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Gerald Edgar
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Ian Morris
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Let $X=\{0,1\}^{\mathbb{N}}$ with the infinite product topology (which is metrisable). For each $n \geq 1$, define $x_n$ to be the sequence given by $x_i=0$ for $1 \leq i \leq n$, $x_i=1$ for $n+1 \leq i \leq 2n$, and $x_{2n+i}=x_i$ for all $i$. Let $T \colon X \to X$ be the shift transformation $T[(x_n)]= (x_{n+1})$. We have $T^{2n}x_n=x_n$ for every $n \geq 1$, so the measure $\mu_n$ defined by $\mu_n:=(2n)^{-1}\sum_{j=0}^{2n-1}\delta_{T^jx_n}$ is an ergodic invariant Borel probability measure for $T$. Let $\overline{0}$ denote the element of $X$ corresponding to an infinite sequence of zeroes, and similarly let $\overline{1}$ denote the infinite sequence of ones; we have $\lim_{n \to \infty} \mu_n = \frac{1}{2}(\delta_{\overline{0}}+\delta_{\overline{1}})$, and this limit is not ergodic (since the set containing only the point $\{\overline{0}\}$ has measure 1/2 but is invariant).

There is a nice paper by Parthasarathy - called, I think, "On the category of ergodic measures" - which shows that for this particular dynamical system and some of its generalisations, the set of all ergodic measures and the set of all non-ergodic measures are both weak-* dense in the set of all invariant measures, so this phenomenon can actually happen quite a lot.

(Hmm, the definition of $X$ above is supposed to have curly set brackets in it, but I can't get them to appear for some reason. Anyway, it's supposed to be the set of all one-sided infinite sequences of zeroes and ones.)

Let $X=\{0,1\}^{\mathbb{N}}$ with the infinite product topology (which is metrisable). For each $n \geq 1$, define $x_n$ to be the sequence given by $x_i=0$ for $1 \leq i \leq n$, $x_i=1$ for $n+1 \leq i \leq 2n$, and $x_{2n+i}=x_i$ for all $i$. Let $T \colon X \to X$ be the shift transformation $T[(x_n)]= (x_{n+1})$. We have $T^{2n}x_n=x_n$ for every $n \geq 1$, so the measure $\mu_n$ defined by $\mu_n:=(2n)^{-1}\sum_{j=0}^{2n-1}\delta_{T^jx_n}$ is an ergodic invariant Borel probability measure for $T$. Let $\overline{0}$ denote the element of $X$ corresponding to an infinite sequence of zeroes, and similarly let $\overline{1}$ denote the infinite sequence of ones; we have $\lim_{n \to \infty} \mu_n = \frac{1}{2}(\delta_{\overline{0}}+\delta_{\overline{1}})$, and this limit is not ergodic (since the set containing only the point $\{\overline{0}\}$ has measure 1/2 but is invariant).

There is a nice paper by Parthasarathy - called, I think, "On the category of ergodic measures" - which shows that for this particular dynamical system and some of its generalisations, the set of all ergodic measures and the set of all non-ergodic measures are both weak-* dense in the set of all invariant measures, so this phenomenon can actually happen quite a lot.

(Hmm, the definition of $X$ above is supposed to have curly set brackets in it, but I can't get them to appear for some reason.)

Let $X=\{0,1\}^{\mathbb{N}}$ with the infinite product topology (which is metrisable). For each $n \geq 1$, define $x_n$ to be the sequence given by $x_i=0$ for $1 \leq i \leq n$, $x_i=1$ for $n+1 \leq i \leq 2n$, and $x_{2n+i}=x_i$ for all $i$. Let $T \colon X \to X$ be the shift transformation $T[(x_n)]= (x_{n+1})$. We have $T^{2n}x_n=x_n$ for every $n \geq 1$, so the measure $\mu_n$ defined by $\mu_n:=(2n)^{-1}\sum_{j=0}^{2n-1}\delta_{T^jx_n}$ is an ergodic invariant Borel probability measure for $T$. Let $\overline{0}$ denote the element of $X$ corresponding to an infinite sequence of zeroes, and similarly let $\overline{1}$ denote the infinite sequence of ones; we have $\lim_{n \to \infty} \mu_n = \frac{1}{2}(\delta_{\overline{0}}+\delta_{\overline{1}})$, and this limit is not ergodic (since the set containing only the point $\{\overline{0}\}$ has measure 1/2 but is invariant).

There is a nice paper by Parthasarathy - called, I think, "On the category of ergodic measures" - which shows that for this particular dynamical system and some of its generalisations, the set of all ergodic measures and the set of all non-ergodic measures are both weak-* dense in the set of all invariant measures, so this phenomenon can actually happen quite a lot.

