Covergence of discretizations to the distribution Let $\mu$ be some probability distribution on $\Bbb R$ with a continuous density function $f$. Let $X = (x_i)$ be a finite subset of $\Bbb R$ and define a discrete distribution $\mu_X$ that gives to $x_i$ probability of $$\frac{f(x_i)}{\sum_j f(x_j)}$$ Are there any conditions on the choice of $X$ to make sure that $\mu_X$ weakly converges to $\mu$ if I add more and more points to $X$?
 A: If you allow that the $(x_i)$ are random variables distributed according to $\mu$ and independent, then it follows for every $g \in L^1(\mu)$ that
$$ \frac{1}{N} \sum_{i=1}^N g(x_i) \to \int g \mu $$
converges in probability. For details, see Law of large numbers.
Edit: Maybe if you take more and more rational points from $\mathbb{Q}$ you might get convergence for bounded, continuous functions.
Edit2: How is this related to your question?
You want to know if $\int g d\mu_n \to \int g d \mu $ for all bounded continues g. In your case we have
$$ \int g d\mu_n  = \sum_{i=1}^n g(x_i) \frac{f(x_i)}{\sum_{j=1}^n f(x_j)}.$$
So your question is what restriction on the sequence $(x_i)$ is necessary so that for every continuous bounded function g we get
$$  \sum_{i=1}^n g(x_i) \frac{f(x_i)}{\sum_{j=1}^n f(x_j)} \to \int g \mu .$$
In the case where $\mu$ is simply the Lebesgue Measue, we have $f=1$ and thus $$\int g d\mu_n = \frac{1}{N} \sum_{i=1}^N g(x_i) .$$
In the case where $\mu$ is arbitrary, notice that  if you pick the points $(x_i)$ all over $\mathbb{R}$, and the points $(y_i)$ distributed according to $\mu$, then you have
$$ \sum_{i=1}^n g(x_i) \frac{f(x_i)}{\sum_j f(x_j)} \approx \frac{1}{n} \sum_{i=1}^n g(y_i).$$ 
