My title - Departamento de Matemática da Universidade do Minho

16 ERGODICITY AND CONVERGENCE OF TIME MEANS 103
different. To prove the converse, assume that the invariant measure µ is not ergodic, hence there
exists an invariant event C such that 0 < µ (C) < 1. Let µ 0 and µ 1 be the ”conditional probability
measures” defined as µ 1 (A) = µ (A ∩ C) /µ (C) and µ 0 (A) = µ (A ∩ C c ) /µ (C c ). Clearly they are
different, both are invariant, and µ = µ (C) µ 1 + (1 − µ (C)) µ 0 . □
Ergodic decomposition. In the first lines of the above proof, we actually showed that any
two ergodic invariant measure µ and ν are either equal or ”mutually singular”, namely, if µ ≠ ν
then there exists a measurable set A such that µ (A) = ν (A c ) = 1 and µ (A c ) = ν (A) = 0. This
suggests that maybe any invariant measure could be ”disintegrated” along a partition whose atoms
are the support of all the different ergodic measure, in other word that µ is a ”convex combination”,
namely an integral, of the ergodic measures. This is true, sometimes, but both its statement and
proof are quite technical: we just quote the result.
Ergodic decomposition theorem. Let f : X → X be a continuous transformation of the
compact metrizable space X. There exists a partition P = {P e } e∈E
of X (modulo sets of zero
measure) into invariant measurable sets indexed by a Lebesgue space E, and a measurable map
E ∋ e ↦→ µ e ∈Prob f with values in the space of ergodic Borel probability measures and with the
property that µ e (P e ) = 1 for any P e ∈ P, such that any invariant Borel probability measure µ can
be written as an integral
∫
µ = µ e dµ (e)
where µ is some probability measure on E.
E
Observe that the above theorem contains the statement that any continuous transformation of
a compact space admits at least one ergodic Borel probability measure.
16.2 Examples of ergodic maps
Bernoulli shift. Let σ : Σ + → Σ + be the Bernoulli shift over the alphabet X = {1, 2, ..., z},
and p any probability on X. The Bernoulli invariant measure µ defined by p is ergodic. First
observe that, given two centered cilinders C α and C β , the definition of µ implies that there exists
a time n ≥ 1 such
µ ( C α ∩ σ −k (C β ) ) = µ (C α ) · µ ( σ −k (C β ) ) = µ (C α ) · µ (C β )
whenever k ≥ n. Indeed, one can take n = |α| + 1, and the above reflect the ”independence” of
the different trials encoded in the construction of the Bernoulli measure. By aditiviity, the same
holds true for any couple of elements of A, the algebra made of finite unions of centered cilinders.
Now, assume that A ∈ B is invariant. Since any Borel set A ∈ B can be aproximated in measure
by an elements of A, given any ε > 0 one can find an A ε ∈ A such that µ (A∆A ε ) < ε. Using the
above result, we can find an n ≥ 1 such that
µ ( A ε ∩ σ −n (A ε ) ) = µ (A ε ) · µ ( σ −n (A ε ) ) = µ (A ε ) 2
where the last equality comes from invariance of µ. Then, observe that the symmetric difference
between A ∩ σ −n (A) and A ε ∩ σ −n (A ε ) is contained in (A∆A ε ) ∪ σ −n (A∆A ε ). This gives
∣ µ
(
A ∩ σ −n (A) ) − µ ( A ε ∩ σ −n (A ε ) )∣ ∣ ≤ µ (A∆A ε ) + µ ( σ −n (A∆A ε ) )
≤
2 · µ (A∆A ε ) < 2ε
which, together with
∣ ∣∣µ (A) 2 − µ (A ε ) 2∣ ∣ ∣ ≤ 2 · µ (A∆Aε ) < 2ε
gives
∣ ∣∣µ (A) − µ (A)
2 ∣ ∣ ∣ < 4ε
Since ε > 0 was arbitrary, we just showed that the measure of any invariant Borel set A satisfies
µ (A) = µ (A) 2 , hence it is either 0 or 1. Observe that this proof is very similar to the argument
in the Kolmogorov zero-one law for tail events in the theory of stochastic processes.