I know that a Riemannian manifold of dimension $n$ is naturally endowed with a volume form induced by the Riemannian metric $\omega=\sqrt{|g|}dx_1\wedge\dots\wedge dx_n$.

Is the same thing true with only topological assumptions? With this I mean: does a metric induce something similar to a volume form in a metric space?

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    $\begingroup$ In a purely topological space there's no way to measure orthogonality. With a metric you have that possibility so there a volume form might exist, at least if the space is finite dimensional. $\endgroup$ – md2perpe Jul 29 '17 at 17:59
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    $\begingroup$ Do you mean that you want to define a measure using a metric? $\endgroup$ – Moishe Kohan Jul 29 '17 at 19:28

Let me first try to make sense of your question. First of all, the notion of a differential form is meaningless for general metric spaces $(X,d)$. However, one can still talk about Borel measures $\mu$ on the topological space $X$ (topologized using the metric $d$). The Borel condition still leaves too much freedom since we did not use the metric $d$ (only the topology). The most common condition these days is the one of a metric measure space which ties nicely $d$ and $\mu$ and allows one to do quite a bit of analysis on $(X,d,\mu)$ similarly to the analysis on the Euclidean $n$-space $E^n$. (The literature on this subject is quite substantial, just google "metric measure space".)

Definition. A triple $(X,d,\mu)$ (where $d$ is a metric on $X$ and $\mu$ is a Borel measure on $X$) is called a metric measure space if the measure $\mu$ is doubling with respect to $d$, i.e. there exists a constant $D<\infty$ such that for every $a\in X, r>0$, we have $$ \mu(B(a, 2r))\le D \mu(B(a,r)), $$ where $B(a,R)$ denotes the closed ball of radius $R$ centered at $a$.

Example. Every closed subset of $E^n$ (equipped with the restriction of the Euclidean metric) is a complete doubling metric space.

The basic existence result for doubling measures $\mu$, proven in "Every complete doubling metric space carries a doubling measure", by J. Luukainen, E. Saksman, Proc. Amer. Math. Soc. 126 (1998) p. 531–534, is the following:

Theorem. A complete metric space $(X,d)$ carries a doubling measure $\mu$ if and only if the metric $d$ is doubling, i.e. there exists a constant $C$ such that every ball of radius $2r$ in $X$ is covered by at most $C$ balls of radius $r$.

The proof of this theorem is not long but by no means trivial.


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