Understanding an empirical distribution defined by a Dirac measure There is a definition in this paper (equation 2.3) which I cannot seem to grasp.
I'm writing it here in a simpler form:
Let $\delta_a$ be the Dirac measure which assigns probability 1 to the point $a$. Denote measure $\mu \in \mathcal{P}_{ac}(R^d)$ where $\mathcal{P}_{ac}(R^d)$ is the family of all Lebesgue absolutely continuous probability measures on $R^d$. Let $\boldsymbol{X}$ be a random vector, and its empirical observations are $\boldsymbol{X}_1, \boldsymbol{X}_2, ..., \boldsymbol{X}_n\  \stackrel{i.i.d.}{\sim}\ \mu \in \mathcal{P}_{ac}(R^d)$.
Now, in the paper they define the empirical distribution of $\boldsymbol{X}$ as: $$\mu_n^{\boldsymbol{X}} := \frac{1}{n} \sum_{i=1}^n \delta_{\boldsymbol{X}_i},$$
and this is the part I don't get. How does the above sum represent an empirical distribution? is it actually like saying the observations are uniformly distributed because for each observation the value of the above sum will be $1/n$?

It's easiest to think backwards: integrate this PDF (measure) to get to CDF.
For instance, consider PDF $f(X)=(\delta(x)+\delta(x-1))/2$, then integrate to get CDF $$F(x)=\int_{-\infty}^x f(s)ds$$
This CDF is step function with $F(X<0)=0$, $F(0\le x<1)=1/2$ and $F(1\le X)=1$. This is your empirical CDF.
Once you get this one dimensional case then it's easy to extend the intuition to multi-dimensional case that you have. It's a mechanical application of integration, even if it's Lebesgue, the idea's the same.
If you were a physicist then you'd know how Dirac came up with his function: a continuous analogue of the Kroneker delta function, i.e. spiritually the same idea of representing discrete points in continuous function analysis.
