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--- world:measure [2023/10/07 12:28]
rdouc
+++ world:measure [2023/10/07 12:32] (current)
rdouc [Proof]
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-{{Pages>:defs}}
+{{page>:defs}}
-This result is taken from Billiingsley, // Probability and measure //.
+This result is taken from Billiingsley, // Probability and measure //, 3rd edition, page 46.
 ====== Statement ======
-There exists no probability measure  $P$  such that $P(x) = 0 $for each$ x \in \mathbb{R}$. Since $A(x) = 0 $, this implies that it is impossible to extend$ A $to$ \mathbb{R}$ at all.
+There exists no probability measure $P $on$ \mathcal{P}(\rset) $such that$ P\{x\} = 0 $for each$ x \in \mathbb{R}$.
+===== Proof =====
-====== Proof ======
+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  $(s(1), s(2), \ldots)$  of positive integers. Then  $S$  has the power of the continuum. (Let the  $n$ th partial sum of a sequence in  $S$  be the position of the  $n$ th  $1$  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 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  $(s(1), s(2), \ldots)$  of positive integers. Then  $S$  has the power of the continuum. (Let the  $n$ th partial sum of a sequence in  $S$  be the position of the  $n$ th  $1$  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  $s, t \in S$ , write  $s < t$  if  $s(i) < t(i)$  for all  $i > 1$ . Say that  $t$  rejects  $s$  if  $t <_w s$  and  $s < t$ ; this is a transitive relation. Let  $T$  be the set of unrejected elements of  $S$ . Let  $V_s$  be the set of elements that reject  $s$ , and assume it is nonempty. If  $t$  is the first element (with respect to  $<_w$ ) of  $V_s$ , then  $t \in T$  (if  $t'$  rejects  $t$ , then it also rejects  $s$ , and since  $t <_w t'$ , there is a contradiction). Therefore, if  $s$  is rejected at all, it is rejected by an element of  $T$ .

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