The standard probability space $(I, \mathcal B, \lambda)$ consists of the interval $I = [0,1]$, its Borel $\sigma$-algebra $\mathcal B := \mathcal B(I)$ and Lebesgue measure $\lambda$. In applications, most probability spaces of interest are measure isomorphic to the standard probability space.

Let $\Gamma = \operatorname{Aut}(I,\mathcal B, \lambda)$ denote the automorphism group of the standard probability space. Recall that a measurable automorphism is a bijection $f : I \to I$ which pulls back and pushes forward measurable sets. A measure automorphism additionally preserves measure: $\lambda = \lambda \circ f^{-1} = \lambda \circ f$.

Let $\Gamma_{erg}$ be the subgroup of ergodic automorphisms. i.e., $f \in \Gamma_{erg}$ if $f \in \Gamma$ and $$\lambda(A \, \triangle \, f A) = 0 \mathrm{~implies~} \lambda(A) = 0 \mathrm{~or~} \lambda(A) = 1$$ for all Borel sets $A \in \mathcal B(I)$, where $\triangle$ denotes the symmetric difference.

Is there a nice characterization of $\Gamma$ or $\Gamma_{erg}$? These are both very, very large groups, and hard for me to conceive of. What structure do they satisfy? Does every Lie group embed as a subgroup of $\Gamma$ or $\Gamma_{erg}$? Does the quotient $\Gamma / \Gamma_{erg}$ have any meaningful structure?

This is an open-ended question, so I've marked it as big list (and community wiki). I would be happy with a list of some non-trivial subgroups of $\Gamma$ and $\Gamma_{erg}$.

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    $\begingroup$ If $G$ is a semisimple connected Lie group, then it admits a lattice $\Lambda$ and the action of $G$ on $G/\Lambda$ is faithful, this probably embeds $G$ into your group $\Gamma$ (since $G/\Lambda$ is a standard probability space). $\endgroup$
    – YCor
    Oct 18, 2013 at 20:00
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    $\begingroup$ This paper seems relevant. ams.org/journals/proc/1990-110-02/S0002-9939-1990-1009997-6/… $\endgroup$ Oct 18, 2013 at 21:17
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    $\begingroup$ The ergodic transformations do not form a group... $\endgroup$ Oct 18, 2013 at 21:18
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    $\begingroup$ I don't know what you are looking for. $\Gamma$ has a natural metric structure on it under the Ky-Fan metric which is the metric of convergence in probability. The space $\Gamma$ is a closed sunspace (under this metric) of the space of measurable maps from the standard probability space to [0,1]. The ergodic maps are a closed subspace of $\Gamma$. $\endgroup$
    – Jason Rute
    Oct 18, 2013 at 23:14
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    $\begingroup$ @TomLaGatta: I'm confused; the identity map is not ergodic. $\endgroup$ Oct 19, 2013 at 2:58

1 Answer 1


I can't explain the group theoretic structure of $\Gamma$, but I can explain the topological structure. (Warning, this post is mostly a continuous stream of thoughts. I hope it is well organized and truthful, but you should check the details.)

The space of measurable functions $f\colon(I,\mathcal{B},\lambda) \rightarrow \mathbb{R}$

There are a number of metrics and norms on spaces of measurable functions $f\colon(I,\mathcal{B},\lambda) \rightarrow \mathbb{R}$. Of course there are

  • The $L^p$ norms (on the subspace of $p$-integrable functions).

Also, there are lesser-known metrics which give the topology of convergence in probability (a.k.a. convergence in measure). These two metrics are equivalent:

  • The Ky-Fan metric $$\rho_\textit{Ky-Fan} (f,g) = \inf\left\{\varepsilon > 0 : \lambda \left\{x : |f(x) - g(x)| \geq \varepsilon\right\} \leq \varepsilon\right\}.$$ This definition makes more sense when you consider the definition of convergence in probability.

  • The metric $$\rho (f,g) = \int \min \left\{|f-g|,1\right\} \, d\lambda$$ (If you know the name for this metric, please answer this MO question!) Notice the similarity between this metric and the $L^1$ metric. Also notice, that for indicator functions, this metric becomes the familiar metric $\rho(\mathbf{1}_A,\mathbf{1}_B) = \lambda(A \triangle B)$.

The space of measurable functions $f\colon(I,\mathcal{B},\lambda) \rightarrow I$

For the subspace of functions $f\colon(I,\mathcal{B},\lambda) \rightarrow I$, it is easy to see that this last metric is exactly the same as the $L^1$ metric. Moreover, one can show on this space that all the $L^p$ metrics are equivalent (easy exercise).

