My title - Departamento de Matemática da Universidade do Minho

16 ERGODICITY AND CONVERGENCE OF TIME MEANS 104
Now, let ϕ k be the the characteristic function of {x ∈ Σ + s.t. x 1 = k}. The observables ϕ k ◦σ n
form a sequence of independent and identically distributed random variables with mean p k . One
can interprete the event {ϕ k ◦ σ n = 1} = {x ∈ Σ + s.t. x n = k} as ”sucess in the n-th trial”, where
the probability of sucess in each trial is p k . The Birkhoff-Khinchin ergodic theorem, together with
the ergodicity of µ, gives the result that
µ
{
x ∈ Σ + s.t.
1 (
ϕk + ϕ k ◦ σ 1 + ϕ k ◦ σ 2 + ... + ϕ k ◦ σ n) }
(x) → p k = 1
n + 1
which is the Kolmogorov strong law of large numbers.
Expanding endomorphisms of the circle. Let ×λ : x+Z ↦→ λ·x+Z, with λ ∈ Z and |λ| > 1,
be an expanding endomorphism of the circle. Lebesgue probability measure l is an ergodic measure
for ×λ. To prove ergodicity, let A be an invariant Borel set, and assume that l (A) < 1. We
must show that the complement B = (R/Z) \A, that has positive measure, has indeed probability
one. The argument goes as follows: if l (B) > 0, then, according to Lebesgue density theorem, B
contains nearly all the mass of some nonempty interval. Namely, given any ε > 0, we can find an
open interval I n with lenght l (I n ) = |λ| −n and centered at a density point of B such that
l (B ∩ I n ) > (1 − ε) · l (I n )
Now observe that the restriction (×λ) n | In is an injective map sending I n onto the circle minus
one point, in particular, l ((×λ) n (I n )) = 1. Since ×λ uniformly dilatates lenghts by a factor |λ|,
there follows that
l ((×λ) n (B ∩ I n ))
l ((×λ) n (I n ))
= l (B ∩ I n)
l (I n )
Since, moreover, A is invariant, its complement B is +invariant, and this implies that the left-hand
side above is equal to l (B). There follows that
and, since ε was arbitrary, that l (B) = 1.
16.3 Normal numbers
l (B) = l (B ∩ I n)
l (I n )
> (1 − ε)
Normal numbers. In particular, Lebesgue measure l is ergodic w.r.t. multiplication by 10
in the unit circle. Identify the circle with the interval [0, 1[, and let x = 0, x 1 x 2 x 3 ... be the base
10 expression of a point of the circle, which is unique outside a subset of Lebesgue measure zero.
For k = 0, 1, 2, ..., 9, let ϕ k be the characteristic function of the interval [k/10, (k + 1) /10[, i.e. the
observable which is equal to ϕ k (x) = 1 if x 1 = k and ϕ k (x) = 0 otherwise. The time mean of ϕ k
is
1
n + 1
n∑
j=0
)
ϕ k
((×10) j (x) = 1
n + 1 · card {1 ≤ j ≤ n + 1 s.t. x j = k}
that is the number of k’s within the first n + 1 digits of the decimal expansion of x. The limit as
n → ∞, if it exists, is the “asymptotic frequency” of k’s contained in the expansion of x. Ergodicity
of µ implies that there exists a set A k ⊂ [0, 1[ of Lebesgue measure one where the limit ϕ k (x)
exists and is equal to ∫ ϕ k dl = 1/10. Since the intersection A 0 ∩ A 1 ∩ ... ∩ A 9 has still probability
one, the result is that Lebesgue almost any number x ∈ [0, 1[ contains in its decimal expansion any
of the letters 0, 1, 2, ..., 9 with asymptotic frequency 1/10.
Actually, one could repeat the same argument considering any finite word b = b 1 b 2 ...b n in the
alphabeth {0, 1, 2, ..., 9}, and show that there is a set A b ⊂ [0, 1[ of probability one such that the
base 10 expansion of any x ∈ A b contains the word b with asymptotic frequency 10 −n . A real
number x whose base 10 expansion contains any finite word with the right asymptotic frequency is
called 10-normal (meaning “normal in base 10”). Since finite words in the alphabeth {0, 1, 2, ..., 9}
are countable, and a countable union of zero measure sets still has zero measure, we just showed
that Lebesgue almost any real number x is 10-normal. This observation, and indeed the stronger