We continue our study of basic ergodic theorems, establishing the maximal and pointwise ergodic theorems of Birkhoff. Using these theorems, we can then give several equivalent notions of the fundamental concept of ergodicity, which (roughly speaking) plays the role in measure-preserving dynamics that minimality plays in topological dynamics. A general measure-preserving system is not necessarily ergodic, but we shall introduce the *ergodic decomposition*, which allows one to express any non-ergodic measure as an average of ergodic measures (generalising the decomposition of a permutation into disjoint cycles).

— The maximal ergodic theorem —

Just as we derived the mean ergodic theorem from the more abstract von Neumann ergodic theorem in the previous lecture, we shall derive the maximal ergodic theorem from the following abstract maximal inequality.

Theorem 1. (Dunford-Schwartz maximal inequality) Let be a probability space, and let be a linear operator with P1=1 and (i.e. for all . Assume also that P maps non-negative functions to non-negative functions. Then the maximal function obeys the inequality(1)

for any .

**Proof.** We can rewrite (1) as

. (2)

Since , we thus see (by replacing f with ) that we can reduce to proving (2) in the case .

For every , consider the modified maximal function . Observe that if and only if for all sufficiently large m. By the dominated convergence theorem, it thus suffices to show that

(3)

for all m. But observe from definition of (and the positivity preserving nature of P) that we have the pointwise recursive inequality

. (4)

Integrating this on the region and using the non-negativity of , we obtain

. (6)

Since and , the claim follows.

Applying this in the case when P is a shift operator, and replacing f by |f|, we obtain

Corollary 1.(Maximal ergodic theorem) Let be a measure-preserving system. Then for any and one has. (7)

Note that this inequality implies Markov’s inequality

. (8)

as a special case. Applying the real interpolation method, one also easily deduces the maximal inequality

(9)

for all , where the constant depends on p (it blows up like in the limit ).

**Exercise 1** (Rising sun inequality). If , and , establish the *rising sun inequality*

(10)

for any . (*Hint*: one can either adapt the proof of Theorem 1, or else partition the set appearing in (10) into disjoint intervals. The latter proof also leads to a proof of Corollary 1 which avoids the Dunford-Schwartz trick of introducing the functions . The terminology “rising sun” comes from seeing how these intervals interact with the graph of the partial sums of f, which resembles the shadows cast on a hilly terrain by a rising sun.)

**Exercise 2.** (Transference principle) Show that Corollary 1 can be deduced directly from (10). (*Hint*: given , apply (10) to the functions for each (truncating the integers to a finite set if necessary), and then integrate in x using Fubini’s theorem.) This is an example of a *transference principle* between maximal inequalities on and maximal inequalities on measure-preserving systems.

**Exercise 3** (Stein-Stromberg maximal inequality). Derive a continuous version of the Dunford-Schwartz maximal inequality, in which the operators are replaced by a semigroup acting on both and , in which the underlying measure space is only assumed to be -finite rather than a probability space, and the averages are replaced by . Apply this continuous version with equal to the heat operator on for to deduce the *Stein-Stromberg maximal inequality*

(11)

for all and , where m is Lebesgue measure, is the Euclidean ball of radius R centred at x, and the constant C is absolute (independent of d). This improves upon the Hardy-Littlewood maximal inequality, which gives the same estimate but with replaced by . It is an open question whether the dependence on d can be removed entirely; the estimate (11) is still the best known in high dimension. For d=1, the best constant C is known to be , a result of Melas.

**Remark 1.** The study of maximal inequalities in ergodic theory is, of course, a subject in itself; a classical reference is this monograph of Stein.

— The pointwise ergodic theorem —

Using the maximal ergodic theorem and a standard limiting argument we can now deduce

Theorem 2(Pointwise ergodic theorem). Let be a measure-preserving system, and let . Then for -almost every , converges to .

**Proof.** By subtracting from f if necessary, it suffices to show that

(12)

a.e. whenever . By telescoping series, (12) is already true when f takes the form for some . So by the arguments used to prove the von Neumann ergodic theorem from the previous lecture, we have already established the claim for a dense class of functions f in with , and thus also for a dense class of functions in with (since the latter space is dense in the former, and the norm controls the norm by the Cauchy-Schwarz inequality).

