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 $\mathcal{P}(\rset)$such that $P\{x\} = 0$ for each $x \in \mathbb{R}$. | 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 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. | + | 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. |
| 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 $T$. | 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 $T$. | ||
world/measure.1696674721.txt.gz · Last modified: 2023/10/07 12:32 by rdouc
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