Given a metric space $X$, let $B_r(x):=\{y\in X\mid d(x,y)\leq r\}$ be the closed balls. Fix a real number $p>0$. My question now is: Is there a Borel-measure $\mu$ on $X$ such that \begin{align*} \mu(B_r(x))=\big(\operatorname{diam} B_r(x)\big)^p,\qquad r>0,x\in X? \end{align*} I was thinking about the $p$-dimensional Hausdorff measure, but that has the property that if $\mathcal H^s(A)<\infty$ for some set $A$, then $\mathcal H^t(A)=0$ for $t>s$, and this would contradict the above property, right?
Any help is highly appreciated. Thanks in advance.
In general, no. Think of some infinite-dimensional Hilbert space. Choose linearly independent unit vectors $(e_n) _{n \in \mathbb N}$. Inside the ball of center $0$ and radius $1$ (which would have measure $2^p$) you can place an infinity of smaller, identical open balls: each one with center at $\frac 1 2 e_n$ and radius $1 \over 4$. These balls would all be disjoint and have measure $({1 \over 2}) ^p$ (your measure, as it is defined, is translation-invariant). So the union of these balls would have infinite measure, but being a subset of the larger ball of radius $1$ would have measure $\leq 2^p$, which is a contradiction. (This is in fact a classic result: there is no non-trivial locally-finite translation-invariant measure on infinite-dimensional vector spaces.)
As you see, your space should have some concept of dimension, and it should be of finite such dimension. In a purely topological context there are several such notions of dimension, but once you have this it makes more sense to use the Hasudorff measures.