# Proof of Krylov-Bogoliubov theorem

Where can I find a proof (in English) of the Krylov-Bogoliubov theorem, which states if $$X$$ is a compact metric space and $$T\colon X \to X$$ is continuous, then there is a $$T$$-invariant Borel probability measure? The only reference I've seen is on the Wikipedia page, but that reference is to a journal that I cannot find.

Of course, feel free to answer this question by providing your own proof.

• I searched google books for the following words: Krylov Bogolyubov compact metrizable invariant borel measure. The first result was books.google.com/… - p.8 from Bunimovich, Sinai: Dynamical systems, ergodic theory and applications, which contains the proof. (Hopefully, it will be viewable for you in google books.) Jun 7, 2011 at 5:47
• There is another Krylov-Bogolubiov theorem which is about an invariant measure for a family of functions- Does anyone know about that one? May 28, 2018 at 8:09

First, fix $x \in X$ and let $\mu_1 := \delta_x$ be the Dirac measure supported at $x$. Then define a sequence of probability measures $\mu_n$ such that for any $f \in C^0 (X)$, $$\int_X f(y) \mathrm{d} \mu_n (y) = \frac{1}{n} \sum_{k=0}^{n-1} \int_X f \circ T^k (y) \mathrm{d} \mu_1 (y).$$ Apply the Banach-Alaouglu Theorem to deduce there exists a subsequence $\mu_{n_j}$ which converges in the weak-$\star$ topology. It is then very easy to prove that this limit measure is in fact T-invariant, using the formulation that $\mu$ is T-invariant if and only if $$\int_X f \circ T \mathrm{d} \mu = \int_X f \mathrm{d}\mu$$ for all continuous $f$.

There are two pretty simple proofs. Both rely on studying the action $T_* \colon \mathcal{M} \to \mathcal{M}$, where $\mathcal{M}$ is the space of Borel probability measures on $X$ and the action is given by $(T_* \mu)(E) := \mu(T^{-1}(E))$. A measure $\mu$ is $T$-invariant if and only if $T_* \mu = \mu$.

One proof is the one given by Michael Coffey in his answer: start with any measure $\mu$, not necessarily invariant, such as the $\delta$-measure sitting at an arbitrary point, and then consider the sequence of measures $\mu_n = \frac 1n \sum_{k=0}^{n-1} T^k_* \mu$. Because $\mathcal{M}$ is weak* compact, some subsequence $\mu_{n_j}$ converges to a measure $\nu\in \mathcal{M}$, and it's not hard to show that $\nu$ is invariant.

An alternate proof is to observe that $\mathcal{M}$ is a compact convex subset of the locally convex vector space $C(X)^*$, and that $T_*$ acts continuously on $\mathcal{M}$, whence by the Schauder-Tychonoff fixed point theorem it has a fixed point $\nu=T_* \nu$.

• Does the Krylov-Bogoliubov theorem hold if $T$ is only assumed measurable? Dec 29, 2011 at 18:26
• @Quinn: No, measurability alone isn't enough. Define $f\colon [0,1] \to [0,1]$ by $f(x) = x/2$ when $x>0$, and $f(0)=1$. Then if $\mu$ is invariant and $\mu(E)>0$ for some $E\subset [0,1]$ with $0\notin E$, you can look at the images $f^n(D)$ for a suitable $D\subset E$ and conclude that $\mu([0,1]) = \infty$. The only probability measure that remains to consider is $\delta_0$, which isn't invariant because $0$ is not fixed. Jan 13, 2012 at 18:46
• Can someone point out to me where we have used the fact that T is continuous ? Dec 16, 2013 at 20:59
• Continuity of T is needed when you show that the limiting measure $\nu = \lim \mu_{n_j}$ is invariant. For this part of the proof one can let $D$ be a metric compatible with the weak* topology and then observe that $D(\nu,T_*\nu) \leq D(\nu,\mu_n) + D(\mu_n,T_*\mu_n) + D(T_*\mu_n, T_*\nu)$. The first term goes to zero along $n_j$ by choice of the subsequence, the second goes to zero by the construction of $\mu_n$, and the third goes to zero because $T$ is continuous. Dec 17, 2013 at 17:03

In addition to the excellent answers above, I also suggest the nice survey Oxtoby, Ergodic Sets (Zbl 0046.11504, MR47262, DOI: 10.1090/S0002-9904-1952-09580-X).

Introduction. Ergodic sets were introduced by Kryloff and Bogoliouboff in 1937 in connection with their study of compact dynamical systems [16]. The purpose of this paper is to review some of the work that has since been done on the theory that centers around this notion, and to present a number of supplementary remarks, applications, and simplifications. For simplicity were shall confine attention to systems with a discrete time. Continuous flows present no difficulty, but the development of a corresponding theory for general transformation groups is still in an incomplete stage. An example due to Kolmogoroff (see [5]) shows that such an extension cannot be made without sacrificing either the invariance or the disjointness of ergodic sets.

In §§1 and 2 we give a brief, but self-sufficient, development of the basic theorems of Kryloff and Bogoliouboff. In §3 we collect some auxiliary results for later use. In §4 a simple characterization of transitive points is obtained. In §5 the distinctive properties of some special types of systems and subsystems are discussed, and in §6 these results are used to discover conditions under which the ergodic theorem holds uniformly. In §7 a generalization to noncompact systems is considered, and in §§8 and 9 some known representation theorems are obtained as an application of ergodic sets. In §10 there is given an example of a minimal set that is not strictly ergodic, similar to one constructed by Markoff.

You can see the famous book by Peter Walter:An introduction to ergodic theory, Pages 151-152

Then the theorem also doesn't hold. Take $X = \mathbb{R}$, and $T\colon x \mapsto x+1$. Then $T$ has no invariant probability measure. (all the measures "escape to infinity").

• @Quinn: this was the answer to your other question. Nov 7, 2012 at 17:40
• Yes, thanks. I think this answer was essentially given here: math.stackexchange.com/questions/94981/…. Nov 8, 2012 at 3:19

If you still need a reference:

You can find a clear exposition of the result and the proof of the theorem in "Ergodic Theory" by Einsiedler 2013 (free PDF here).

Or see the original work by Kryloff and Bogoliouboff, here.