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I'm reading Probability, Random Processes & Ergodic properties by Gray.

$(\Omega, \mathcal F, P, T)$ is a random dynamical system where $T$ is a $\mathcal F$-measurable map (i.e. $T^{-1}(F)$ is measurable for every $F \in \mathcal F$).

An invariant event is defined as any event $F$ such that $T^{-1}(F)=F$.

Now my question is: does the above imply also $T(F)=F$ for any invariant event $F$ ?

Note that there is not an explicit constraint on $T$ to be onto (surjective).

Thank you.

Carlo C
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2 Answers2

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Now my question is: does the above imply also $T(F)=F$ for an invariant event $F$ ?

In general no: for a measurable selfmap $T: \Omega \to \Omega$, it is not necessarily true that the image $T(F)$ of a measurable subset $F$ is measurable. (See e.g. the discussion at measurable functions. Why defined like this? for more on this).

On the other hand, if $T$ is a bimeasurable selfmap, so that there is a selfmap $T^{-1}$ that is also measurable and $T\circ T^{-1} = \text{id}_{\Omega}$, then the property becomes true. Often even when $T$ is not bimeasurable one can consider image sets as measurable by way of e.g. the Lusin-Souslin Theorem (see e.g. Measurable invariance of domain) or passing to the natural invertible extension (see e.g. Proving a.e surjectivity of suggested factor map in Natural extension of Standard Borel dynamical system or Exr.2.1.7 and 2.1.8 on p.20 of Einsiedler & Ward's Ergodic Theory).

Alp Uzman
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  • I don't understand the following claim there p.142: "Since the limiting sample averages of all indicator functions of invariant events (in particular) are constants with probability one". Why is true in general ? – Carlo C Nov 12 '23 at 17:59
  • @CarloC As the sentence preceding the claim you are citing alludes to, that's due to an ergodicity assumption. – Alp Uzman Nov 12 '23 at 18:13
  • Since $T^{-1}(F)=F$, any point $\omega \in F$ will be mapped by $T$ in $F$. Any other point $\omega$ in $F^c$ will be mapped by $T$ in $F^c$ where the event $F$ indicator function is zero. Then the sample average of the invariant event $F$'s indicator function will be 1 or 0 with probability one. – Carlo C Nov 12 '23 at 20:19
  • I think the following part "the indicator functions themselves are constant with probability one" is due actually to the ergodicity assumption, since by this the limiting sample average of invariant event's indicator function (that's always the indicator function itself) is assumed to be a constant number with probability one. Does it make sense? – Carlo C Nov 12 '23 at 21:52
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I think the following is independent from the ergodicity assumption:

Since by definition $T^{-1}(F)=F$, any point $\omega \in F$ will be mapped by $T$ in $F$. Any other point $\omega$ in $F^c$ will be mapped by $T$ in $F^c$ where the event $F$'s indicator function is zero. Therefore the sample average of the invariant event $F$'s indicator function will be 1 or 0 with probability one (it is the indicator function itself).

Carlo C
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