Is the weak* limit of the average of Dirac measures a continuous measure? Let $X$ be a metric space and $X_0=\{x_1,x_2,...,x_n,... \}$ a countable dense subset of $X$.
Assume that the following weak* limit exists $$\lim\limits_{n \rightarrow \infty}\frac{1}{n}\sum\limits_{i=1}^n\delta_{x_i}=\mu$$ 
Is $\mu$ a continuous (that is, non-atomic) measure?
EDIT: "If $X=[0,1]$ and $X_0$ are the rationals $\mu$ is the Lebesgue measure (it can be prooven with the Portmantea theorem)" is in fact wrong...
 A: First we should note that "continuous" is not the same as "non-atomic". By definition $\mu$ is continuous if $\mu(\{x\})=0$ for every $x$, while $\mu$ is non-atomic if there are no atoms, where $E$ is an atom if $\mu(E)>0$ and every (measurable) subset of $E$ has measure $0$ or $\mu(E)$. So a non--atomic measure must be continuous, but for example the Dieudonne measure on $\omega_1$ is continuous but not non-atomic.
In any case the answer to your question is no. Say $(x_n)$ and $\mu$ are as in the question. Suppose $y_n\to y$ and let $(z_n)$ be the sequence $$x_1,y_1,x_2,y_2,\dots.$$Then $(z_n)$ is dense, but$$\frac1n\sum_{i=1}^n\delta_{z_i}\to\frac12(\mu+\delta_y).$$
New Question, regarding continuous measures on compact metric spaces: If $X$ is a countable compact metric space then there is no continuous probability measure. If $X$ also has no isolated points then of course there is. One could construct a continuous probability measure of the form $\lim\frac1n\sum_1^n\delta_{x_i}$, but the details involved in ensuring that $\mu$ is continuous would be, at least in the construction I have in mind, more or less a copy of the proof that $X$ contains  a Cantor set.
Edit: Just saw a nice clean way to show there is a continuous Borel probability measure on a compact metric space  with no isolated points. Not exactly of the form $\lim\frac1n\sum_1^n\delta_{x_i}$, but in the same spirit: It is the limit of averages of point masses:
Since $X$ has no  isolated points you can define closed balls $B_\alpha=\overline{B(x_\alpha,r_\alpha)}$ for each finite sequence of zeroes and ones $\alpha$ such that $$B_{\alpha,0},B_{\alpha,1}\subset B_\alpha$$and $$B_{\alpha,0}\cap B_{\alpha,1}=\emptyset.$$ Letting $|\alpha|$ denote the length of $\alpha$, define $$\mu_n=2^{-n}\sum_{|\alpha|=n}\delta_{x_\alpha}.$$
If you assume in addition that $r_\alpha\to0$ as $|\alpha|\to\infty$ then $\mu=\lim\mu_n$ exists; if you don't want to worry about that, just let $\mu$ be a weak-* accumulation point of $(\mu_n)$.
Now since $B_\alpha$ is closed it follows that $$\mu(B_\alpha)\ge 2^{-|\alpha|},$$and since $\mu(X)=1$ this shows that in fact $$\mu(B_\alpha)=2^{-|\alpha|}.$$Hence $$\lim_{r\to0}\mu(B(x,r))=0$$for every $x$, so $\mu$ is continuous. (Let $K=\bigcap_n\bigcup_{|\alpha|=n}B_\alpha$, and consider the cases $x\in K$ and $x\notin K$ separately...)
