MHB A question on ergodic theory: topological mixing and invariant measures

[Alex V]

Hi All,

This is a question on ergodic theory - not quite analysis, but as close as you can get to it, so I decided to post it here.

Suppose I have a compact metric space $X$, with $([0,1], B, \mu)$ a probability space, with $B$ a (Borel) sigma algebra, and $\mu$ the probability measure. Suppose also that $\mu(A) > 0$ for any nonempty set $A \subset X$. If we take a measure invariant transformation $T:X->X$ and assume it is NOT topologically mixing, how do we show it CANNOT be mixing with respect to $\mu$?

This is how I would attempt it. Take two sets nonempty open sets$A$ and $B$ in $X$.

Since we know that T is NOT topologically mixing, there are infinitely many natural numbers $n \in \mathbb{N}$ such that $T^{n}(A) \cap B = \emptyset$.

The preimage of $T^{n}(A)$ is $T^{n-1}(A)$. By measure preservation, we have $\mu (T^{n-1}(A)) = \mu(T^{n}(A))$. By repeated argument, we eventually have $\mu (T^{-n}(A)) = \mu(A)$.

Now we have $T^{-n}(A) \cap B = \emptyset$ for infinitely many $n$. Hence $\mu(T^{-n}(A) \cap B) = \emptyset$

This contradicts the requirement that for mixing, we need $\mu(T^{-n}(A) \cap B) = \mu(A)\mu(B)$ in the limit of $n$ tending to infinity, as the our assumption was that the measure of any nonempty set is bigger than zero.

Proof complete.

Is this correct, or are there any gaps or errors in my logic? If it is faulty, I'd be grateful to see the correct version. Thanks!

 

[Opalg]

Hi All,

This is a question on ergodic theory - not quite analysis, but as close as you can get to it, so I decided to post it here.

Suppose I have a compact metric space $X$, with $([0,1], B, \mu)$ a probability space, with $B$ a (Borel) sigma algebra, and $\mu$ the probability measure. Suppose also that $\mu(A) > 0$ for any nonempty set $A \subset X$. If we take a measure invariant transformation $T:X->X$ and assume it is NOT topologically mixing, how do we show it CANNOT be mixing with respect to $\mu$?

This is how I would attempt it. Take two sets nonempty open sets $\color{red}{A}$ and $\color{red}{B}$ in $\color{red}{X}$.

Since we know that T is NOT topologically mixing, there are infinitely many natural numbers $\color{red}{n \in \mathbb{N}}$ such that $\color{red}{T^{n}(A) \cap B = \emptyset}$.

The preimage of $T^{n}(A)$ is $T^{n-1}(A)$. By measure preservation, we have $\mu (T^{n-1}(A)) = \mu(T^{n}(A))$. By repeated argument, we eventually have $\mu (T^{-n}(A)) = \mu(A)$.

Now we have $T^{-n}(A) \cap B = \emptyset$ for infinitely many $n$. Hence $\mu(T^{-n}(A) \cap B) = \emptyset$

This contradicts the requirement that for mixing, we need $\mu(T^{-n}(A) \cap B) = \mu(A)\mu(B)$ in the limit of $n$ tending to infinity, as the our assumption was that the measure of any nonempty set is bigger than zero.

Proof complete.

Is this correct, or are there any gaps or errors in my logic? If it is faulty, I'd be grateful to see the correct version. Thanks!

Hi Alex, and welcome to MHB!

Your proof looks good. The only thing that needs adjusting is the part I have highlighted in red. The definition of topological mixing is that for all pairs of open sets $A$ and $B$, $T^{n}(A) \cap B \ne \emptyset$ for all sufficiently large $n$. If you want to negate a "for all" statement, the negation must always be a "there exists" statement. In this case, the definition of NOT topologically mixing is that there exist open sets $A$ and $B$ such that $T^{n}(A) \cap B = \emptyset$ for infinitely many $n$. In other words, you can't randomly choose $A$ and $B$ before mentioning the definition of not topologically mixing. The existence of such sets is actually included in that definition.