world:measure
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world:measure [2023/10/07 12:32] rdouc [Proof] |
world:measure [2023/10/07 12:32] (current) rdouc [Proof] |
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| There exists no probability measure P on P(\rset)such that P{x}=0 for each x∈R. | There exists no probability measure P on P(\rset)such that P{x}=0 for each x∈R. | ||
| - | ======= Proof ======= | + | ===== Proof ===== |
| - | The proof of this second impossibility theorem requires the well-ordering principle (equivalent to the axiom of choice) and also the continuum hypothesis. Let S be the set of sequences (s(1),s(2),…) of positive integers. Then S has the power of the continuum. (Let the nth partial sum of a sequence in S be the position of the nth 1 in the nonterminating dyadic representation of a point in (0,1]; this gives a one-to-one correspondence.) By the continuum hypothesis, the elements of S can be put in a one-to-one correspondence with the set of ordinals preceding the first uncountable ordinal. Carrying the well-ordering of these ordinals over to S by means of the correspondence gives S a well-ordering relation <w with the property that each element has only countably many predecessors. | + | The proof of this impossibility theorem requires the well-ordering principle (equivalent to the axiom of choice) and also the continuum hypothesis. Let S be the set of sequences (s(1),s(2),…) of positive integers. Then S has the power of the continuum. (Let the nth partial sum of a sequence in S be the position of the nth 1 in the nonterminating dyadic representation of a point in (0,1]; this gives a one-to-one correspondence.) By the continuum hypothesis, the elements of S can be put in a one-to-one correspondence with the set of ordinals preceding the first uncountable ordinal. Carrying the well-ordering of these ordinals over to S by means of the correspondence gives S a well-ordering relation <w with the property that each element has only countably many predecessors. |
| For s,t∈S, write s<t if s(i)<t(i) for all i>1. Say that t rejects s if t<ws and s<t; this is a transitive relation. Let T be the set of unrejected elements of S. Let Vs be the set of elements that reject s, and assume it is nonempty. If t is the first element (with respect to <w) of Vs, then t∈T (if t′ rejects t, then it also rejects s, and since t<wt′, there is a contradiction). Therefore, if s is rejected at all, it is rejected by an element of T. | For s,t∈S, write s<t if s(i)<t(i) for all i>1. Say that t rejects s if t<ws and s<t; this is a transitive relation. Let T be the set of unrejected elements of S. Let Vs be the set of elements that reject s, and assume it is nonempty. If t is the first element (with respect to <w) of Vs, then t∈T (if t′ rejects t, then it also rejects s, and since t<wt′, there is a contradiction). Therefore, if s is rejected at all, it is rejected by an element of T. | ||
world/measure.1696674721.txt.gz · Last modified: 2023/10/07 12:32 by rdouc
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