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Suppose, $X$ is some measurable space, $\mu$ is a probability measure on it, could anyone explain to me how the following is a map or transformation on $\mathcal M_1(X)$ (space of all probability measure on $X$)? Also, in this case, what is $T$ on Suppose, $X$ is some measurable space, $\mu$ is a probability measure on it, could anyone explain to me how the following is a map or transformation on $\mathcal M_1(X)$ (space of all probability measure on $X$)?

$T(\mu)(A)=\mu(T^{-1}(A)$? now my doubt is $T$ acting as a function on $\mathcal M_1(X) $, which one is varying here as an input? $A$ or $\mu$? or both? where $A\in B(X)$, $B(X)$ is all Borel sets of $X$. Thanks! ?

Also, what is the transformation $T$ on $\mathcal M_1(X)$ , in this case? $T(A)=\mu(T^{-1}(A)$?

enter image description here

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$T$ is assumed to be a measurable transformation in the sense $T^{-1}(A)$ is a measurable set whenever $A$ is a measurable set. Given any probability measure $\mu$ you define a new probability measure $\nu$ by $\nu(A) =\mu (T^{-1}(A))$. The induced transformation is defined by $T_*(\mu)=\nu$. It maps probability measures to probability measures.

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  • $\begingroup$ now from this definition of $\nu$ what is the definition of invariant probability measures? $\nu(A) =\mu (T^{-1}(A))=\mu(A)\forall A$, something like this? $\endgroup$ – miosaki Dec 3 at 0:19
  • $\begingroup$ @miosaki Exactly. Invariance means $\mu =\nu$. $\endgroup$ – Kabo Murphy Dec 3 at 0:21
  • $\begingroup$ I stil dont understand the role of $T_*$ in the discussion above screenshot. $\endgroup$ – miosaki Dec 3 at 0:25
  • $\begingroup$ did you mean $T_*$ as an induced transformation? $\endgroup$ – miosaki Dec 3 at 0:34
  • $\begingroup$ @miosaki By an abuse of notation they are writing $T$ for $T_{*}$ also. $\endgroup$ – Kabo Murphy Dec 3 at 0:36

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