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Let $\tau: E \to E$ be a measure-preserving transformation of the measure space $(E, \mathcal{E}, \mu)$, i.e. $\mu(\tau^{-1}(A)) = \mu(A)$ for all $A \in \mathcal{E}$. Let $\mathcal{E}_\tau = \{ A \in \mathcal{E} : \tau^{-1}(A) = A \}$. In my lecture notes, it is claimed that a measurable function $f$ is invariant (i.e. $f \circ \tau = f$) if and only if it is measurable with respect to $\mathcal{E}_\tau$.

This is evidently false: Let $E = \{ 0, 1 \}$, and let $\mathcal{E} = \mathcal{P}(E)$ be the power set. Let $\mu$ be the uniform distribution. Let $\tau$ act on $E$ by transposing 0 and 1. Let $f: (E, \mathcal{E}) \to (E, \mathcal{E}_\tau)$ be the identity map on $E$. Clearly, $f$ is measurable and not invariant. Yet it's also measurable with respect to $\mathcal{E}_\tau$: the preimage of every measurable set in its codomain is in $\mathcal{E}_\tau$, by construction.

Have I misinterpreted the claim? If not, is there a similar claim which is true, e.g. by fixing the codomain of $f$?

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Unless I've misinterpreted what you've written, that $f$ you give is invariant. Surely any bijection on a finite set preserves the uniform distribution on that set. – Brad Dec 29 '10 at 4:25
@Brad: Dear Brad, I think that you have misinterpreted what is written. The identity function is not invariant under composition with $\tau$, since $\tau$ is not the identity. Regards, – Matt E Dec 29 '10 at 4:52
up vote 3 down vote accepted

I suspect that measurable function in the statement should be understood to be real-valued measurable function, or something similar.

More generally, if each point $y$ in the codomain $Y$ of $f$ is measurable, and if $f$ is measuarble w.r.t. $\mathcal E_{\tau}$, then we see that $f^{-1}(y)$ should be invariant under $\tau$, which is to say that $f\circ\tau = f$.

The problem with your counter-example is that the points of $E$ are not in your particular $\mathcal E_{\tau}$, and so the preceding argument breaks down (as you implicitly noted!).

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That does work, although I was hoping for something more interesting than requiring that singletons be measurable in the codomain! (For instance, could we obtain a weaker result where $f \circ \tau = f$ almost-everywhere with weaker conditions on the codomain?) – Zhen Lin Dec 29 '10 at 8:02

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