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It is well known that if $T$ is a nonsingular transformation of a standard probability space $(X,\mu)$ and there exists an ergodic measure preserving transformation $S$ of $(X,\mu)$ such that $T$ commutes with $S$, then $T$ is measure preserving.

I would like to know of there is an approximate version of this fact. More explicitly, let $T$ be a nonsingular transformation of $(X,\mu)$ and suppose that for some small $\epsilon$ there exists an ergodic nonsingular tranformation $S$ which commutes with $T$ and such that there is a set $A \subseteq X$ with $\mu(A) > 1-\epsilon$ and $\frac{\mathrm{d}S \mu}{\mathrm{d}\mu}(x) = 1$ for all $x \in A$. Does this imply that $T$ is close to being measure preserving in some sense, for example do we have that the quantity $||\frac{\mathrm{d}T \mu}{\mathrm{d}\mu} - 1||_1$ or the quantity $- \int_X \log \frac{\mathrm{d}T \mu}{\mathrm{d}\mu}(x) \hspace{2 pt} \mathrm{d}\mu(x)$ is small?

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    $\begingroup$ I didn't know the result you're quoting. Do you have a reference? $\endgroup$
    – Anthony Quas
    Jan 28, 2017 at 15:58
  • $\begingroup$ @AnthonyQuas: I don't think this is right. What if $T$ is a shift on $\{ 0, 1 \}^{\mathbb Z}$, $S=T^2$, and $\mu$ is supported by the sequences that are $0$ at the odd digits (and is the usual product measure on the other half)? $\endgroup$
    – Christian Remling
    Jan 28, 2017 at 17:15
  • $\begingroup$ @ChristianRemling: Then the map $T$ is singular (I had already thought about cases like this: another nice example is $\mu$ is a $(\frac 13,\frac 23)$ Bernoulli measure and $T$ swaps 0's and 1's; of course $T$ is singular in that case too). $\endgroup$
    – Anthony Quas
    Jan 28, 2017 at 17:52
  • $\begingroup$ @AnthonyQuas: Well, that demonstrates that I don't know what "singular" means here. $\endgroup$
    – Christian Remling
    Jan 28, 2017 at 18:01
  • $\begingroup$ Singular means $\mu(A)=0$ implies $\mu(T^{-1}A)=0$. $\endgroup$
    – Anthony Quas
    Jan 28, 2017 at 18:10

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No. I don't think there can be any result of that type. Let $X=\{0,1\}^{\mathbb N}$ and let $S$ be the dyadic odometer (I like to write elements of $X$ so they are infinite on the left; in this case, the map $S$ is just add 1 with carry to the left -- the way it should be!). Equip $X$ with the measure that is the product measure of $(\frac 12,\frac 12)$ on each coordinate except the $N$th; and $(\epsilon,1-\epsilon)$ on the $N$th coordinate. Let $T$ be $S^{2^N}$. Then $S$ is measure preserving everywhere except on a set of measure $2^{-N}$: the cylinder set $[111111111]$, while $T$ is very far from measure-preserving everywhere.

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  • $\begingroup$ Yes, this makes intuitive sense. I was mostly concerned with the situation where $S$ and $T$ generate a free action of $\mathbb{Z}^2$, but upon further reflection I doubt the desired statement holds even in that case. $\endgroup$
    – burtonpeterj
    Jan 28, 2017 at 22:50
  • $\begingroup$ Right. It's not hard to make $T$ be a different rotation in the odometer group that has the same property, but where the action of $S$ and $T$ is free. $\endgroup$
    – Anthony Quas
    Jan 28, 2017 at 23:05

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