Take the function $p: \mathcal{Y}\times \mathcal{V}\rightarrow \mathbb{R}$, where $\mathcal{Y}$ is finite and $\mathcal{V}$ is not finite. This function is the unknown of a linear program and needs to be approximate to get a tractable finite-dimensional linear program.
Suppose $\mathcal{V}\equiv [0,1]$. Then, we can consider the following sieve approximation of $p(y,v)$ using Bernstein polynomials of order $K$: $$ (*) \quad p(y,v)\approx \sum_{k=0}^K \theta_{k,K}^y b_{k,K}(v), $$ where $b_{k,K}(v)\equiv {K \choose k} v^k (1-v)^{K-k}$ is a univariate Bernstein basis, $\theta_{k,K}^y \equiv p(y, \frac{k}{K})$ is its coefficient, and $K$ is finite. Some properties of such approximation are:
(a) $p(y,v)\geq 0$ for each $v$ if and only if $\theta_{k,K}^y\geq 0$ for each $k$
(b) $\sum_{y\in \mathcal{Y}} p(y,v) =1$ is approximatively equal to $\sum_{y\in \mathcal{Y}} \theta_{k,K}^y=1$ for each $k$
Questions:
(1) Uni-dimensional case with arbitrary support: Suppose $\mathcal{V}\equiv [c,d]$ for some arbitrary $c,d$ with $c<d$. How does $(*)$ become (so that properties a,b are maintained)? What if $\mathcal{V}$ is unbounded, e.g., $\mathcal{V}\equiv \mathbb{R}$?
(2) Multi-dimensional case with arbitrary support: Suppose $\mathcal{V}\subseteq \mathbb{R}^\ell$ with $\ell>1$. How does $(*)$ become (so that properties a,b are maintained)?
The question discusses approximations, so in order to get a better answer one would usually first come up with a way to measure how good a given answer is. E.g. suppose we came up with 2 families of functions $b^{1}_{k,K}(v)$ and $b^{2}_{k,K}(v)$. How do we decide which one is better? If there is a class (probability distribution) of functions and a measure of an error we could check which one of $b^1$ and $b^2$ achieves smaller error on average across the given class of functions. Without that one can come up with multiple options and it would be hard to tell which one is better. That said, below I list an option for each of your questions.
Answer to Q(1).1: If $\mathcal{V}=[c,d]$ we can use a map $[c,d]\to [0,1]: \nu \mapsto (v-c)/(d-c)$. Now $p(y,v)$ can be written as $\widetilde p(y,(v-c) / (d-c))$, where $\widetilde p(y, v) = p(y, c + v (d-c))$. Now $\widetilde p$ is a function from $\mathcal{Y}\times [0,1] \to \mathbb{R}$, hence your formula (*) applies. That is, the approximation takes the form
$$p(y,v) \approx \sum_{k=1}^K \theta_{k,K}^y b_{k,K}\left(\frac{v-c}{d-c}\right). \tag{1}$$
Answer to Q(1).2: One option is to use some map $\mathcal{V} \to [0,1]$ and reuse your solution for $[0,1]$. E.g. for $\mathcal{V} = \mathbb{R}$ one such map is $f$ defined by $f(v) = \frac{v-2+\sqrt{4+v^2}}{2v}$, $f(0) = 1/2$. Notice that $f^{-1}(v) = \frac{1}{1-v} - \frac{1}{v}$.
$$p(y,v) \approx \sum_{k=1}^K \theta_{k,K}^y b_{k,K}\left(f(v)\right). \tag{2}$$
Note that in this case the map $v \mapsto b_{k,K}\left(f(v)\right)$ is no longer a polynomial. For half-infinite interval $\mathcal{V} = [c,+\infty)$ you can take $f$ defined by $f(v) = 1 / (v - c + 1)$.
Answer to Q(2): If $\mathcal{V} = \mathcal{V}_1 \times \mathcal{V}_2 \times \cdots \times \mathcal{V}_{l}$, then you can use the basis consisting of $K^l$ functions $a_{k_1,\dots,k_l}(v_1,\dots,v_l) = \prod_{i=1}^{l} a^i_{k_i}(v_i)$. You could also use the same if $\mathcal{V} = \mathcal{V}_1\times\mathcal{V}_2 \times \cdots \times \mathcal{V}_{l}$: just pick the closest point in $\mathcal{V}$ when evaluating $\theta_{k_1,\dots,k_l;K}^{y}$.