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edited Mar 2 2011 at 15:55 |
...as soon as you have a non-standard number, you get a non-measurable set.
Every nonstandard natural number $N$ gives rise to a nonprincipal ultrafilter $U$ on $\mathbb{N}$, by saying that a set $X\subset\mathbb{N}$ is in $U$ if and only if $N\in X^*$, the nonstandard analogue of $X$. In other words, the ultrafilter says $X$ is large if it expresses a property that the nonstandard number $N$ has. We may regard $U$ as a subset of $2^\mathbb{N}$, which carries a natural probability measure. But a nonprincipal ultrafilter cannot be measurable there, since the full bit-flipping operation, which is measure-preserving, carries $U$ exactly to its complement, so $U$ would have to have measure $\frac12$, but $U$ is invariant by the operation of flipping any finite number of bits, and so must have measure $0$ or $1$ by Kolmogorov's zero-one law. (See also this article by Blackwell and Diaconis proving the same fact.)
...every set you can name is measurable.
Another way of saying that a set is easily described is to say that it lies low in the descriptive set-theoretic hierarchy, and the lowest such sets are necessarily measurable. For example, every set in the Borel hierarchy is measurable, and the Borel context is often described as the domain of explicit mathematics. Under stronger set-theoretic axioms, such as large cardinals or PD, the phenomenon rises to higher levels of complexity, for under these hypotheses it follows even that all sets in the projective hierarchy are Lebesgue measurable. This would include any set that you can define by quantifying over the reals and the integers and using any of the basic mathematical operations.
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edited Mar 2 2011 at 15:32 |
...as soon as you have a non-standard number, you get a non-measurable set.
Every nonstandard natural number $N$ gives rise to a nonprincipal ultrafilter $U$ on $\mathbb{N}$, by saying that a set $X\subset\mathbb{N}$ is in $U$ if and only if $N\in X^*$, the nonstandard analogue of $X$. In other words, the ultrafilter says $X$ is large if it expresses a property that the nonstandard number $N$ has. We may regard $U$ as a subset of $2^\mathbb{N}$, which carries a natural probability measure. But a nonprincipal ultrafilter cannot be measurable there, since the full bit-flipping operation, which is measure-preserving, carries $U$ exactly to its complement, so $U$ would have to have measure $\frac12$, but $U$ is invariant for by the operation of flipping only finitely many any finite number of bits, which is ergodic, and so the measure must be $0$ or $1$. .
...every set you can name is measurable.
Another way of saying that a set is easily described is to say that it lies low in the descriptive set-theoretic hierarchy, and the lowest such sets are necessarily measurable. For example, every set in the Borel hierarchy is measurable, and the Borel context is often described as the domain of explicit mathematics. Under stronger set-theoretic axioms, such as large cardinals or PD, the phenomenon rises to higher levels of complexity, for under these hypotheses it follows even that all sets in the projective hierarchy are Lebesgue measurable. This would include any set that you can define by quantifying over the reals and the integers and using any of the basic mathematical operations.
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edited Mar 2 2011 at 15:24 |
...as soon as you have a non-standard number, you get a non-measurable set.
Every nonstandard natural number $N$ gives rise to a nonprincipal ultrafilter $U$ on $\mathbb{N}$, by saying that a set $X\subset\mathbb{N}$ is in $U$ if and only if $N\in X^*$, the nonstandard analogue of $X$. In other words, the ultrafilter says $X$ is large if it expresses a property that the nonstandard number $N$ has. We may regard $U$ as a subset of $2^\mathbb{N}$, which carries a natural probability measure. But a nonprincipal ultrafilter cannot be measurable there, since the full bit-flipping operation, which is measure-preserving, carries $U$ exactly to its complement, so $U$ would have to have measure $\frac12$, but $U$ is invariant for the operation of flipping only finitely many bits, which is ergodic, and so the measure must be $0$ or $1$.
...every set you can name is measurable.
Another way of saying that a set is easily described is to say that it lies low in the descriptive set-theoretic hierarchy, and the lowest such sets are necessarily measurable. For example, every set in the Borel hierarchy is measurable, and the Borel context is often described as the domain of explicit mathematics. Under stronger set-theoretic axioms, such as large cardinals or PD, the phenomenon rises to higher levels of complexity, for under these hypotheses it follows even that all sets in the projective hierarchy are Lebesgue measurable. This would include any set that you can define by quantifying over the reals and the integers and using any of the basic mathematical operations.
