Let $X$ be a metric space, complete and separable or even compact if needed. Fix some index set $I$, possibly uncountable. For each $i\in I$ consider a sequence $(\mu^i_n)$ of Borel probability measures on $X$. If this does matter, one can assume that for some Borel non-negative locally finite measure $\lambda$ for each $i\in I$ and for each $n\in \mathbb{N}$ the measure $\mu^i_n$ is absolutely continuous with respect to $\lambda$. Suppose for each $i\in I$ the sequence $(\mu_n^i)$ converges weakly to a Borel probability measure $\mu^i$. Suppose there is a Borel non-negative finite measure $\nu$ on $X$ that is an upper bound of the family $(\mu^i)$, that is, \begin{equation*} \mu_i(V)\leq \nu(V) \end{equation*} for each $i\in I$ and for each Borel set $V\subseteq X$.
My question is if there are some transparent sufficient conditions for this to hold:
\begin{equation*} \liminf\limits_{n\to +\infty} \bigg(\bigvee\limits_{i\in I} \mu^i_n\bigg)(X)\leq \nu(X), \end{equation*} where $\bigvee$ denotes the lattice supremum of measures. If matters, one can assume also that $I$ is countable. One more possible relaxation is that one can, before taking the supremum of measures, for each fixed $i\in I$ pass first to some subsequence of $(\mu^i_n)$.
The only thought on this I have is that the condition above reminds me reverse Fatou's lemma from the theory of Lebesgue integral, which is about a sufficient condition for this inequality to hold: \begin{equation*} \limsup\limits_{k\to +\infty} \int\limits g_k\leq \int\limits \limsup\limits_{k\to +\infty} g_k, \end{equation*} where $(g_k)$ is a sequence of non-negative measurable functions. The lemma states that such an inequality holds if there is a measurable function $g$ such that $g_k\leq g$ almost everwhere. Is there some analogue of this to the described case?
Will be greatful for any help!