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 serts 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 Lebesque 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.