Evaluation of "concentration of measure function" at zero. Let $(X,d)$ be a metric space equipped with a probability measure $\mu$ (defined on the Borel $\sigma$-algebra on the topology induced by the metric $d$). We define the concentration function of the triple $X,d$ and $\mu$ as follows:
$\alpha_{(X,d,\mu)}(r)=\sup\{1-\mu(A_r):A\subset X\text{ is measurable and }\mu(A)\geq 1/2\}$, $r\geq0$,
where we define $A_r=\{x\in X:d(x,A)<r\}$ (EDIT: I adopt the convention $A_0=A$).
I am interested in the value of the following (sparing you the elementary details that leads to the equality below):
$\alpha_{(X,d,\mu)}(0)=\sup\{\mu(B):B\subset X\text{ is measurable and }\mu(B)\leq 1/2\}$
If there exists a measurable set $A\subset X$ such that $\mu(A)=1/2$,
then it clearly is the case that $\alpha_{(X,d,\mu)}(0)=1/2$.
If there is no set with measure $1/2$, then the situation is not so clear.
I can easily find examples where there is no set of measure $1/2$ and $\alpha_{(X,d,\mu)}(0)$ is not $1/2$ (Dirac probability measure comes to mind), but I am unable to either prove (by finding an example ideally) or disprove that there exists a probability measure on some metric space such that no set has a measure of $1/2$ but $\alpha_{(X,d,\mu)}(0)=1/2$.
 A: It turns out that the image of any finite measure is closed (see here), which implies that $1/2$ (or any other constant $\alpha\in[0,1]$ for that matter) cannot be a limit point of the image of $\mu$ without being itself in the image of $\mu$.
A: Based on your clarification, I think that the question is equivalent to saying the following:
Is it possible to have a sequence of sets whose measure tends upwards to $\frac{1}{2}$, without there being a set who measure is $\frac{1}{2}$?
Clearly, if the sets are nested, this is not possible, not just in a Borel $\sigma$ algebra generated by a metric, but in any measure space.
If we don't have a probability measure, we can take a sequence of disconnected points whose measure tends upwards to $\frac{1}{2}$. If they all have measure above $\frac{1}{4}$, then there could be no set which has measure exactly $\frac{1}{2}$.
Now, maybe you can consider a similar situation, and try and show that a sequence of disconnected points whose total measure is 1, where there are finite subsets whose measure tends upwards towards $\frac{1}{2}$, must have a subset of measure $\frac{1}{2}$.
This is a fact about sums of a sequence of real numbers, rather than about measure theory. I didn't think it through quite yet, so it might be trivial or non-trivial (or false).
But if you can show that there must be a set of measure $\frac{1}{2}$ for the sequence of disconnected points, this might imply the general case. Post back if you manage to carry any of the above out, otherwise I will let you know anything further I think of.