(Hmm, the definition of $X$ above is supposed to have curly set brackets in it, but I can't get them to appear for some reason. Anyway, it's supposed to be the set of all one-sided infinite sequences of zeroes and ones.)

added 372 characters in body; added 26 characters in body
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Ian Morris
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Let $X=\{0,1\}^{\mathbb{N}}$ with the infinite product topology (which is metrisable). For each $n \geq 1$, define $x_n$ to be the sequence given by $x_i=0$ for $1 \leq i \leq n$, $x_i=1$ for $n+1 \leq i \leq 2n$, and $x_{2n+i}=x_i$ for all $i$. Let $T \colon X \to X$ be the shift transformation $T[(x_n)]= (x_{n+1})$. We have $T^{2n}x_n=x_n$ for every $n \geq 1$, so the measure $\mu_n$ defined by $\mu_n:=(2n)^{-1}\sum_{j=0}^{2n-1}\delta_{T^jx_n}$ is an ergodic invariant Borel probability measure for $T$. Let $\overline{0}$ denote the element of $X$ corresponding to an infinite sequence of zeroes, and similarly let $\overline{1}$ denote the infinite sequence of ones; we have $\lim_{n \to \infty} \mu_n = \frac{1}{2}(\delta_{\overline{0}}+\delta_{\overline{1}})$, and this limit is not ergodic (since the set containing only the point $\{\overline{0}\}$ has measure 1/2 but is invariant).

There is a nice paper by Parthasarathy - called, I think, "On the category of ergodic measures" - which shows that for this particular dynamical system and some of its generalisations, the set of all ergodic measures and the set of all non-ergodic measures are both weak-* dense in the set of all invariant measures, so this phenomenon can actually happen quite a lot.

(Hmm, the definition of $X$ above is supposed to have curly set brackets in it, but I can't get them to appear for some reason.)

Let $X=\{0,1\}^{\mathbb{N}}$ with the infinite product topology (which is metrisable). For each $n \geq 1$, define $x_n$ to be the sequence given by $x_i=0$ for $1 \leq i \leq n$, $x_i=1$ for $n+1 \leq i \leq 2n$, and $x_{2n+i}=x_i$ for all $i$. Let $T \colon X \to X$ be the shift transformation $T[(x_n)]= (x_{n+1})$. We have $T^{2n}x_n=x_n$ for every $n \geq 1$, so the measure $\mu_n$ defined by $\mu_n:=(2n)^{-1}\sum_{j=0}^{2n-1}\delta_{T^jx_n}$ is an ergodic invariant Borel probability measure for $T$. Let $\overline{0}$ denote the element of $X$ corresponding to an infinite sequence of zeroes, and similarly let $\overline{1}$ denote the infinite sequence of ones; we have $\lim_{n \to \infty} \mu_n = \frac{1}{2}(\delta_{\overline{0}}+\delta_{\overline{1}})$, and this limit is not ergodic (since the set $\{\overline{0}\}$ has measure 1/2 but is invariant).

(Hmm, the definition of $X$ above is supposed to have curly set brackets in it, but I can't get them to appear for some reason.)

Let $X=\{0,1\}^{\mathbb{N}}$ with the infinite product topology (which is metrisable). For each $n \geq 1$, define $x_n$ to be the sequence given by $x_i=0$ for $1 \leq i \leq n$, $x_i=1$ for $n+1 \leq i \leq 2n$, and $x_{2n+i}=x_i$ for all $i$. Let $T \colon X \to X$ be the shift transformation $T[(x_n)]= (x_{n+1})$. We have $T^{2n}x_n=x_n$ for every $n \geq 1$, so the measure $\mu_n$ defined by $\mu_n:=(2n)^{-1}\sum_{j=0}^{2n-1}\delta_{T^jx_n}$ is an ergodic invariant Borel probability measure for $T$. Let $\overline{0}$ denote the element of $X$ corresponding to an infinite sequence of zeroes, and similarly let $\overline{1}$ denote the infinite sequence of ones; we have $\lim_{n \to \infty} \mu_n = \frac{1}{2}(\delta_{\overline{0}}+\delta_{\overline{1}})$, and this limit is not ergodic (since the set containing only the point $\{\overline{0}\}$ has measure 1/2 but is invariant).

There is a nice paper by Parthasarathy - called, I think, "On the category of ergodic measures" - which shows that for this particular dynamical system and some of its generalisations, the set of all ergodic measures and the set of all non-ergodic measures are both weak-* dense in the set of all invariant measures, so this phenomenon can actually happen quite a lot.

(Hmm, the definition of $X$ above is supposed to have curly set brackets in it, but I can't get them to appear for some reason.)

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Ian Morris
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