Are Sobolev spaces on non-compact manifolds separable? If $M$ is a Riemannian manifold that is not compact, is it true that the Sobolev spaces on $M$, $W^{k,p}(M)$,  still be separable (for $p < \infty$)?
 A: Yes they are.
Step 1 There exists measurable sections $e_1, e_2, \dotsc, e_m$, where $m = \dim M$, of $TM$ (measurable functions mapping a point $x$ to a vector of its tangent plane $T_xM$) such that for each $x \in M$, $e_1 (x), e_2 (x), \dotsc, e_m (x)$ forms an orthonormal basis of the tangent space $T_xM$.
In a neighbourhood of a given point, it is possible to construct such continuous sections by applying a Gram−Schmidt procedure. Such continuous maps can be patched together to form measurable sections.
Step 2 
For $\alpha \in \mathbb{N}^m$, define to be $\partial^\alpha u$ to be the partial derivative in the basis $e_1, e_2, \dotsc, e_m$ 
The map
$$
  A: u \in W^{k, p} (M) \longmapsto A (u) = (\partial^\alpha u)_{\vert \alpha \vert \le k} \in L^p (M)^\nu,
$$
(where $\nu = \# \{\alpha \in \mathbb{N}^n : \vert \alpha \vert \le k\}$)
is continuous and its inverse defined on $A (W^{k, p} (M))$ is also continuous.
Step 3 The subspace $A (W^{k, p} (M))$ is separable as a subset of the separable space $L^p (M)^\nu$ (since $M$ is a separable measure space because it is a second countable topological space). Since $A (W^{k, p} (M))$ is isomorphic to the space $W^{k, p} (M)$, the space  $W^{k, p} (M)$ is separable.