Now we use a standard limiting argument. Let with . Then we can find a sequence in the above dense class which converges in to f. For almost every x, we thus have

(13)

for all j, and so by the triangle inequality we have

. (14)

But by Corollary 1 we see that the right-hand side of (14) converges to zero in measure as . Since the left-hand side does not depend on j, it must vanish almost everywhere, as required.

**Remark 2. ** More generally, one can derive a pointwise convergence result on a class of rough functions by first establishing convergence for a dense subclass of functions, and then establishing a maximal inequality which is strong enough to allow one to take limits and establish pointwise convergence for all functions in the larger class. Conversely, principles such as Stein’s maximal principle indicate that in many cases this is in some sense the *only* way to establish such pointwise convergence results for rough functions.

**Remark 3. ** Using the dominated convergence theorem (starting first with bounded functions f in order to get the domination), one can deduce the mean ergodic theorem from the pointwise ergodic theorem. But the converse is significantly more difficult; pointwise convergence for various ergodic averages is often a much harder result to establish than the corresponding norm convergence result (in particular, many of the techniques discussed in this course appear to be of sharply limited utility for pointwise convergence problems), and many questions in this area remain open.

**Exercise 4** (Lebesgue differentiation theorem). Let with Lebesgue measure dm. Show that for almost every , we have , and in particular that .

— Ergodicity —

Combining the mean ergodic theorem with the pointwise ergodic theorem (and with Exercises 7, 8 from the previous lecture) we have

Theorem 3(Characterisations of ergodicity) Let be a measure-preserving system. Then the following are equivalent:

- Any set which is invariant (thus TE=E) has either full measure or zero measure .
- Any set which is almost invariant (thus TE differs from E by a null set) has either full measure or zero measure.
- Any measurable function f with a.e. is constant a.e.
- For any and , the averages converge in norm to .
- For any two , we have .
- For any two measurable sets E and F, we have .
- For any , the averages converge pointwise almost everywhere to .

A measure-preserving system with any (and hence all) of the above properties is said to be *ergodic*.

**Remark 4. ** Strictly speaking, ergodicity is a property that applies to a measure-preserving system . However, we shall sometimes abuse notation and apply the adjective “ergodic” to a single component of a system, such as the measure or the shift T, when the other three components of the system are clear from context.

Here are some simple examples of ergodicity:

**Example 1.** If X is finite with uniform measure, then a shift map is ergodic if and only if it is a cycle.

**Example 2.** If a shift T is ergodic, then so is . However, from Example 1 we see that it is not necessarily true that is ergodic for all n (this latter property is also known as *total ergodicity*).

**Exercise 5.** Show that the circle shift (with the usual Lebesgue measure) is ergodic if and only if is irrational. (*Hint*: analyse the equation for (say) using Fourier analysis. *Added*, Feb 21: As pointed out to me in class, another way to proceed is to use the Lebesgue density theorem (or Lebesgue differentiation theorem) combined with Exercise 14 from Lecture 6.)

**Exercise 6.** Let be a standard Borel probability space. Show that the Bernoulli shift on the product system is ergodic. (*Hint*: first establish property 6 of Theorem 3 when E and F each depend on only finitely many of the coordinates of .)

**Exercise 7.** Let be an ergodic system. Show that if is an eigenvalue of , then , the eigenspace is one-dimensional, and that every eigenfunction f has constant magnitude |f| a.e.. Show that the the eigenspaces are orthogonal to each other in , and the set of all eigenvalues of T forms an at most countable subgroup of the unit circle .

Now we give a less trivial example of an ergodic system.

Proposition 1.(Ergodicity of skew shift) Let be irrational. Then the skew shift is ergodic.

**Proof.** Write the skew shift system as . To simplify the notation we shall omit the phrase “almost everywhere” in what follows.

We use an argument of Parry. If the system is not ergodic, then we can find a non-constant such that Tf = f. Next, we use Fourier analysis to write , where . Since f is T-invariant, and the vertical rotations commute with T, we see that the are also T-invariant. The function depends only on the x variable, and so is constant by Exercise 5. So it suffices to show that is zero for all non-zero m.

Fix m. We can factorise . The T-invariance of now implies that . If we then define for , we see that , thus is an eigenfunction of the circle shift with eigenvalue . But this implies (by Exercise 7) that is orthogonal to for close to zero. Taking limits we see that is orthogonal to itself and must vanish; this implies that and hence vanish as well, as desired.

