Are these two definitions of "uniformly distributed" equivalent?

Question

For an article I am writing, I would like to know that two somewhat different looking conditions are in fact equivalent. Here is the setting. $X$ is a compact (and first countable) metric space and $\mu$ is a Radon probability measure on $X$. That is: $\mu$ is a measure on the $\sigma$-algebra of Borel sets of $X$ (the $\sigma$-algebra generated by the open sets), has total measure one, and is inner regular, that is the measure of any Borel set $B$ is the sup of the measures of the compact subsets of $B$.

Now let $\{x_n\}$ be a sequence in $X$. Let's say that this sequence is "$\mu$-uniformly-distributed-A" if for any open subset $O$ of $X$ $$ \mu(O) =\lim_{N \to \infty} { \#(O,N) \over N}$$ where $\#(O,N)$ is the number of $x_k$ in $O$ for $k = 1, \ldots, N$. (Or, in other words, the measure of $O$ is the "average number of the $x_n$ that are in $O$").

On the other hand, let's say that the sequence is "$\mu$-uniformly-distributed-B" if for any continuous real valued function $f : X \to R$, $$ \int f(x) \, d\mu = \lim_{N \to \infty} {1\over N}\sum_{k = 1}^N f(x_k)$$ (in other words the integral of $f$ is the "average value of $f$ on the $x_n$"). (Note that if we assume this equality not for all continuous functions but rather for all the characteristic functions of open sets, then it reduces to the definition of "$\mu$-uniformly-distributed-A".) So, as you have no doubt guessed, what I want to know is if "$\mu$-uniformly-distributed-A" and "$\mu$-uniformly-distributed-B" are in fact always equivalent.

It is well-known that for the special case where $X = [0,1]$ and $\mu$ is Lebesgue measure the two are equivalent---see for example Theorem B of section 3.5 of Volume 2 of Knuth's "Art of computer programming" --- but I don't see how to generalize the argument there. So does anyone know if this equivalence always does hold, and if so can they direct me to a proof in the literature.

Answer 1

They are not equivalent. Suppose $X = [0,1]$, $\mu$ is a unit mass at 0, and $x_n = 1/n$. This sequence is $\mu$-uniformly-distributed-B, because for any continuous $f$, $\int f(x) \, d\mu = f(0) = \lim_{N \to \infty} \frac{1}{N} \sum_{k=1}^N f(1/k)$.

However, it is not $\mu$-uniformly-distributed-A: take $O = (0,1]$

Answer 2

Actually, for Lebesgue measure on the unit interval, the definition of "uniformly distributed A" does not seem to make sense. Consider any sequence $x_n$ and define $O$ to be the union of all intervals centered at $x_n$ with length $\epsilon_n$. Then all $x_n$ are in $O$ by construction, but we can make $\sum \epsilon_n$ as small as we want!

Answer 3

A sequence is $\mu-$ uniformly distributed -B iff the limit relation A holds for each Borel set $M$ whose boundary has $\mu-$ measure $0.$

Edit

In the meantime I found the following books which have proofs of this theorem:

"Uniform distribution of sequences" by L. Kuipers and H. Niederreiter, Wiley 1974

and P. Billingsley, "Convergence of probability measures", Wiley 1999.

Answer 4

See discussion of "weak convergence" or "narrow convergence" of measures. Let $\mu_n$ be the measure with mass $1/n$ at each of $x_1,\dots,x_n$. Your condition B is what can be taken as the definition for: $\mu_n$ converges narrowly to $\mu$. The equivalent condition in terms of open sets is not your condition A, but rather $$ \mu(O) \le \liminf_{N \to \infty} \frac{\#(O,N)}{N} $$ for all open sets $O$.

Reference: Gilman & Jerison, Rings of Continuous Functions.

Answer 5

On the other hand, A implies B. If A holds for open sets, it also holds for closed sets (simply consider complements on both sides of the equation). Every continuous function can be approximated uniformly by a linear combination of characteristic functions of open and closed sets (just divide up the range of the function into a disjoint union of open and closed intervals).