For a metric space $(X,d)$ and a dimension function $\varphi:[0,\infty)\to[0,\infty)$ we can define a metric outer measure $H_\varphi$ which is $\varphi$-Hausdorff measure. Since it is a metric outer measure, all Borel subsets of $(X,d)$ are further $H_\varphi$-measurable in the sense of the Caratheodory. For example, if $\varphi(r) = r^n$ we would obtain an $n$-dimensional Hausdorff measure.
Let us say that $H_\varphi$ is non-trivial on $(X,d)$ if the following condition holds $$ 0<H_\varphi(B(x,r))<\infty $$ for any $0<r<\infty$, where $B(x,r) = \{y\in X:d(x,y)<r\}$ are open balls in $X$. I wonder which metric spaces admit a non-trivial measure $H_\varphi$, in particular does any separable metric space admit it?
As an example of the metric space which does not admit any non-trivial $\varphi$-Hausfordd measure, let $X$ be an uncountable set and let $d$ be the discrete metric on $X$. If $r>0\implies H_\varphi(B(x,r))>0$, then $H_\varphi$ is equivalent to the counting measure, hence $H_\varphi(B(x,2)) = H_\varphi(X) = \infty$.
The set $[0,1]\cup \{2\}$, with the metric induced from $\mathbb R$, does not admit a nontrivial Hausdorff measure. To obtain a connected, and less trivial example, stick a line segment into a square in $\mathbb R^2$. No matter what $\varphi$ is, either the line segment gets zero measure, or the square (containing infinitely many segments) gets infinite measure.
To rule out the counterexamples given above, one can assume that the space is somehow "homogeneous". Under appropriate homogeneity assumptions a nontrivial Hausdorff measure indeed exists: see Inversion invariant bilipschitz homogeneity by David Freeman. In fact, the conclusion in this paper is stronger: there is $Q>0$ such that $$cr^Q\le H^Q(B(x,r))\le Cr^Q \tag1$$ where $H^Q$ corresponds to $\varphi(d)=d^Q$ and the constants $c,C$ are positive and independent of $x,r$. The property (1) is variously called Ahlfors regularity, David-Ahlfors regularity, or $Q$-regularity. I haven't seen a version of (1) with more general $\varphi$.