Weak convergence of measures on non-metrizable spaces

Question

(ZF + Countable Choice)




Let $\langle X,\mathcal{T} \hspace{.06 in} \rangle$ be a second-countable Hausdorff space. Let $\mu$ be a Borel measure on $X$.
Let $\langle I,\leq_I \rangle$ be a directed set, and let $\{\mu_i : i\in I\}$ be a collection of Borel measures on $X$.
Are the following are equivalent?


$1. \qquad$ For all open subsets $U$ of $X$, $\hspace{.12 in} \mu(U) \hspace{.08 in} \leq \hspace{.08 in} \displaystyle\liminf_i \hspace{.06 in} \mu_i(U)$

$2. \qquad$ For all closed subsets $C$ of $X$, $\hspace{.12 in} \displaystyle\limsup_i \hspace{.06 in} \mu_i(C) \hspace{.08 in} \leq \hspace{.08 in} \mu(C)$



Assume $\delta$ is a proximity relation on $X$ that induces $\mathcal{T} \hspace{.05 in}$. $\hspace{.04 in}$ Are the following equivalent?


For all Borel subsets $A$ and $B$ of $X$, if $\quad A \; \; \delta \; \; (X-B) \quad$ is false, then

$3. \qquad \mu(A) \hspace{.08 in} \leq \hspace{.08 in} \displaystyle\liminf_i \hspace{.06 in} \mu_i(B)$

$4. \qquad \displaystyle\limsup_i \hspace{.06 in} \mu_i(A) \hspace{.08 in} \leq \hspace{.08 in} \mu(B)$


Is (1 and 2) equivalent to (3 and 4) ?

Answer 1

Allowing infinite measures will defeat equivalence of 1 and 2. Say Hausdorff measures of various dimensions $<1$ on $\mathbb R$. All nonempty open sets have measure $\infty$. But closed sets can have interesting values.

On the other hand, for finite measures, then of course 1 and 2 are equivalent, by taking complements.