- #1
Alex V
- 1
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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!
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!