Ergodic Induced Transformations

[Juanriq]

Homework Statement

Salutations, all. I'm trying to show that if $T_A$ is ergodic, then so is $T$. This was an iff, and I have the other inplication. I'm a little lost with how to proceed, so any help would be appreciated!

Homework Equations

$T(X, \mathcal{B}, m)$ to itself is an invertible measure preserving transformation of a probabilitiy space. Also, $T_A: (A, \mathcal{B}|A, m|a )$ to itself is the induced transformation.

The Attempt at a Solution

I'm not sure if I'm on the right track, but $T_A$ is the set of minimum iteration which bring us back to $A$. Also, if it is ergodic, it can't be decomposed into sets of measure 0 or 1. So if we iterate ourselves further $r_A$ (the minimum distance to get back to A) then we are going to still have an ergodic set. Since $A \subset \mathcal{B}$, then $T$ must be ergodic as well since we can't have an ergodic and a non-ergodic subset.



Does this make any sense? Thanks in advance!

 

[mighty2000]

Thank you for your post. Your approach seems to be on the right track. To show that T is ergodic, we can use the definition of ergodicity: for any measurable set B in the probability space, if T(B) = B, then the measure of B must be either 0 or 1.

Since T_A is ergodic, it means that for any measurable set B in A, if T_A(B) = B, then the measure of B must be either 0 or 1. Now, since A is a subset of \mathcal{B}, we can say that for any measurable set B in \mathcal{B}, if T_A(B) = B, then the measure of B must be either 0 or 1.

But since T_A is induced by T, it follows that for any measurable set B in \mathcal{B}, if T(B) = B, then the measure of B must be either 0 or 1. This shows that T is also ergodic, as any set B in \mathcal{B} that satisfies T(B) = B must also satisfy T_A(B) = B.

I hope this helps. Let me know if you have any further questions. Good luck with your work!