**Exercise 8**. Show that for any irrational and any , the iterated skew shift system is ergodic.

— Generic points —

Now let us suppose that we have a topological measure preserving system , i.e. a measure-preserving system which is also a topological dynamical system , with the Borel -algebra of T. Then we have the space C(X) of continuous (real or complex-valued) functions on X, which is dense inside . From the Stone-Weierstrass theorem we also see that C(X) is separable.

A sequence in X is said to be *uniformly distributed* with respect to if we have

(15)

for all . A point x in X is said to be *generic* if the forward orbit is uniformly distributed.

**Exercise 9. **Let be a compact metrisable space with a Borel probability measure , and let be a sequence in X. Show that this sequence is uniformly distributed if and only if for all open sets U in X with . What happens if the hypothesis that the boundary of has measure zero is removed?

From Theorem 3 and the separability of C(X) we obtain

Proposition 2.A topological measure-preserving system is ergodic if and only if almost every point is generic.

A topological measure-preserving system is said to be uniquely ergodic if *every* point is generic. The following exercise explains the terminology:

**Exercise 10.** Show that a topological measure-preserving system is uniquely ergodic if and only if the only T-invariant Borel probability measure on T is . (Hint: use Lemma 1 from Lecture 7.) Because of this fact, one can sensibly define what it means for a topological dynamical system to be uniquely ergodic, namely that it has a unique T-invariant Borel probability measure.

It is not always the case that an ergodic system is uniquely ergodic. For instance, in the Bernoulli system (with uniform measure on , say), the point is not generic. However, for more algebraic systems, it turns out that ergodicity and unique ergodicity are largely equivalent. We illustrate this with the circle and skew shifts:

**Exercise 11.** Show that the circle shift (with the usual Lebesgue measure) is uniquely ergodic if and only if is irrational. (*Hint*: first show in the circle shift system that any translate of a generic point is generic.)

Proposition 3.(Unique ergodicity of skew shift) Let be irrational. Then the skew shift is uniquely ergodic.

**Proof.** We use an argument of Furstenberg. We again write the skew shift as . Suppose this system was not uniquely ergodic, then by Exercise 10 there is another shift-invariant Borel probability measure . If we push and down to the circle shift system by the projection map , then by Exercises 10, 11 we must get the same measure. Thus and must agree on any set of the form .

Let E denote the points in which are generic with respect to ; note that this set is Borel measurable. By Proposition 2, this set has full measure in . Also, since the vertical rotations commute with T and preserve , we see that E must be invariant under such rotations; thus they are of the form for some A. By the preceding discussion, we conclude that E also has full measure in . But then (by the pointwise or mean ergodic theorem for ) we conclude that -almost everywhere for every continuous f, and thus on integrating with respect to we obtain for every continuous f. But then by the Riesz representation theorem we have , a contradiction.

Corollary 2.If is irrational, then the sequence is uniformly distributed in (with respect to uniform measure).

**Exercise 11a.** Show that the systems considered in Exercise 8 are uniquely ergodic. Conclude that the exponent 2 in Corollary 2 can be replaced by any positive integer d.

Note that the topological dynamics theory developed in Lecture 6 only establishes the weaker statement that the above sequence is dense in rather than uniformly distributed. More generally, it seems that ergodic theory methods can prove topological dynamics results, but not vice versa. Here is another simple example of the same phenomenon:

**Exercise 12. **Show that a uniquely ergodic topological dynamical system (with the support of the measure equal to the whole space) is necessarily minimal. (The converse is not necessarily true, as already mentioned in Remark 6 of Lecture 7.)

— The ergodic decomposition —

Just as not every topological dynamical system is minimal, not every measure-preserving system is ergodic. Nevertheless, there is an important decomposition that allows one to represent non-ergodic measures as averages of ergodic measures. One can already see this in the finite case, when X is a finite set with the discrete -algebra, and is a permutation on X, which can be decomposed as the disjoint union of cycles on a partition of X. In this case, all shift-invariant probability measures take the form

(16)

where is the uniform probability measure on the cycle , and are non-negative constants adding up to 1. Each of the are ergodic, but no non-trivial linear combination of these measures is ergodic. Thus we see in the finite case that every shift-invariant measure can be uniquely expressed as a convex combination of ergodic measures.

