Example of a Random Measure on $[-a,a]$ which is non-degenerate I just started reading about random measures and I'm trying to get a concrete example going for illustrative purposes.  
Let $a>0$ and let $\mathcal{P}([a,a]^d)$ be the collection of Borel probability measure on the cube $[-a,a]^d$ in $\mathbb{R}^d$; where the topology is induced by the Levi-Prokhorov metric.  
What is an example of a random measure $$\nu:([-a,a]^d,\mathcal{B}([-a,a]^d),P)\rightarrow (\mathcal{P}([-a,a]^d),\mathcal{B}(\mathcal{P}([-a,a]^d))$$ of Borel-measures on the cube $[-a,a]^d$ which satisfies
$$
\begin{align}
& P\left( \omega \in [-a,a]^d: \,
\nu(\omega) \in U
\right)>0;\qquad &(\mbox{for every non-empty open subset $U$ of $\mathcal{P}([-a,a]^d)$})
\end{align}
$$
Here $P$ is a uniform probability measure on the cube $[-a,a]^d$ and $\mathcal{B}([-a,a]^d)$ is the corresponding Borel $\sigma$-algebra (and we are viewing $\nu$ as a random element in $\mathcal{P}([-a,a]^d)$).  
 A: This proves that every distribution $z$ over $[0,1]^d$ is arbitrarily close (in the Levi-Prokhorov metric) to a point mass distribution with a finite number of points and rational probability masses. This is claimed in the other answer. I use $[0,1]^d$ for simplicity, the proof is the same for $[-a,a]^d$.
Fix $d$ as a positive integer and fix $z$ as a distribution over $[0,1]^d$, that is, $z \in \mathcal{P}([0,1]^d)$. Fix $\epsilon>0$.  
For each set $A\subseteq[0,1]^d$ define $A^{\epsilon}$ by 
$$ A^{\epsilon} = \{x \in [0,1]^d : ||x-a||< \epsilon \quad \mbox{for some $a \in A$}\}$$
where $||x|| = \sqrt{\sum_{i=1}^d x_i^2}$ is the standard Euclidean norm. We want to construct a point mass distribution $s$ over $[0,1]^d$ (with a finite number of points and rational probability masses) such that for all Borel measurable sets $A \subseteq [0,1]^d$ we have: 
\begin{align}
z[A] &\leq s[A^{\epsilon}]+\epsilon \\
s[A] &\leq z[A^{\epsilon}] + \epsilon
\end{align}
Construction
Define $i$ as an integer that satisfies $1/i \leq \frac{\epsilon}{2\sqrt{d}}$. Chop $[0,1]^d$ into $i^d$ disjoint subcubes, each with edge size $1/i$, and place a point $x_n$ inside each subcube $n \in \{1, ..., i^d\}$. Let $subcube_n$ denote the set of points in subcube $n \in \{1, ..., i^d\}$. Notice that the open ball of radius $\epsilon$ centered at any point in $subcube_n$  completely contains $subcube_n$. Define 
$$ q_n = z[subcube_n] \quad \forall n \in \{1, ..., i^d\}$$
Let $r_1, ..., r_{i^d}$ be nonnegative rational numbers that sum to 1 and that satisfy 
$$ |r_n-q_n|\leq \frac{\epsilon}{2i^d} \quad \forall n \in \{1, ..., i^d\} \quad (Eq. 1)$$
Let $s$ be the point mass distribution on $[0,1]^d$ defined by points 
$\{x_1, x_2, ..., x_{i^d}\}$ with corresponding masses $(r_1, r_2, ..., r_{i^d})$.
Proof of closeness
Fix a Borel measurable set $A\subseteq [0,1]^d$. Define $\mathcal{I}$ as the set of indices $n\in \{1, ..., i^d\}$ for which $A \cap subcube_n \neq \phi$ (where $\phi$ is the empty set). Define 
$$ \tilde{A} = \cup_{n \in \mathcal{I}} subcube_n$$
We observe that $\tilde{A}$ is Borel measurable (it is just a union of subcubes) and: 
$$ A \subseteq \tilde{A} \subseteq A^{\epsilon} \quad (Eq. 2)$$
Then
\begin{align}
z[A] &\overset{(a)}{\leq} z[\tilde{A}] \\
&= \sum_{n \in \mathcal{I}}z[subcube_n] \\
&\overset{(b)}{\leq} \left[\sum_{n \in \mathcal{I}}r_n\right] + \epsilon/2 \\
&= s[\tilde{A}]+\epsilon/2 \\
&\overset{(c)}{\leq} s[A^{\epsilon}]+\epsilon/2
\end{align}
where inequalities (a) and (c) hold by (Eq. 2);  inequality (b) holds by (Eq. 1). 
Similarly
$$ s[A] \leq s[\tilde{A}] = \sum_{n \in \mathcal{I}}r_n \leq \left[\sum_{n \in \mathcal{I}} z[subcube_n]\right]+\epsilon/2 = z[\tilde{A}]+\epsilon/2 \leq z[A^{\epsilon}]+\epsilon/2$$
$\Box$
A: An English version of the link is here:Levi-Prokhorov metric
Here is an example. 
Fix $d$ as a positive integer. 
For simplicity, we shall use the set $[0,1]^d$ rather than $[-a,a]^d$.  Let $\mathcal{P}([0,1]^d)$ be the set of all probability measures for the sample space $[0,1]^d$ and using the Borel sigma algebra. Let $P \in \mathcal{P}([0,1]^d)$ be the uniform distribution over $[0,1]^d$.

