Balls of minimal variance Consider a metric space $(X, \rho)$ and let's say that $\mu$ is a Borel $\sigma$-finite measure on this space. For every measurable set $A$ its variance is defined as 
$$
V_{\mu, \rho}(A) := \inf_{a'\in A}\int\limits_{A}\rho(a,a')\mu(\mathrm da').
$$
Here if minimum is attained at some $a'$ we can think of the latter as a mass center of $A$ with respect to $\mu$ and $\rho$. Let's call a finite measure set $A$ good if there's no set with the at least that measure and lower variance. That is, $A$ is good iff
$$
V_{\mu,\rho}(A) = \inf_{A':\mu(A') \geq \mu(A)}V_{\mu, \rho}(A').
$$
Intuitively it seems that if $\mu$ and $\rho$ are consistent in some way, then $A$ is good iff it is a $\rho$-ball a.e. that is there exists $x\in X, r\geq 0$ such that $\mu(A\triangle B_\rho(x,r)) = 0$ where $\triangle$ denotes the symmetric difference between two sets.
Are there any results of the latter kind known? In particular, let's say $X = \Bbb R^2$ endowed with the Euclidean metric $\rho$ and Lebesgue measure $\mu$. Some particular questions would be


*

*Is that true that every ball is good?

*Are there any good sets that are not balls a.e.?

*What if $\rho(a,a')$ is a $q$-norm for $q\neq 2$?

*What if $\rho$ is a Euclidean metric, but $\mu$ has a density of a standard normal random variable?
 A: I think I have a fairly general answer to this question. Let's say $A$ is some measurable set of finite measure and variance and there exists $\hat a$ at which the minimum of the variance is attained. Consider $r$ satisfying the following condition $\mu(B_\rho(\hat a,r)) = \mu(A)$ and denote this ball simply by $B$. Then for $A' := A\setminus B$ and $B':=B\setminus A$ we have $\mu(A') = \mu(B')$ and hence
$$
\begin{align}
V_{\mu, \rho}(A) &= \int_A\rho(\hat a, a)\mu(\mathrm da) = \int_{A\cap B}\rho(\hat a, a)\mu(\mathrm da) + \int_{A'}\rho(\hat a, a)\mu(\mathrm da) 
\\
&\geq \int_{A\cap B}\rho(\hat a, a)\mu(\mathrm da) + r\mu(A') = \int_{A\cap B}\rho(\hat a, a)\mu(\mathrm da) + r\mu(B')
\\
&\geq \int_{A\cap B}\rho(\hat a, a)\mu(\mathrm da) + \int_{B'}\rho(\hat a, a)\mu(\mathrm da) = \int_A\rho(\hat a, a)\mu(\mathrm da) = V_{\mu, \rho}(B).
\end{align}
$$
Here we show that for any set $A$ there exists a ball of the same measure with at most the same variance, and we only need the following assumption: existence of $\hat a$ and existence of $r$. The latter can be just covered by a stronger condition that there are balls of any measure from $0$ to $\mu(X)$, and I think existence of $\hat a$ - which does not necessarily has to be an element of $A$ should always be true.
One more thing, regarding existence of non-a.e.-ball good sets: for them $\mu(A') > 0$, and since for this set $\rho(\hat a,a)>r$ we get all inequalities in the big formula above to be strict, hence $A$ is not good.
