Let $(\Omega,\mathcal{A},\mu,T)$ be a dynamic system in meaasure theory. Show that the following statements are equivalent:

(1) $(\Omega,\mathcal{A},\mu,T)$ is ergodic

(2) $\forall A\in\mathcal{A}: T^{-1}(A)\subset A\implies\mu(A)\in\left\{0,1\right\}$

(3) $\forall A\in\mathcal{A}: \mu(A)>0\implies \mu(\bigcup_{k=1}^{\infty} T^{-k}A)=1$

I know it that way:

A dynamic system in measure theory consists of a measurable space $(\Omega,\mathcal{A})$, a measurable function $T\colon\Omega\to\Omega$ and a non-singular measure $\mu$ (i.e. for $A\in\mathcal{A}$ it is $\mu(A)0=0$ if and only if $\mu(T^{-1}(A))=0$). A measure $\mu$ is especially non-singular if it is invariant, i.e. for every $A\in\mathcal{A}$ it is $\mu(T^{-1}(A))=\mu(A)$.

A dynamic system is called ergodic if: $A\in\mathcal{A}: T^{-1}(A)\subset A\implies \mu(A)=0\text{ or }\mu(A^C)=0$

But then $(1)\implies (2)$ is trivial, isn't it?

Do you know a definition of "ergodic dynamic system" which does not use (2)?

  • 1
    $\begingroup$ Well, the equivalence between (1) and (2) is only true if $\mu$ is a probability measure (or, with the obvious modification, if $\mu$ is finite). In this case, it is trivial. The equivalence with (3) is more interesting, and I think it is the goal of the exercise. $\endgroup$
    – D. Thomine
    Jan 8, 2014 at 12:22
  • $\begingroup$ Is it a good strategy to show $(2)\implies (3)$ and $(3)\implies (1)$? If yes, how can i first show that $(2)\implies (3)$? My idea is to show for $C:=\bigcup_{k=1}^{\infty}T^{-k}B$ that $T^{-1}(C)\subset C$, then it follows $\mu(C)\in\left\{0,1\right\}$. Does then follow from $\mu(B)>0$ that $\mu(C)=1$? $\endgroup$
    – user34632
    Jan 8, 2014 at 12:35
  • $\begingroup$ In this case, the equivalence between (1) and (2) is so obvious that I would show it at first. Then I take care of (3) separately. $\endgroup$
    – D. Thomine
    Jan 8, 2014 at 13:15
  • $\begingroup$ To prove (2)-> (3): your idea works, and you only need to prove that $\mu (C) > 0$. Note that $T^{-1} B \subset C$. $\endgroup$
    – D. Thomine
    Jan 8, 2014 at 13:18
  • $\begingroup$ Would do that per contradiction: Assume that $\mu(C)=0$, then because of $T^{-1}(B)\subset C$ and the invariance of $\mu$ it is $\mu(B)=\mu(T^{-1}(B))\leq\mu(C)=0$. From this it follows $\mu(B)=0$ in contradiction to $\mu(B)>0$. So $(2)\implies (3)$ is shown. Now I have $(1)\implies (2)$ and $(2)\implies (3)$. What would be the next step? $\endgroup$
    – user34632
    Jan 8, 2014 at 13:24

1 Answer 1


First, I'll recall the broadest definition of ergodicity.

Let $(\Omega, \mathcal{A})$ be a measurable space. Let $T : \Omega \to \Omega$ be a measurable transformation. Let $\mu$ be a nonnegative, non-zero, $\sigma$-finite measure on $\Omega$.

We say that the dynamical system $(\Omega, \mathcal{A}, \mu, T)$ is non-singular if, for any Borel set $A$, we have $\mu (T^{-1}A) = 0 \Leftrightarrow \mu (A) = 0$. By Radon-Nikodym theorem, a system is non-singular if and only if $T_* \mu \ll \mu$, that is, if $T$ does not send some mass to a subset of measure zero.

We say that the dynamical system $(\Omega, \mathcal{A}, \mu, T)$ is measure-preserving if, for any Borel set $A$, we have $\mu (T^{-1}A) = \mu (A)$. That is, a system preserves the measure if and only if $T_* \mu = \mu$.

We say that the dynamical system $(\Omega, \mathcal{A}, \mu, T)$ is ergodic if it is non-singular and, for any Borel set $A$ such that $T^{-1}A \subset A$, we have $\mu (A) = 0$ or $\mu (A^c) = 0$. Ergodicity does not require that the measure be preserved, just that it be non-singular.

