TITLE: On Alexandrov embedding theorem
QUESTION [9 upvotes]: Consider a complete $C^\infty$ Riemannian metric on $\mathbb R^2$ of positive sectional curvature. 

Is the metric embeddable as the boundary of a convex subset of $\mathbb R^3$?
Is the embedding unique? 
Are there generalizations of 1-2 to complete noncompact surfaces of nonnegative sectional curvature?
What are good references for these matters?

UPDATE: 
$\bullet\ $ after doing some reading on the subject I found that the assertion 1 is true in the sense that the surface is isometric, as a metric space, to the boundary of a convex body in $\mathbb R^3$ (as proved by Alexandrov back in 1942). The matter of uniqueness is well-understood.
$\bullet\ $ However, one should not expect the boundary to be smooth, e.g. there are examples of $C^\infty$ metrics of nonnegative curvature on $S^2$ which cannot be isometrically $C^3$-embedded into $\mathbb R^3$. 
$\bullet\ $ If the curvature is positive, then smoothness can be achieved as proved by Pogorelov and Nirenberg (independently in the 1950s). 
$\bullet\ $ Local smooth isometric embedding for nonnegatively curved surfaces was established by Lin in 1985. 
$\bullet\ $ A more recent reference for these matters is the book by Burago and Zalgaller, Geometry III, Encyclopedia of Mathematical Sciences.

REPLY [10 votes]: Is the metric embeddable as the boundary of a convex subset of 3?
YES, it is a limit case of standard Alexandrov's theorem. Moreover one can choose any embedding of cone at infinity and construct the embedding. This is a theorem of Olovyanishnikov --- one of three students of Alexandrov who died in the war.
Is the embedding unique?
NO, but I suspect it is unique once you fixed the convex embedding of the cone at infinity. It might follow from the proof of Pogorelov's theorem but I was not able to check his proof.
Are there generalizations of 1-2 to complete noncompact surfaces of nonnegative sectional curvature?
I'm not sure what you mean --- if it has strictly positive curvature at one point then it is automatically $\mathbb R^2$. If it is $\mathbb R^2$ then it is all the same.