Can an arbitrary metric on a surface be approximated in a given conformal class? Let $X$ be a closed Riemann surface, and let $d: X \times X \to \mathbb{R}_{\geq 0}$ be a distance function, i.e. $d$ is symmetric, satisfies the triangle inequality, and $d(x,y)=0$ iff $x=y$. Assume moreover that $d$ is continuous with respect to the manifold topology and induces the manifold topology on $X$. 
Is is true that for all $\epsilon>0$ there a Riemannian metric in the conformal class of $X$ whose distance function differs from $d$ by at most $\epsilon$?
If not, can one characterize the metrics on $X$ that can be approximated by Riemannian metrics in the conformal class of $X$?
 A: Riemannian metrics are intrinsic. This means that for every $x,y$ and every
$t, 0<t<d(x,y)$ there exists $z$ such that $d(x,z)+d(z,y)=d(x,y)$.
For example the standard spherical metric is intrinsic, while the chordal metric is not and thus cannot be approximated by a Riemannian metric.
Intrinsic metrics satisfying certain condition, too complicated to be stated here, can be approximated by Riemannian metrics in the same conformal class.
This is a result of Reshetnyak published only in Russian in 1960.
But here is his detailed survey in English:
Yu. G. Reshetnyak, "Two-dimensional manifolds of bounded curvature”, 3–163, 245–250, in the book:
Geometry. IV.
Nonregular Riemannian geometry. A translation of Geometry, 4 (Russian), Akad. Nauk SSSR, Vsesoyuz. Inst. Nauchn. i Tekhn. Inform., Moscow, 1989 [MR1099201]. Translation by E. Primrose. Encyclopaedia of Mathematical Sciences, 70. Springer-Verlag, Berlin, 1993. vi+250 pp. ISBN: 3-540-54701-0 
EDIT. If the metric is not intrinsic, or intrinsic and does not satisfy the Aleksandrov-Reshetnyak's condition, then what is the conformal class? 
