4 1. Higher order Fourier analysis

for all continuous scalar-valued functions f ∈ C(X). Note (from the Riesz

representation theorem) that any sequence is asymptotically equidistributed

with respect to at most one Borel probability measure μ.

It is also useful to have a slightly stronger notion of equidistribution: we

say that a sequence x: N → X is totally asymptotically equidistributed if

it is asymptotically equidistributed on every infinite arithmetic progression,

i.e. that the sequence n → x(qn + r) is asymptotically equidistributed for

all integers q ≥ 1 and r ≥ 0.

A doubly infinite sequence (x(n))n∈Z, indexed by the integers rather

than the natural numbers, is said to be asymptotically equidistributed rel-

ative to μ if both

halves2

of the sequence x(1),x(2),x(3),... and x(−1),

x(−2),x(−3),... are asymptotically equidistributed relative to μ. Simi-

larly, one can define the notion of a doubly infinite sequence being totally

asymptotically equidistributed relative to μ.

Example 1.1.1. If X = {0, 1}, and x(n) := 1 whenever

22j

≤ n

22j+1

for some natural number j and x(n) := 0 otherwise, show that the sequence

x is not asymptotically equidistributed with respect to any measure. Thus

we see that asymptotic equidistribution requires all scales to behave “the

same” in the limit.

Exercise 1.1.1. If x: N → X is a sequence into a compact metric space

X, and μ is a probability measure on X, show that x is asymptotically

equidistributed with respect to μ if and only if one has

lim

N→∞

1

N

|{1 ≤ n ≤ N : x(n) ∈ U}| = μ(U)

for all open sets U in X whose boundary ∂U has measure zero. (Hint: For

the “only if” part, use Urysohn’s lemma. For the “if” part, reduce (1.1) to

functions f taking values between 0 and 1, and observe that almost all of

the level sets {y ∈ X : f(y) t} have a boundary of measure zero.) What

happens if the requirement that ∂U have measure zero is omitted?

Exercise 1.1.2. Let x be a sequence in a compact metric space X which is

equidistributed relative to some probability measure μ. Show that for any

open set U in X with μ(U) 0, the set {n ∈ N : x(n) ∈ U} is infinite, and

furthermore has positive lower density in the sense that

lim inf

N→∞

1

N

|{1 ≤ n ≤ N : x(n) ∈ U}| 0.

In particular, if the support of μ is equal to X, show that the set {x(n) :

n ∈ N} is dense in X.

2This omits x(0) entirely, but it is easy to see that any individual element of the sequence

has no impact on the asymptotic equidistribution.