Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Let $\Omega$ be a standard atomless probability space, we can assume $\Omega=(0,1)$ with Lebesgue measure. A bijection $f:\Omega/A_1\to\Omega/A_2$ is almost automorphism, if $P(A_1)=P(A_2)=0$, $f(A)$ is measurable if and only if $A$ is, and $P(f(A))=P(A)$ in this case. An almost automorphism preserves a (real-valued) random variable (r.v.) $X$ if $X(\omega)=X(f(\omega))$ for almost all $\omega$.

The question is: Assume that every almost automorphism preserving $Y$ preserves also $X$. Does this imply that $X$ is $Y$-measurable, that is $X=g(Y)$ a.s. for some $g:R \to R$?

share|improve this question

1 Answer 1

up vote 2 down vote accepted

Fix for convenience $\Omega = [0,1)$ with Lebesgue measure $\mu$. We exhibit a family of continuum-many Borel functions $X_\alpha \colon [0,1) \to \mathbb{R}$, indexed by $\alpha \in (0, 1/2)$, such that

  • Each $X_\alpha$ has only trivial almost automorphisms, in the sense that each almost automorphism $f$ preserving $X_\alpha$ has $\mu$-null support (i.e., the set $\{r : r \neq f(r)\}$ is $\mu$-null).

  • When $\alpha \neq \beta$ we have that $X_\alpha$ is not in your sense $X_\beta$-measurable.

In particular we get a negative answer to the question.

For each $\alpha \in (0,1/2)$ put $I_\alpha = [0,\alpha)$ and $J_\alpha = [0,1) \setminus I_\alpha = [\alpha, 1)$. Let $i_\alpha \colon [0,1) \to [0,1)$ be the obvious fixed-point-free involution flipping $I_\alpha$ and $J_\alpha$. Note that since $\alpha < 1/2$ this involution distorts $\mu$ in the following sense: any $\mu$-positive Borel set $A$ contains a $\mu$-positive Borel set $A' \subseteq A$ such that $\mu(A') \neq \mu(i_\alpha[A'])$. In particular, no restriction of $i_\alpha$ to a set of $\mu$-positive measure can be a $\mu$-preserving map.

Now define $X_\alpha$ by

  • $X_\alpha(r) = r$ if $r \in I_\alpha$,

  • $X_\alpha(r) = i_\alpha(r)$ if $r \in J_\alpha $.

Note that $X_\alpha(r) = X_\alpha(s)$ iff $r = s$ or $r = i_\alpha(s)$. It follows that any almost automorphism preserving $X_\alpha$ is almost everywhere some restriction of $i_\alpha$, but by the previous paragraph it must in fact be almost everywhere the identity. So each $X_\alpha$ has only trivial almost automorphisms.

Finally, suppose that $\alpha < \beta$ in $(0,1/2)$. Put $R = [\alpha, \beta)$. Note that for each $r \in R \subseteq J_\alpha$ we have $i_\beta(r) \in J_\alpha$ and in particular $X_\alpha(r) \neq X_\alpha(i_\beta(r))$. Of course by construction we have $X_\beta(r) = X_\beta(i_\beta(r))$. So if $g \colon \mathbb{R} \to \mathbb{R}$ is any function, for each $r\in R$ we must have at least one of $g(X_\beta(r)) \neq X_\alpha(r)$ or $g(X_\beta(i_\beta(r)) \neq X_\alpha(i_\beta(r))$ and in particular $g \circ X_\beta$ does not agree $\mu$-a.e. with $X_\alpha$. In other words, $X_\alpha$ is not $X_\beta$-measurable.

A similar argument using $i_\alpha$ shows that $X_\beta$ is not $X_\alpha$-measurable.

share|improve this answer
    
Thank you very much! –  Bogdan Apr 9 '13 at 12:34
    
You're welcome! I'll delete the disclaimer as I suppose it is no longer relevant. –  Clinton Conley Apr 10 '13 at 19:10

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.