# Different uses of the word “ergodic”

There appear to be two definitions of the word ergodic.

The dynamical systems definition says that a measure space $(X,\mathit B, \mu)$ and measure preserving transformation $T: X \mapsto X$ is ergodic if

the only $T$-invariant sets have measure 0 or 1.

However, a Markov chain is ergodic if

there exists $t$ such that for all $x,y \in \Omega, P^t(x,y) >0$

I've used the Markov chain notation and definition found here

I would like to know if these definitions are equivalent.

Of course, I am asking here because it seems to me that they are not. For example, if $X=\{0,1\}$, $\mathit B = \{\emptyset, \{0\},\{1\},X\}$, $\mu(\{0\})=0,\mu(\{1\})=1$ and $T(x) = 1$ for all $x\in X$, then $(X,\mathit B, \mu, T)$ is ergodic as a dynamical system, but the equivalent Markov chain is not ergodic, since the probability of traveling from $0$ to $1$ is zero.

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More precisely, an initial distribution $m$ of a Markov chain on a state space $X$ determines the associated measure $\mathbf P_m$ on the space of sample paths $X^{\mathbb Z_+}$. The measure $\mathbf P_m$ is shift invariant iff the measure $m$ is stationary. Now, if $m$ is finite (this condition is important; otherwise the following claim is false), then ergodicity of the time shift is equivalent to absence of non-trivial partitions of $X$ into non-communicating subsets.
By the way, your example is really too degenerate: the standard example for difference between ergodicity and mixing for Markov chains is presence of so-called periodic classes $A_1\to A_2\to\dots\to A_k\to A_1$ (the only allowed transitions are from $A_i$ to $A_{i+1}$ mod k). For finite chains this is actually the only reason for difference between ergodicity and mixing, but for general state spaces the situation is more complicated.