Let $\mathcal{K}\subseteq 2^{X}$ be a family of subset such that $\emptyset\in \mathcal{K}$ and let $\nu\colon\mathcal{K}\to [0,+\infty]$ an application such that $\nu(\emptyset)=0.$ We denote with $$\mathcal{R}_{\mathcal{K}}\big(E\big)=\bigg\{\{I_k\}_{k\in\mathbb{N}}\subseteq\mathcal{K}\;\bigg|\;E\subseteq\bigcup_{k\in\mathbb{N}}I_k\bigg\}.$$ We define $$\mu^{(\mathcal{K},\nu)}\big(E\big):=\inf\bigg\{\sum_{k\in\mathbb{N}}\nu(I_k)\;\bigg|\;\{I_k\}\in\mathcal{R}_{\mathcal{K}}\big(E\big)\bigg\}\quad\text{if}\quad\mathcal{R}_{\mathcal{K}}\big(E\big)\ne\emptyset$$ and $$\mu^{(\mathcal{K},\nu)}\big(E\big):=+\infty\quad\text{if}\quad \mathcal{R}_{\mathcal{K}}\big(E\big)=\emptyset.$$
Question. If I know that $\mathcal{R}_{\mathcal{K}}\big(E\big)\ne \emptyset$, can I conclude that $\mu^{(\mathcal{K},\mu)}\big(E\big)<+\infty?$
For me the answer is no. Because we do not know that every series is convergent. That is: we suppose that $\mathcal{R}_{\mathcal{K}}\big(E\big)\ne\emptyset$, and we suppose that exists one $\{I_k\}\in\mathcal{R}_{\mathcal{K}}\big(E\big)$, such that $\sum_{k\in\mathbb{N}}\nu(I_k)=+\infty$, in this case what can we say about $\mu^{(\mathcal{K},\nu)}\big(E\big)$?
Thanks!
In general, the answer is indeed no, and you can see this with a very simple example. Just take $\mathcal K = \{\emptyset, X\}$ and $\nu(\emptyset)=0$, $\nu(X)=+\infty$. Then for any non-empty set $E\subset X$ we have $\mathcal R_{\mathcal K}(E)=\big\{ \{X,X,\ldots\} \big\}\neq \emptyset$ and $\mu^{(\mathcal K, \nu)}(E)=+\infty$.
Note that in this case $\mathcal K$ is a $\sigma$-field and $\nu$ is a measure, so there is a lot more structure than in your general setting, but the assertion is still false.