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Question: Could somebody give an example of a nontrivial atomless measure space without measure preserving isomorphisms (except for the identity)?

Background: A measure preserving isomorphism on a measure space $(X,\Sigma,\mu)$ is a bijection $\phi$ such that

$$\forall A\in\Sigma:\mu(\phi^{-1}(A))=\mu(\phi(A))=\mu(A)$$

Edit: By 'except for the identity' I obviously meant to exclude all $\phi$ such that $\mu(A\triangle\phi(A))=0$ for all $A\in\Sigma$.

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    $\begingroup$ I would guess that the answer is that there is no such space, although I don't have a proof. Note that if one-point sets are measurable (with measure zero), then a bijection that simply switches two such points is measure-preserving. More generally, if any two disjoint sets of measure zero have the same cardinality, then switching them is a measure-preserving transformation. Self-bijections of a set of measure zero are also measure preserving as long as they send measurable subsets to measurable subsets. $\endgroup$ – Jim Belk Jun 1 '16 at 19:26
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    $\begingroup$ I think Jim's comment really shows that you want to identify maps that agree a.e. to (perhaps) get an interesting question. $\endgroup$ – user138530 Jun 1 '16 at 19:50
  • $\begingroup$ You might also want to exclude the trivial measure $\mu(A)=0$ while you're making clarifications; obviously, if we take that, every map $\phi$ has $\mu(A\Delta \phi(A))=0$. $\endgroup$ – Milo Brandt Jun 2 '16 at 1:02
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    $\begingroup$ @JimBelk I'm not sure if there exist such spaces that admit an atomless measure, but there exist measurable spaces that do not admit any bimeasurable bijection onto themselves but the identity, see Section 3.2 of Borel Spaces II by Rao and Shortt. $\endgroup$ – Michael Greinecker Jun 2 '16 at 13:43
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Didn't know the answer to the original version, but only because of stupid/boring examples. Here's an example for the modified version: An atomless measure space such that if $\phi$ is any measurable bijection with a measurable inverse, measure-preserving or not, then $\phi(A)=A$ for every measurable set $A$.

Let $<$ be a well-ordering of $[0,1]$ (so in particular $<$ has nothing to do with the standard order). By transfinite recursion we can "construct" a strictly increasing map $\kappa$ from $[0,1]$ to the class of infinite cardinals, for example by taking $\kappa(x_0)=\aleph_0$ if $x_0$ is the smallest element of $[0,1]$ and $$\kappa(x)=2^{\bigcup_{x'<x}\kappa(x')}\quad(x>x_0).$$Let $(S_x)_{x\in[0,1]}$ be a pairwise disjoint collection of sets with $$|S_x|=\kappa(x).$$Set $$X=\bigcup_{x\in[0,1]}S_x.$$For $E\subset[0,1]$ let $$A_E=\bigcup_{x\in E}S_x.$$ Say $A\subset X$ is measurable if and only if $A=A_E$ for some Lebesgue-measurable $E\subset[0,1]$, and define $$\mu(A_E)=m(E),$$where $m$ is Lebesgue measure. No atoms ($S_x$ has no non-trivial measurable subsets but it's not an atom because it has measure zero).

Say $\phi:X\to X$ is a measurable bijection with a measurable inverse. Then induction on $x$ shows that $\phi(S_x)=S_x$. First, since $\phi(S_x)$ is measurable it must be a union of $S_y$ for some set of $y$'s. But $$|S_y|>|S_x|\quad(y>x),$$hence $$\phi(S_x)\subset\bigcup_{y\le x}S_y.$$By induction we have $\phi(S_y)=S_y$ for all $y<x$. Hence $\phi(S_x)\subset S_x$. Similarly $\phi^{-1}(S_x)\subset S_x$, so $\phi(S_x)=S_x$.

And hence $\phi(A)=A$ for every measurable set $A$.


Not an answer to the original version because there do exist plenty of measure-preserving isomorphisms other than the identity, in fact such that $\phi(x)\ne x$ for every $x$.

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