It turns out that a similar decomposition is available in general, at least if the underlying measure space is a compact topological space (or more generally, a Radon space). This is because of the following general theorem from measure theory.

Definition 1(Probability kernel). Let and be measurable spaces. Aprobability kernelis an assignment of a probability measure on X to each in such a way that the map is measurable for every bounded measurable .

**Example 3.** Every measurable map induces a probability kernel . Every probability measure on X can be viewed as a probability kernel from a point to X. If and are two probability kernels from Y to Z and from X to Y respectively, their composition is also a probability kernel, where is the measure that assigns to any measurable set E in Z. Thus one can view the class of measurable spaces and their probability kernels as a category, which includes the class of measurable spaces and their measurable maps as a subcategory.

Definition 2.(Regular space) A measurable space is said to beregularif there exists a compact metrisable topology on X for which is the Borel -algebra.

**Example 4.** Every topological measure-preserving system is regular.

**Remark 5. **Measurable spaces in which is the Borel -algebra of a topological space generated by a separable complete metric space (i.e. a Polish space) are known as *standard Borel spaces*. It is a non-trivial theorem from descriptive set theory that up to measurable isomorphism, there are only three types of standard Borel spaces: finite discrete spaces, countable discrete spaces, and the unit interval [0,1] with the usual Borel -algebra. From this one can see that regular spaces are the same as standard Borel spaces, though we will not need this fact here.

Theorem 4(Disintegration theorem). Let and be probability spaces, with regular. Let be a morphism (thus ). Then there exists a probability kernel such that(17)

for any bounded measurable and . Also, for any such g, we have

-a.e. (18)

for -a.e. y.

Furthermore, this probability kernel is unique up to -almost everywhere equivalence, in the sense that if is another probability kernel with the same properties, then for -almost every .

We refer to the probability kernel generated by the above theorem as the *disintegration* of relative to the factor map .

**Proof.** We begin by proving uniqueness. Suppose we have two probability kernels with the above properties. Then on subtraction we have

(19)

for all bounded measurable , . Specialising to for some measurable set , we conclude that for -almost every y. Since is regular, it is separable and we conclude that for -almost every y, as required.

Now we prove existence. The pullback map defined by has an adjoint , thus

(20)

for all and . It is easy to see from duality that we have for all (where we select a compact metrisable topology that generates the regular -algebra ). Recall that is not quite a measurable function, but is instead an equivalence class of measurable functions modulo -almost everywhere equivalence. Since C(X) is separable, we find a measurable representative of to every which varies linearly with f, and is such that for all y outside of a set E of -measure zero and for all . For all such y, we can then apply the Riesz representation theorem to obtain a Radon probability measure such that

(21)

for all such y. We set equal to some arbitrarily fixed Radon probability measure for . We then observe that the required properties (including the measurability of ) are already obeyed for . To generalise this to bounded measurable f, observe that the class of f obeying the required properties is closed under dominated pointwise convergence, and so contains the indicator functions of open or compact sets (by Urysohn’s lemma). Applying dominated pointwise convergence again and inner and outer regularity, we see that the indicator functions of any Borel set lies in . Thus all simple measurable functions lie in , and on taking uniform limits we obtain the claim.

Finally, we prove (18). From two applications of (17) we have

(22)

for all bounded measurable and . The claim follows (using the separability of the space of all f).

**Exercise 13. **Let the notation and assumptions be as in Theorem 4. Suppose that is also regular, and that the map is continuous with respect to some compact metrisable topologies that generate and respectively. Then show that for -almost every y, the probability measure is supported in .

Proposition 4(Ergodic decomposition). Let be a regular measure-preserving system. Let be the system defined by , , , and , and let be the identity map. Let be the disintegration of with respect to the factor map . Then for -almost every y, the measure is T-invariant and ergodic.

**Proof.** Observe from the T-invariance of (and of ) that the probability kernel would also be a disintegration of . Thus we have for -almost every y.

Now we show the ergodicity. As the space of bounded measurable is separable, it suffices by Theorem 3 and a limiting argument to show that for any fixed such f, the averages converge pointwise -a.e. to for -a.e. y.