The space of measure preserving automorphisms $\textrm{Aut}(I,\mathcal{B},\lambda)$

The measure preserving automorphisms form a subspace of the previous space. It is closed. This is because the push-forward map $f \mapsto \lambda_f$ is continuous in any of the above metrics, where the topology on the codomain is given by the Levy-Prokohorov metric, that is the metric of convergence in distribution.

This space is therefore a complete separable metric space (Polish space) under any of the above metrics. However, the usual candidates for a countable dense set (e.g. polynomials with rational coefficients) don't work. Instead, the following functions form a nice dense set: For each $n$ choose, consider a permutation $\pi$ on $\{0,\ldots,2^n - 1\}$. Then let $f^n_\pi \colon [0,1] \rightarrow [0,1]$ be as follows. Break up $[0,1]$ into $2^n$ equally spaces dyadic intervals and let $f^n_\pi$ rearrange the intervals according to $\pi$.

(Actually, consider the $L^1$ metric on this subset of basic functions. Take two such "basic functions" $f^n_\pi$ and $f^n_\sigma$. (WLOG, they break up $[0,1]$ into the same number of intervals.) Then the distance $\| f -g\|_1$ is $2^{-n}\sum_{i=0}^{2^n} (\pi(i) - \sigma(i))$.

In this way, one can think of this space as a continuum sized extension of the countable group $G = \bigcup_n S_{2^n}$ where we embed $S_{2^n}$ into $S_{2^{n+1}}$. (Although, our metric necessarily breaks the symmetry of $S_{2^n}$.)

This space is not compact. (One can find a sequence of such basic functions which does not have a convergence subsequence.)

The space of ergodic measure preserving automorphisms $\textrm{Aut}_\textrm{Ergodic}(I,\mathcal{B},\lambda)$

This is dense in the previous space. To see this, consider an irrational shift $g_\alpha(x) = x + \alpha \mod 1$. Then compose it with a basic function. It only changes the $L^1$ norm of the basic function slightly, but this composed function is now ergodic. (This takes a little thought.)

  • $\begingroup$ There exist compact spaces with sequences with no converging subsequences (e.g. the sequence $(n)_{n\ge 0}$ in the Stone-Cech compactification of $\mathbf{N}$). Also one usually identifies functions coinciding outside null sets in the automorphism group of a measure space, so we'd expect the topology to be well-defined on the quotient. $\endgroup$
    – YCor
    Oct 19, 2013 at 10:05
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    $\begingroup$ @Jason When you said "this space is not compact", did you mean the dense set you just described? What about the whole space of automorphisms? $\endgroup$ Nov 5, 2013 at 17:52
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    $\begingroup$ @HenriquedeOliveira: I meant the space of automorphisms is not compact. There is a sequence of basic functions which does not converge to an automorphism (in the metrics I mentioned). One such sequence is $(f^n_{\pi_n})_n$ where $\pi_{n-1}(i)=2i$ and $\pi_{n-1}(2^n+i)=2i+1$ for $0\leq i \leq 2^{n}$. While there is a limit in the space of measurable functions, namely the map $f(x)=2x \mod 1$, this is not an automorphism. With a little more tweaking we can adjust the sequence so it does not converge to any measurable function. (I'll leave that as an easy exercise.) $\endgroup$
    – Jason Rute
    Nov 5, 2013 at 21:23
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    $\begingroup$ @YvesCornulier: On a metric space, compact and sequentially compact are the same: (en.wikipedia.org/wiki/Sequentially_compact_space). Also, technically the metrics I defined are pseudo-metrics. One needs to work up to a.e. equivalence for them to be metrics. $\endgroup$
    – Jason Rute
    Nov 5, 2013 at 21:28
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    $\begingroup$ @HenriquedeOliveira (1) Nhu's metric is equivalent to mine. (Note, this means his/her metric is equivalent to $L^1$ and $L^2$.) (2) This topology seems "right" because it is the topology which makes the map $A \mapsto f^{-1}(A)$ continuous (where the measurable sets have the topology give by the metric $d(A,B) = \lambda (A \triangle B)$). (3) However, you bring up a good point about compactness. Possibly the weak $L^2$ norm would make the group compact. One would need to check that (a) the space is closed (and hence compact) in weak-$L^2$ and (b) the group is continuous in weak-$L^2$. $\endgroup$
    – Jason Rute
    Nov 15, 2013 at 4:43

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