)= (1/2)^nIt should be clear that one obtainsm_oo(x)=lim_n(m_oo(x|n))=0 as n->0so that denying a completed infinityis equivalent to assigning non-zeromeasures to the points of the product.=================To see that the fat Cantor set measureis unrelated to the Lebesgue measureon the interval 0<=p<1, observe firstthat the measure on the set of pointsfor which only finitely many of thecoordinates is different from one, iszero.That is, for each such point, m_oo(x)=0.There are only countably many finitesubsets and measures are countablyadditive set functions.  So, onecan form the set consisting of theconstant sequence,x=(1,1,1,...)and those sequences that differ fromx at finitely many indices.  This sethas m_oo=0.Next, take the points in the complementof this set.  Identify each of thesepoints with the sum of coordinates,p(x)=Sigma_n(x_n/2^(n+1))  for n=0 to n=ooand make the usual identification betweeneventually constant sequences correspondingto rational numbers.Then, for each A such thatA={p(x)|0<=a<=p(x)<b<=1}A is Lebesgue measurable and the Lebesguemeasure ism_L(A)=(b-a)=================To see that fat Cantor set measure,with respect to foundational considerationsarises from finitism, one must lookat the structure of Lebesgue measurablesets in relation to the Borelhierarchy.Unlike many other measures, Lebesguemeasure has an invariance propertythat permits its product measuresto be defined without the generaltheory of product measures.  Tosee why, consider the binaryexpansions on the interval0<=y<1taking the eventually constantsequences ending in constant 0as the representation for rationalnumbers.  Lety=(y_0, y_1, y_2, y_3, y_4, y_5, y_6, y_7, y_8, y_9, y_10, y_11, ...)be one such number. If one nowwritesx_1=(y_0, y_3, y_6, y_9, ...)x_2=(y_1, y_4, y_7, y_10, ...)x_3=(y_2, y_5, y_8, y_11, ...)one obtains either three eventuallyconstant sequences or three sequencesthat never become eventually constant.Moreover, by the invariance of theLesbegue measure in relation to thisprocess, the transformation of aLebesgue measurable set yields threeLebesgue measurable sets having thesame measure under the product.The transformation to an infinitaryproduct is done with the usualdiagonal strategy,x_1=(y_0, y_1, y_5, y_6, y_14, ...)x_2=(y_2, y_4, y_7, y_13, ...)x_3=(y_3, y_8, y_12, ...)x_4=(y_9, y_11, ...)x_5=(y_10, ...)x_6=(...)so that each Lebesgue measurable setcorresponds to an infinitely countablecollection of Lebesgue measurable setswhose union has the same measure asthe original.Now, the measurable spaces overwhich Lebesgue measures are definedare those spaces whose sigma-algebrais generated by the open sets of thetopology and whose atoms correspondwith singletons.  That is, Lebesguemeasures are defined with respectto the Borel hierarchy.Each Lebesgue measurable set differsfrom a Borel set by a set of measurezero.  So, in fact, it is the Borelsets that are responsible for theinvariance demonstrated by theLebesgue measure in this transformation.But, one does not capture theindividuation of the Borelsigma-algebra without invokingcompleted infinities.  Consequently,as is seen with the non-zero measureof the fat Cantor sets, "set of measurezero" has no meaning without thecompleted infinity invoked at eachstage of definition for the Borelhierarchy.Thus, the claim that the non-zeromeasure of fat Cantor sets arisesfrom relation with an atomicmeasurable space whose atoms arenot singletons has been verified.
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