Let $(\Xi,\mathscr{E})$ be a measurable space, $(\mathbb{R_{+}},\mathfrak{B})$ other measurable space where $\mathfrak{B}$ a $\sigma$-algebra. We consider the measurable space $(\Xi\times\Xi\times\mathbb{R}_{+},\mathscr{E}\otimes\mathscr{E}\otimes \mathfrak{B})$ and $\mathcal{P}(\Xi\times\Xi\times\mathbb{R}_{+})$ the set of all probability measures in $\Xi\times\Xi\times\mathbb{R}_{+}$ and $\mathcal{M}(\Xi\times\Xi\times\mathbb{R}_{+})$ the set of all signed measures generated by $\mathcal{P}(\Xi\times\Xi\times\mathbb{R}_{+})$ and $\mathcal{M}^{+}(\Xi\times\Xi\times\mathbb{R}_{+})$ the set of all positive measures. We assume that $\Xi \subseteq\mathbb{R}^{m}$ and there exists a metric $d$ such that $(\Xi,d)$ is a metric space.
Let $v\geq 0$, $\left\{\widehat{\xi}_{1},\ldots, \widehat{\xi}_{N}\right\}\subseteq \Xi$ y $\varepsilon>0$. We consider the set $$ \mathcal{G}:=\left\{\left. \left(\Pi(\Xi\times\Xi\times\mathbb{R}_{+}),\int g_{1}(\xi,\zeta,\varrho)\Pi(d\xi,d\zeta,d\varrho),\ldots,\int g_{N+2}(\xi,\zeta,\varrho)\Pi(d\xi,d\zeta,d\varrho) \right) \right| \: \Pi\in \mathcal{M}^{+}(\Xi\times\Xi\times\mathbb{R}_{+}) \right\}. $$ where
$$g_{i}(\xi,\zeta,\varrho):=\left\{\begin{array}{ll} \mathbb{1}_{\left\{\zeta=\widehat{\xi}_{i}\right\}}(\xi,\zeta,\varrho) &\mbox{para }i=1,\ldots,N \\ \mathbb{1}_{\left\{\varrho=v\right\}}(\xi,\zeta,\varrho) & \mbox{para }i=N+1\\ d^{p}(\xi,\zeta)+\varrho &\mbox{para }i=N+2. \end{array}\right.$$
The question: I need to show that $b=(1,1/N,\ldots,1/N,1,\varepsilon)$ is inner point of $\mathcal{G}$.
Remark: This problem arises when I try to understand the ideas of the article in this link (you can download it on that web page, it is also in arxiv but this page is not working), specifically the proof of Proposition 4 on page 45, the author says simply that $ \mathcal{G}=\mathbb{R}^{N+3}_{+}$ but the author does not show it, for me is important to prove it, I have tried show it so many times that I am already doubting that this fact not is true. I appreciate any help you can give me on this subject.
Note: The author of the article handles a different notation where he calls $ \mathcal{G}$ as $ \mathcal{Q}_{f} $.
