Borel measures on $\mathbb{R}$ that satisfy $\mu(A)=\mu(\bar{A})$ for every $A$ 
Describe all Borel measures $\mu : B(\mathbb{R})\rightarrow [0, \infty])$ that satisfy $\mu(A)=\mu(\bar{A})$ for every $A$. ($\bar{A}$ is the closure of $A$).

Well, there are several results I already have but none of them really solved the question. First of all, $\mu(\mathbb{Q})=\mu(\mathbb{R}-\mathbb{Q})=\mu(\mathbb{R})$ so $\mu(\mathbb{R})$ is either $0$ or $\infty$.
The $0$ case is trivial. The other case, tho, is much more intersting. 
The main trick I used now is dealing with sets that are sequences, where the closure contains the limit:$\overline{\left \{ a_n : n\in\mathbb{N} \right \}}={\left \{ a_n : n\in\mathbb{N} \right \} }\cup \left \{ \lim a_n \right \}$
This allowed me to prove the following results (you can use them of course, but it will take a lot of time to explain all of them):
Let's call a real number $x$ "heavy" if $\mu(\left \{ x \right \})>0$. then:


*

*The set of heavy rationals is infinite.

*If $x$ is heavy, then any open set that contains $x$ has measure $\infty$.


Any suggestions?
 A: Neat question!  Here's a first observation: every subset $A\subseteq\mathbb{R}$ has a countable dense subset $B\subseteq A$, and then $\mu(B)\leq \mu(A)\leq \mu(\overline{B})$ forces $\mu(A)=\mu(B)=\sum_{b\in B}\mu(\{b\})$.  It follows that in fact $$\mu(A)=\sum_{a\in A}\mu(\{a\})$$ for any set $A$.  That is, $\mu$ is a "weighted counting measure", weighted by some function $f:\mathbb{R}\to[0,\infty]$ such that $\mu(\{x\})=f(x)$.
However, $f$ can't be just any function.  Indeed, following your idea of using sequences, whenever $(x_n)$ is a sequence of distinct points converging to a limit $x$, we must have either $\mu(\{x\})=0$ or $\mu(\{x_n:n\in\mathbb{N}\})=\infty$.  In terms of $f$, this says either $f(x)=0$ or $\sum_n f(x_n)=\infty$.
That is, whenever $f(x)>0$, any sequence converging to $x$ must have a non-summable sequence of $f$-values.  This is equivalent to saying that whenever $f(x)>0$ there exists an $\epsilon>0$ and a neighborhood $U$ of $x$ such that $f(y)>\epsilon$ for all $y\in U$ (if no such $\epsilon$ and $U$ existed, then by taking $\epsilon=2^{-n}$ and $U$ to be smaller and smaller neighborhoods of $x$ you could find a sequence converging to $x$ whose $f$-values were summable).
Conversely, I claim that if $f:\mathbb{R}\to[0,\infty]$ satisfies the condition above, then if $\mu$ is counting measure weighted by $f$, $\mu(A)=\mu(\overline{A})$ for all $A$.  Indeed, it suffices to consider $A$ such that $\mu(A)<\infty$, which means $\sum_{a\in A}f(a)<\infty$.  But by our hypothesis on $f$, this means that if $x$ is any accumulation point of $A$, $f(x)=0$.  In particular, $f(x)=0$ for any $x\in\overline{A}\setminus A$, so $\mu(\overline{A})=\mu(A)$.
