6Ergodic theory
II Probability and Measure
6.1 Ergodic theorems
The proofs in this section are nonexaminable.
Instead of proving the ergodic theorems directly, we first start by proving
the following magical lemma:
Lemma (Maximal ergodic lemma). Let f be integrable, and
S
∗
= sup
n≥0
S
n
(f) ≥ 0,
where S
0
(f) = 0 by convention. Then
Z
{S
∗
>0}
f dµ ≥ 0.
Proof. We let
S
∗
n
= max
0≤m≤n
S
m
and
A
n
= {S
∗
n
> 0}.
Now if 1 ≤ m ≤ n, then we know
S
m
= f + S
m−1
◦ Θ ≤ f + S
∗
n
◦ Θ.
Now on A
n
, we have
S
∗
n
= max
1≤m≤n
S
m
,
since S
0
= 0. So we have
S
∗
n
≤ f + S
∗
n
◦ Θ.
On A
C
n
, we have
S
∗
n
= 0 ≤ S
∗
n
◦ Θ.
So we know
Z
E
S
∗
n
dµ =
Z
A
n
S
∗
n
dµ +
Z
A
C
n
S
∗
n
dµ
≤
Z
A
n
f dµ +
Z
A
n
S
∗
n
◦ Θ dµ +
Z
A
C
n
S
∗
n
◦ Θ dµ
=
Z
A
n
f dµ +
Z
E
S
∗
n
◦ Θ dµ
=
Z
A
n
f dµ +
Z
E
S
∗
n
dµ
So we know
Z
A
n
f dµ ≥ 0.
Taking the limit as
n → ∞
gives the result by dominated convergence with
dominating function f.
We are now going to prove the two ergodic theorems, which tell us the
limiting behaviour of S
n
(f).
Theorem
(Birkhoff’s ergodic theorem)
.
Let (
E, E, µ
) be
σ
finite and
f
be
integrable. There exists an invariant function
¯
f such that
µ(
¯
f) ≤ µ(f),
and
S
n
(f)
n
→
¯
f a.e.
If Θ is ergodic, then
¯
f is a constant.
Note that the theorem only gives
µ
(

¯
f
)
≤ µ
(
f
). However, in many cases,
we can use some integration theorems such as dominated convergence to argue
that they must in fact be equal. In particular, in the ergodic case, this will allow
us to find the value of
¯
f.
Theorem
(von Neumann’s ergodic theorem)
.
Let (
E, E, µ
) be a finite measure
space. Let
p ∈
[1
, ∞
) and assume that
f ∈ L
p
. Then there is some function
¯
f ∈ L
p
such that
S
n
(f)
n
→
¯
f in L
p
.
Proof of Birkhoff’s ergodic theorem. We first note that
lim sup
n
S
n
n
, lim sup
n
S
n
n
are invariant functions, Indeed, we know
S
n
◦ Θ = f ◦ Θ + f ◦ Θ
2
+ ··· + f ◦ Θ
n
= S
n+1
− f
So we have
lim sup
n→∞
S
n
◦ Θ
n
= lim sup
n→∞
S
n
n
+
f
n
→ lim sup
n→∞
S
n
n
.
Exactly the same reasoning tells us the lim inf is also invariant.
What we now need to show is that the set of points on which
lim sup
and
lim inf do not agree have measure zero. We set a < b. We let
D = D(a, b) =
x ∈ E : lim inf
n→∞
S
n
(x)
n
< a < b < lim sup
n→∞
S
n
(x)
n
.
Now if
lim sup
S
n
(x)
n
6
=
lim inf
S
n
(x)
n
, then there is some
a, b ∈ Q
such that
x ∈ D
(
a, b
). So by countable subadditivity, it suffices to show that
µ
(
D
(
a, b
)) = 0
for all a, b.
We now fix
a, b
, and just write
D
. Since
lim sup
S
n
n
and
lim inf
S
n
n
are both
invariant, we have that
D
is invariant. By restricting to
D
, we can assume that
D = E.
