You misstated Riemann's (original) theorem:
a crucial assumption is that your open subset
is simply connected.

Both theorems can be considered as classification theorems
of Riemann surfaces. The Riemann original theorem says that every simply connected domain in the sphere, whose complement contains at least 2 points
is conformally equivalent to the
unit disk.

"Measurable Riemann theorem" says that a sphere equipped with any Riemannian metric, subject to certain condition (that the Beltrami coefficient $\mu$
has norm <1) is conformally equivalent to the Riemann sphere.

It has simple corollaries that a plane or a disk equipped with a Riemannian metric
satisfying the same condition are conformally equivalent to the plane and disk respectively with the standard metric. (A disk with arbitrary Riemannian metric is a generalization of a simply connected domain in
the plane with the usual metric).

The old, classical name of the "Measurable Riemann theorem" was "Existence and uniqueness theorem for Beltrami equation", or it was called simply by the name of an
author (Korn and Lichtenstein, or Morrey or Boyarski, depending on exact conditions, and the taste of the person who refers). Boyarski's contribution is the very important fact that properly normalized $f$ depends on $\mu$ analytically.

The modern name comes from the paper of Ahlfors and Bers, Riemann's mapping theorem for variable metrics, Ann. Math., 72 2 (1960), 385-404, where they restated the result of Boyarski in the spirit that I outlined above, and emphasized this analytic dependence on $\mu$. Besides $\|\mu\|_\infty<1$,
no condition on $\mu$ is imposed except that it is Lebesgue measurable, and the word "measurable" in the name of the theorem comes from this fact.