Let $s(i,j)$ for $i \in \{1, 2, 3, …\}$, $j \in \{1, 2, 3, …\}$ represent a countably infinite 
number of probability measures with the following properties: 


*

*$s(i,j) \in \mathcal{P}([0,1]^d)$ for all $i, j \in \{1, 2, 3, …\}$.

*For each $z \in \mathcal{P}([0,1]^d)$ and each $\epsilon>0$ we can find an index $(i,j)$ such that $d(z, s(i,j)) < \epsilon$, where distance is measured in the Levi-Prokhorov metric. 


We shall construct such $s(i,j)$ shortly. 
Now for each $\omega \in [0,1]^d$, construct $v(\omega) \in \mathcal{P}([0,1]^d)$ as follows: 
Construct independent random variables $X=X(\omega)$, $Y=Y(\omega)$ for $\omega \in [0,1]^d$ that are geometrically distributed so that 
$$ P[X=i] = P[Y=i] = 1/2^i \quad \forall i \in \{1, 2, 3, …\}$$
This is easy to do since each $\omega =(\omega_1, ..., \omega_d)\in [0,1)^d$ has infinite precision in its decimal expansion of its first coordinate $\omega_1$: 
$$ \omega_1= 0.\omega_{11}\omega_{12}\omega_{13}... = \sum_{k=1}^{\infty} \omega_{1k} 10^{-k}$$ 
Define
$$ v(\omega) = s(X(\omega), Y(\omega)) \quad \forall \omega \in [0,1]$$
To see that this works, fix $U\subseteq\mathcal{P}([0,1]^d)$  as a nonempty open subset. Fix $z \in U$.  Then there is an $\epsilon>0$ such that $\{r \in \mathcal{P}([0,1]^d) : d(z,r) < \epsilon\}\subseteq U$. 
We know there exists an $(i^*,j^*)$ with $i^*,j^*\in \{1, 2, 3, …\}$ 
such that $d(z,s(i^*,j^*)) < \epsilon$ and so 
$$ (X(\omega),Y(\omega))=(i^*,j^*) \implies v(\omega)=s(i^*,j^*) \in U$$ 
Thus 
\begin{align}
P[\omega \in [0,1]^d: v(\omega) \in U] &\geq P[\omega \in [0,1]^d: X(\omega)=i^*, Y(\omega)=j^*]\\
&=P[X(\omega)=i^*]P[Y(\omega)=j^*] \\
&= (1/2)^{i^*}(1/2)^{j^*}\\
&>0
\end{align}
As desired. 

It remains to construct $s(i,j)$ for each $i \in \{1, 2, 3, …\}$ and $j \in \{1, 2, 3, …\}$.  Given $i \in \{1, 2, 3, ...\}$, chop $[0,1]^d$ into $i^d$ disjoint subcubes, each of edge size $1/i$. Place a point $x_n$ in the center of each subcube $n \in \{1, ..., i^d\}$.  We can enumerate all possible $i^d$-dimensional probability mass functions $(p_1, ..., p_{i^d})$ that have nonnegative rational components that add to 1.  Enumerate these as $\{\vec{P}_{ij}\}_{j=1}^{\infty}$.  For each $j \in \{1, 2, 3, ...\}$ define $s(i,j)$ as the "point mass" probability measure on $[0,1]^d$ with points $x_1, ..., x_{i^d}$ with corresponding  probability masses $(p_1, ..., p_{i^d}) = \vec{P}_{ij}$, where $\vec{P}_{ij}$ is the $j$th vector in our enumeration of $i^d$-dimensional vectors with nonnegative rational components that sum to 1.
Clearly $s(i,j) \in \mathcal{P}([0,1]^d)$ for all $i,j$ and so property 1 holds. To see that these $s(i,j)$ distributions have the desired property 2, for each $z \in \mathcal{P}([0,1]^d)$ and each $\epsilon>0$ we can choose a positive integer $i$ such that $1/i$ is sufficiently small.  The Levi-Prokhorov metric has the property that every distribution in $\mathcal{P}([0,1]^d)$ can be approximated arbitrarily closely by a point mass distribution with $i^d$ points and with rational probability masses (see other answer for a proof of this claim).  $\Box$
