On the space $S=\{ 0,1,\ldots,m \}^{\mathbb{N}}$ for some $m\in \mathbb{Z}_{+}.$ And given a probability $\mu$ on it. Is it true that $\mu$ is fully supported if and only if it is ergodic for the action of the shift?
1
-
1$\begingroup$ It seems from this and your previous post that you have some wrong idea in mind of the notion of ergodic. To give a really simple example (the simplest of Vaughn's examples below), imagine a measure supported at the point 0000...... That is $\mu(A)$ is 1 if $A$ contains $000\ldots$ and 0 otherwise. It's not hard to check that $\mu$ is invariant: $A$ contains $0\ldots$, if and only if $S^{-1}A$ contains $0\ldots$. It's also immediate that $\mu$ is ergodic. You have to show that every invariant set has measure 0 or 1. -- But every set has measure 0 or 1. $\endgroup$– Anthony QuasCommented Feb 12, 2013 at 22:34
Add a comment
|
1 Answer
$\begingroup$
$\endgroup$
No, there are many fully-supported non-ergodic measures. Just take a convex combination of a fully-supported measure and anything else. (Recall that an invariant measure is ergodic if and only if it cannot be writen as a non-trivial convex combination of two distinct invariant measures.)
Also there are many ergodic measures that are not fully supported. For example, given any periodic sequence consider the atomic measure that gives equal weight to each of the shifts of this sequence. This measure is ergodic but its support is a finite set.
answered Feb 12, 2013 at 21:10