Let $\mu$ be a *finite* Borel measure on $\mathbb{R}^n$. I am interested in how does $\mu(B(x,r))$ behave, where $B(x,r)$ is the open ball of radius $r$ centered at $x$. For instance, as far as I recall, for each $\alpha \in [0,n]$, there exists finite $\mu$ so that $\mu(B(x,r)) \sim r^{\alpha}$ for $\mu$-a.e. $x$, which are called dimensionally-regular measures (with constant local dimension $\alpha$).


Here is my question: does there exist *finite* Borel measure$\mu$ such that that $\mu(B(x,r))$ vanishes superpolynomially fast, say, $\sim e^{-\frac{1}{r}}$, i.e., 


$\mu\left(\left\{x: \liminf\_{r \to 0} r |\log \mu (B(x,r)) | > 0 \right\}\right) > 0$?



local behavior of a finite Borel measure
If I am not wrong, the answer is no (even for $\sigma$-finite measures). We can assume we are working with a finite measure $\mu$ in $[0,1]^d$. 


Consider the sets $A\_k =${$ x \ : \ \liminf\_{r\to 0} |\log \mu(B(x,r))| r > 1/k$}. It is enough to show that the sets $A\_k$ have zero measure. 


Now, $A\_k = \bigcap\_N \bigcup\_{n>N} B\_n$ where $B\_n =$ {$x \ : \ \mu(B(x,1/n))< e^{-n/k} $}. 


To see this, notice that for every $x\in A\_k$, there exists $n\_0$ such that for every $n>n\_0$ we have that $1/n |\log \mu(B(x,1/n))| > 1/k$. 


So, it is enough to show that $\sum\_{n>N} \mu(B\_n)\to 0$ as $N\to \infty$. 


For every $n>0$ we can divide $[0,1]^d$ in $C n^d$ cubes whose interiors cover $[0,1]^d$ and such that any of them is has diameter smaller than, say, $1/(10n)$.


Now, $\mu(B\_n) \leq \sum\_{i\in I} \mu(Q\_i)$ where the $Q\_i$ are the cubes which intersect $B\_n$. Now, for each $Q\_i$ with $i\in I$ we have that
 $$\mu(Q\_i) \leq e^{-n/k}$$ 


since it is contained in $B(x,1/n)$ of a point $x\in B\_n$. 


So, $\mu(B\_n) \leq C n^d e^{-n/k} \to 0$ and this implies that $\sum\_{n>N} \mu(B\_n) \to 0$ as $N\to \infty$. 