A system which is ergodic is, in some sense, irreducible: you cannot split it into two non-trivial (for the measure $\mu$) sub-systems which are invariant under $T$. For instance, the transformation $x \mapsto x+1/2 [1]$ on $[0, 1)$ is not ergodic with respect to the Lebesgue measure, because the subsets $[1, 1/4) \cup [1/2, 3/4)$ and $[1/4, 1/2) \cup [3/4, 1)$ are invariant and of positive measure.

Ergodicity can also be viewed functionally: the dynamical system $(\Omega, \mathcal{A}, \mu, T)$ is ergodic if and only if, for any measurable function $f: \Omega \to \mathbb{R}$, if $f = f \circ T$ almost everywhere for $\mu$, then there exists a constant $C$ such that $f = C$ almost everywhere for $\mu$.

In the following, I'll assume that $\mu$ is a probability measure, because ergodic theory with infinite measures is something very specific.


The fact that $(1) \Leftrightarrow (2)$ is obvious, since $\mu (A) = 1 \Leftrightarrow \mu(A^c) = 0$ for any Borel subset $A$.

Let us prove that $(2) \Rightarrow (3)$, which you have already done more or less properly in the comments. We don't need to use the contraposition. Assume that the system is ergodic. Let $A$ be a Borel subset, with $\mu (A) > 0$. Let $B := \bigcup_{k=1}^{+ \infty} T^{-k} A$. Since the system is non-singular, $\mu (T^{-1}A) > 0$. But $T^{-1} A \subset B$, so $\mu(B)>0$. In addition,

$$T^{-1} B = T^{-1} \bigcup_{k=1}^{+ \infty} T^{-k} A = \bigcup_{k=1}^{+ \infty} T^{-(k+1)} A = \bigcup_{k=2}^{+ \infty} T^{-k} A \subset B.$$

By the property $(2)$, we have $\mu (B) = 1$.

Now, I'll give you some hints to prove that $(3) \Rightarrow (2)$. This time, I would use the contrapositive. That is, assume that there is a Borel subset $A$ with $T^{-1}A \subset A$ and $\mu (A) \in (0,1)$, and try to prove that $\mu (\bigcup_{k=1}^{+ \infty} T^{-k} A) < 1$. As a middle point, try to prove that $\mu (A^c \cap \bigcup_{k=1}^{+ \infty} T^{-k} A) = 0$.

Edit : Proof of $(3) \implies (2)$.

I'll use the contrapositive : it is enough to prove that $\not (2) \implies \not (3)$. Assume that $(2)$ is false. Then I can find a measurable subset $A$ such that $\mu (A) \in (0, 1)$ and $T^{-1} A \subset A$.

By induction, we show that $T^{-n} A \subset T^{-1}A$ for all $n \geq 1$. This is obviously true for $n = 1$. If it is true for some $n$, then $T^{-(n+1)} A = T^{-n} T^{-1} A \subset T^{-n}A \subset T^{-1} A$, as $ T^{-1} A \subset A$. Hence,

$$T^{-1} A \subset B := \bigcup_{k=1}^{+ \infty} T^{-k} A \subset T^{-1} A,$$

so $B = T^{-1} A$. In addition, $\mu (B) > 0$, as the measure $\mu$ is non-singular, and $B \subset A$, so $\mu (B) \leq \mu (A) < 1$. This contradicts $(3)$.

  • $\begingroup$ I have one question. Why are you calling sets $A\in\mathcal{A}$ Borel sets? $\endgroup$
    – user34632
    Jan 8, 2014 at 15:56
  • $\begingroup$ And the OR in the definition of "ergodic" is meant EXCLUSIVE, right? $\endgroup$
    – user34632
    Jan 8, 2014 at 16:07
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    $\begingroup$ @math12 : I said "Borel sets" because that's usually what they are (most of the time, we work the Borel $\sigma$-algebra associated with some nice topology). You can replace it with "measurable" if you wish. | The OR can be inclusive or exclusive, that doesn't change anything (it is de facto exclusive, so there is no need to specify it logically). $\endgroup$
    – D. Thomine
    Jan 8, 2014 at 17:15
  • $\begingroup$ And then I have a question to your contrapositive. It is to show not (3)$\implies$ not (2), right? But you are proving not (2)$\implies$ not (3)? $\endgroup$
    – user34632
    Jan 8, 2014 at 17:16
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    $\begingroup$ @math12 : I've already proved that $(2) \implies (3)$: that's the paragraph above. The contrapositive is to prove $\not (2) \implies \not (3)$, that is, $(3) \implies (2)$. $\endgroup$
    – D. Thomine
    Jan 8, 2014 at 17:18

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