(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](http://en.wikipedia.org/wiki/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](http://en.wikipedia.org/wiki/Proximity_space) 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) ?



Weak convergence of measures on non-metrizable spaces
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.


