Hello and thanks in advance for your time.
Let $\mathcal{S} \subset \mathbb{R}^d$ (we can also choose $\mathcal{S}$ compact if this helps here) for some integer $d$ and let $M(\mathcal{S})$ be the space of finite signed Borel measures on $\mathcal{S}$. We also define some finite set $\mathcal{H}$ where its elements define a projection of $\mathcal{S}$: for $C \in \mathcal{H}$, $\mathcal{S}_C$ is the projection of $\mathcal{S}$ on some subspace of $\mathbb{R}^d$ which is specified by $C$ (it's not relevant to specify how). Now let:
$$ E_1 = M(\mathcal{S}) \times \prod_{C \in \mathcal{H}} M(\mathcal{S}_C) $$ (cartesian product of finite signed Borel measure on $\mathcal{S}$ and every $\mathcal{S}_C$ (in finite number)).
$$\displaystyle E_2 = \prod_{C \in \mathcal{H}} M(\mathcal{S}_C)$$
Now I want to prove that the following linear operator is weakly continuous:
$$ A: \left| \begin{array}{rcl} & E_1 & \longrightarrow &E_2 \\ & (\mu,(\nu_C)_{C \in \mathcal{H}}) & \longmapsto & (\mu_{\vert C} + \nu_C)_{C \in \mathcal{H}} \\ \end{array} \right.$$
where $\mu_{\vert C}$ stands for the marginalization on the subspace $\mathcal{S}_C$.
It seems to me that it is bounded operator hence continuous (not just weakly continuous) but I am very unfamiliar with norms for Borel measure (total variation I guess) and I am not sure if I am not mistaken.
To clarify, I am unsure I can write the following: $$ \vert \vert A \left( (\mu,(\nu_C)_C) \right) \vert \vert_{E_2} = \vert \vert (\mu_{\vert C} + \nu_C)_C \vert \vert_{E_2} = \sum_{C} \vert \vert \mu_{\vert C} + \nu_C \vert \vert_{E_2} \leq \vert \vert (\mu,(\nu_C)_C) \vert \vert_{E_1} $$
EDIT: I think we can consider the following norm on finite Borel measure on $\mathcal{K}$ ($\sigma$-algebra of a compact in $\mathcal{R}^d$): $\vert \vert \mu \vert \vert = \sup_{\mathcal{\sigma \in K}} \mu(\sigma)$ and thus we would have $\vert \vert \mu \vert \vert = \mu(X)$ for $X = \cup_{\sigma \in \mathcal{K}} \sigma$. So then $\sum_C \vert \vert \mu_{ \vert C} \vert \vert \leq \vert \vert \mu \vert \vert$ because $\cup_C X_C \subset X$ where $X_C$ is the corresponding set for $\mathcal{S}_C$ and $X$ for $\mathcal{S}$. Thus:
\begin{align} \vert \vert A \left( (\mu,(\nu_C)_C) \right) \vert \vert_{E_2} & = \vert \vert (\mu_{\vert C} + \nu_C)_C \vert \vert_{E_2} \\ & = \sum_{C} \vert \vert \mu_{\vert C} + \nu_C \vert \vert_{E_2} \\ & \leq \sum_C \vert \vert \mu_{\vert C} \vert \vert + \vert \vert \nu_C \vert \vert \\ & \leq \vert \vert \mu \vert \vert + \sum_C \vert \vert \nu_C \vert \vert \\ & \leq \vert \vert (\mu,(\nu_C)_C) \vert \vert_{E_{1}} \\ \end{align}
Then $A$ is bounded linear operator and thus continuous. I still need to think about it but it may conclude the post. Thanks for your help.