For a metric space, is the measure determined by the metric? I recently asked another question here Do orthogonal functions map to orthogonal functions under a conformal transformation?, but one of the answers vexed me a little. Given some (possibly pseudo-) Riemannian manifold M of dimension n equipped with a metric $g_{\mu\nu}$ we can designate a volume element (or form) via the square root of the metric determinant:
$$dvol=\sqrt{|g|}dx^{1}\wedge..\wedge dx^{n}$$
Such a quantity is invariant under arbitrary coordinate transformations. From another point of view, one often equips a space with a measure $\mu$ so that integrations may be performed. To be similar to above we could consider a volume integral over our manifold:
$$V=\int_{M}\mu dx^{1}\wedge..\wedge dx^{n}$$
Isn't the measure then completely determined by the metric, via a relationship:
$$\mu=\sqrt{|g|}$$
???
 A: A small remark: your question title makes it seem like you are asking about metric spaces, whereas the body of your question is concerned with (psuedo-)Riemannian manifolds. These two notions are not the same, and the notion of a Riemannian metric (inner product) is different from the usual notion of a metric (distance).
So I think you're confusing the Riemannian volume form with general measures. The Riemannian volume form is just a very special measure associated to a Riemannian manifold, but a measure is just anything you can integrate against!
Let $(M,g)$ be a $d$-dimensional Riemannian manifold. Recall that the Riemannian volume form is the unique nowhere vanishing $d$-form $\operatorname{dvol}_g\in\Omega^d(M)$ satisfying $$(\operatorname{dvol}_g)_{p}(e_1,\dots,e_d)=1,$$ for every positive orthonormal basis $\{e_1,\dots,e_d\}$ of $T_{p}M$. This is exactly the volume form you describe in coordinates in your question. This is a very natural measure to consider on a (oriented) Riemannian manifold. Most notably, if $(M,g)$ is isometrically embedded in some Euclidean space then we find that $\int_{M}\operatorname{dvol}_g=\mathcal{H}^{d}(M)$ where $\mathcal{H}^d$ is the $d$-dimensional Hausdorff measure. If the manifold is not oriented then you would consider the associated density (which intuitively just forgets the orientation naturally associated to a volume form). 
The key thing to take a way is that the notion of a measure has nothing to do with the geometry of the underlying manifold, and it certainly does not need to be represented as $\mu dx^1\wedge\cdots\wedge dx^d$ for some smooth function $\mu\in\mathscr{C}^{\infty}(M)$. So to directly answer your question, no the measure is not uniquely determined by the relationship $\mu = \sqrt{|g|}$. There is a Riesz representation theorem for the dual space of continuous functions that reads: Let $X$ be a topological space, then every linear and continuous functional $L:\mathscr{C}(X;\mathbb{R})\to\mathbb{R}$ is represented by a Radon measure in the following way. There exists a unique radon measure $\mu$ such that
$$L(\varphi)=\int_{X}\varphi~d\mu$$ for all $\varphi\in\mathscr{C}(X;\mathbb{R})$. This characterization of measures can help you build some intuition about what measures can be. For example, fix a point $p\in M$ and consider the linear continuous functional $L_p:\mathscr{C}(M;\mathbb{R})\to\mathbb{R}$ defined by $L_p(\varphi)=\varphi(p)$. From the Riesz representation theorem above we deduce that there exists a unique measure, which we will denote as $\delta_{p}$, such that for every continuous function $\varphi$ we have that
$$\varphi(p)=\int_{M}\varphi(x)~d\delta_p(x).$$
I would recommend reading through an introductory text on measure theory and Lebesgue integration to learn these ideas. They specialize to the case when you are working on a Riemannian manifold... but it would be incorrect to believe that the only measure on a Riemannian manifold is the canonical Riemannian volume form! Let me know if you have any questions or want me to elaborate on anything in my answer.
