Are these two definitions of "uniformly distributed" equivalent? 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.
 A: 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 $O = (0,1]$
A: 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!
A: 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.
A: 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.
A: 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).
