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.
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