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edited Mar 2 2011 at 12:36 |
...as soon as you have a non-standard number, you get a non-measurable set.
Every nonstandard natural number $N$ gives rise to a nonprincipal ultrafilter $U$ on $\mathbb{N}$, by saying that a set $X\subset\mathbb{N}$ is in $U$ if and only if $n\in N\in X^*$, the nonstandard analogue of $X$. In other words, the ultrafilter says $X$ is large if it expresses a property that the nonstandard number $N$ has. We may regard $U$ as a subset of $2^\mathbb{N}$, which carries a natural probability measure. But a nonprincipal ultrafilter cannot be measurable there, since the bit-flipping operation, which is measure-preserving, carries $U$ exactly to its complement, so $U$ would have to have measure $\frac12$, but the operation is ergodic, so the measure must be $0$ or $1$.
...every set you can name is measurable.
Another way of saying that a set is easily described is to say that it lies low in the descriptive set-theoretic hierarchy, and the lowest such sets are necessarily measurable. For example, every set in the Borel hierarchy is measurable, and the Borel context is often described as the domain of explicit mathematics. Under stronger set-theoretic axioms, such as large cardinals or PD, the phenomenon rises to higher levels of complexity, for under these hypotheses it follows even that all sets in the projective hierarchy are Lebesgue measurable. This would include any set that you can define by quantifying over the reals and the integers and using any of the basic mathematical operations.
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edited Mar 2 2011 at 11:59 |
...as soon as you have a non-standard number, you get a non-measurable set.
Every nonstandard natural number $N$ gives rise to a nonprincipal ultrafilter $U$ on $\mathbb{N}$, by saying that a set $X\subset\mathbb{N}$ is in $U$ if and only if $n\in X^*$, the nonstandard analogue of $X$. In other words, the ultrafilter says $X$ is large if it expresses a property that the nonstandard number $N$ has. We may regard $U$ as a subset of $2^\mathbb{N}$, a which carries a natural probability measure. But a nonprincipal ultrafilter cannot be measurable there, since the bit-flipping operation, which is measure-preserving, carries $U$ exactly to it's its complement, so $U$ woud would have to have measure $\frac12$, but the operation is ergodic, so the measure must be $0$ or $1$.
...every set you can name is measurable.
Another way of saying that a set is easily described is to say that it lies low in the descriptive set-theoretic hierarchy, and the lowest such sets are necessarily measurable. For example, every set in the Borel hierarchy is measurable, and the Borel context is often described as the domain of explicit mathematics. Under stronger set-theoretic axioms, such as large cardinals or $AD^{L(\mathbb{R})}$, we know PD, the phenomenon rises to higher levels of complexity, for under these hypotheses it follows even that all sets in the projective hierarchy are Lebesgue measurable. This would include any set that you can define by quantifying over the reals and the integers and using any of the basic mathematical operations.
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answered Mar 2 2011 at 11:51 |
...as soon as you have a non-standard number, you get a non-measurable set.
Every nonstandard natural number $N$ gives rise to a nonprincipal ultrafilter $U$ on $\mathbb{N}$, by saying that a set $X\subset\mathbb{N}$ is in $U$ if and only if $n\in X^*$, the nonstandard analogue of $X$. In other words, the ultrafilter says $X$ is large if it expresses a property that the nonstandard number $N$ has. We may regard $U$ as a subset of $2^\mathbb{N}$, a which carries a natural probability measure. But a nonprincipal ultrafilter cannot be measurable there, since the bit-flipping operation, which is measure-preserving, carries $U$ exactly to it's complement, so $U$ woud have to have measure $\frac12$, but the operation is ergodic, so the measure must be $0$ or $1$.
...every set you can name is measurable.
Another way of saying that a set is easily described is to say that it lies low in the descriptive set-theoretic hierarchy, and the lowest such sets are necessarily measurable. For example, every set in the Borel hierarchy is measurable. Under stronger set-theoretic axioms, such as large cardinals or $AD^{L(\mathbb{R})}$, we know that all sets in the projective hierarchy are Lebesgue measurable. This would include any set that you can define by quantifying over the reals and the integers and using any of the basic mathematical operations.
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