From the pointwise ergodic theorem, we already know that converges to outside of a set of -measure zero. By (17), this set also has -measure zero for -almost every y. Thus it will suffice to show that is -a.e. equal to for -a.e. y. Now observe that , so the claim follows from (18) and (21).

**Exercise 14.** Let be a separable measurable space, and let T be bimeasurable bijection . Let denote the Banach space of all finite measures on with the total variation norm. Let denote the collection of probability measures on which are T-invariant. Show that this is a closed convex subset of , and the extreme points of are precisely the ergodic probability measures (which also form a closed subset of ). We remark that one can use these facts to prove some variants of Proposition 4 using Choquet’s theorem, although I do not know of a way to prove this proposition in full generality just from Choquet.

**Exercise 15.** Show that a topological measure-preserving system is uniquely ergodic if and only if the only ergodic shift-invariant Borel probability measure on X is .

## 80 comments

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27 September, 2018 at 4:16 pm

Abhishek KhetanhekDear Professor Tao,

In the second paragraph of the proof of Proposition 4 (ergodic decomposition theorem) there is a line which reads “it suffices by Theorem 3 and a limiting argument …” I am not able to follow the sufficiency. Can you please provide some details. Thank you.

2 October, 2018 at 4:21 pm

Terence TaoOne can find a countable set of bounded measurable functions that is dense in every ; indeed one can take to be rational linear combinations of indicator functions of the countable generating set of the sigma-algebra.

By hypothesis and countable union, we see that for -almost every , the pointwise ergodic theorem holds in for every . By a limiting argument, that shows that the mean ergodic theorem holds in for every , hence by Theorem 3, is ergodic.

7 October, 2018 at 11:58 am

Abhishek KhetanhekThank you for the detailed response.

6 March, 2019 at 8:30 am

Yuping ZhangMy conclusion: The Dr. Terence Tao’s the blue-eyed islanders puzzle only works out if N=2, N=1, and in condition N=0 ( see below)

The example 1) explain 2 colors in the puzzle .

The case, 10 and N=1, and N= is the only person to whose eyes are not green. Rest of the 9 are greens.

Someone announced: ”At least one of the ten has green eyes.”

In the 9th days, CAN all of the 9 greens figure it out their eye color logically? NO

The example 2) explain 2 colors in the puzzle

The case, 10 and N=0 or the entire ten has no green eyes

Someone announced:”At least one of the ten has no green eyes.

In the 10th days, CAN all of the 10 no greens figure it out their eye color logically? YES.

Why? Again!

the crucial key is the entire 10 of them could see the single color in the same amount of people=9, therefore, it’s not leaving each day that tells more and more works out.

But not in Dr. Terence Tao’s the blue-eyed island puzzle.

15 March, 2019 at 8:16 am

Maths studentDear Prof. Tao,

in the proof of uniqueness in the disintegration theorem, do you really mean to say that is regular? Or do you mean that under a suitable metric, is separable?

15 March, 2019 at 8:22 am

Maths studentI see, what you mean is that regularity is in fact a property of the -algebra. Perhaps this should be pointed out.

[See Definition 2 – T.]11 January, 2020 at 7:45 am

Radu ZaharopolDear Professor Tao,

Have you had a chance to see Daniel Worm’s Ph.D. thesis (Leiden 2010) and my 2014 Springer monograph?

In these works we discuss the KBBY (Kryloff-Bogoliouboff-Beboutoff-Yosida) ergodic decomposition. The KBBY decomposition is obtained in a constructive manner, while the decomposition based on the Choquet-Phelps approach is highly nonconstructive.

4 January, 2021 at 4:21 pm

Fausto Di BiaseIn Remark 5 you say that ”regular spaces are the same as standard Borel spaces”, but a countably infinite discrete space is a standard Borel space but it is not compact, hence it is not a regular space. I am confused. Could you please help? Thanks.

4 January, 2021 at 7:35 pm

Terence TaoA countably infinite discrete space is measurably (though not topologically) equivalent to the countable compact set , so it (or more precisely, the measurable space associated to it) is a regular space. (The topology used in the definition of a regular space is not necessarily the one that one may initially wish to assign to the measurable space in question.)

5 January, 2021 at 7:50 am

Fausto Di BiaseThanks a lot!