Suppose that B ∈ E and µ(G) < ∞. We let
g = f − b1
B
.
Then
g
is integrable because
f
is integrable and
µ
(
B
)
< ∞
. Moreover, we have
S
n
(g) = S
n
(f − b1
B
) ≥ S
n
(f) − nb.
Since we know that
lim sup
n
S
n
(f)
n
> b
by definition, we can find an
n
such that
S
n
(g) > 0. So we know that
S
∗
(g)(x) = sup
n
S
n
(g)(x) > 0
for all x ∈ D. By the maximal ergodic lemma, we know
0 ≤
Z
D
g dµ =
Z
D
f − b1
B
dµ =
Z
D
f dµ − bµ(B).
If we rearrange this, we know
bµ(B) ≤
Z
D
f dµ.
for all measurable sets
B ∈ E
with finite measure. Since our space is
σ
finite, we
can find B
n
% D such µ(B
n
) < ∞ for all n. So taking the limit above tells
bµ(D) ≤
Z
D
f dµ. (†)
Now we can apply the same argument with (
−a
) in place of
b
and (
−f
) in place
of f to get
(−a)µ(D) ≤ −
Z
D
f dµ. (‡)
Now note that since
b > a
, we know that at least one of
b >
0 and
a <
0 has to
be true. In the first case, (
†
) tells us that
µ
(
D
) is finite, since
f
is integrable.
Then combining with (‡), we see that
bµ(D) ≤
Z
D
f dµ ≤ aµ(D).
But
a < b
. So we must have
µ
(
D
) = 0. The second case follows similarly (or
follows immediately by flipping the sign of f).
We are almost done. We can now define
¯
f(x) =
(
lim S
n
(f)/n the limit exists
0 otherwise
Then by the above calculations, we have
S
n
(f)
n
→
¯
f a.e.
Also, we know
¯
f
is invariant, because
lim S
n
(
f
)
/n
is invariant, and so is the set
where the limit exists.
Finally, we need to show that
µ(
¯
f) ≤ µ(f).
This is since
µ(f ◦ Θ
n
) = µ(f)
as Θ
n
preserves the metric. So we have that
µ(S
n
) ≤ nµ(f) < ∞.
So by Fatou’s lemma, we have
µ(
¯
f) ≤ µ
lim inf
n
S
n
n
≤ lim inf
n
µ
S
n
n
≤ µ(f)
The proof of the von Neumann ergodic theorem follows easily from Birkhoff’s
ergodic theorem.
Proof of von Neumann ergodic theorem.
It is an exercise on the example sheet
to show that
kf ◦ Θk
p
p
=
Z
f ◦ Θ
p
dµ =
Z
f
p
dµ = kfk
p
p
.
So we have
S
n
n
p
=
1
n
kf + f ◦ Θ + ···+ f ◦ Θ
n−1
k ≤ kfk
p
by Minkowski’s inequality.
So let ε > 0, and take M ∈ (0, ∞) so that if
g = (f ∨ (−M)) ∧ M,
then
kf − gk
p
<
ε
3
.
By Birkhoff’s theorem, we know
S
n
(g)
n
→ ¯g
a.e.
Also, we know
S
n
(g)
n
≤ M
for all n. So by bounded convergence theorem, we know
S
n
(g)
n
− ¯g
p
→ 0
as n → ∞. So we can find N such that n ≥ N implies
S
n
(g)
n
− ¯g
p
<
ε
3
.
Then we have
¯
f − ¯g
p
p
=
Z
lim inf
n
S
n
(f − g)
n
p
dµ
≤ lim inf
Z
S
n
(f − g)
n
p
dµ
≤ kf − gk
p
p
.
So if n ≥ N, then we know
S
n
(f)
n
−
¯
f
p
≤
S
n
(f − g)
n
p
+
S
n
(g)
n
−
¯
f
p
+
¯g −
¯
f
p
≤ ε.
So done.