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EDIT: Let $S$ be a closed orientable 2-dimensional surface equipped with a metric with curvature $\geq \kappa$ in the sense of Alexandrov.

Questions 1. Can one define a measure $K$ on $S$ (thought to be an analogue of the Gauss curvature) satisfying the following properties:

(a) if the metric on $S$ is smooth then $K$ is the usual Gauss curvature times the Lebesgue measure.

(b) If a sequence of orientable surfaces $S_i$ with such metrics (with the same lower bound $\kappa$ on the curvatures) converges to $S$ in the Gromov-Hausdorff sense then $K_i\to K$ weakly (what is weak convergence of measures on different spaces should be made more precise, but I guess it is well known to experts).

(c) Gauss-Bonnet formula: $\int_S K=2\pi \chi(S)$.

(It is very likely that (c) follows from (a)+(b).)

Question 2. If the answer to Question 1 is positive, it seems likely that if $S$ has non-negative curvature which in some open subset is $\geq \kappa>0$ (and $S$ is orientable) then $S$ is homeomorphic to the 2-sphere. Is this consequence known to be true in the context of Alexandrov spaces? (For smooth Riemannian metrics it is well known.)

UPDATE:. The answer to both questions is YES as follows from the answer below by Thomas Richard and the comment by Anton Petrunin.

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    $\begingroup$ For question 2, you need to require that the curvature is not everywhere 0 to exclude the torus. $\endgroup$ – Thomas Richard Jun 11 '15 at 10:58
  • $\begingroup$ Actually in this context, just requiring that the curvature measure is not zero and nonnegative should be enough (which is weaker than what you require, there could be dirac masses in the curvature, think about the surface of a cube for instance). $\endgroup$ – Thomas Richard Jun 11 '15 at 11:12
  • $\begingroup$ You are right, but the notion of the curvature measure is a part of my question. Assuming that it exists, one can weaken the condition as you suggest. But if its existence is false or unknown, then Question 2 still makes sense independently of Question 1. $\endgroup$ – MKO Jun 11 '15 at 11:19
  • $\begingroup$ Existence is known, I'm currently writing an answer about it. $\endgroup$ – Thomas Richard Jun 11 '15 at 11:19
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The answer to question 1 is almost yes. This more or less follows from the work of A. D. Alexandrov on convex surfaces. See "The intrinsic geometry of convex surfaces" by Alexandrov, and also "Intrinsic geometry of surfaces" by A. D. Alexandrov and V. Zalgaller (which is written in a more general context).

Note: the curvature measure is a signed measure.

The only thing missing from these reference is continuous dependence with respect to GH convergence (which was not really in the at that time), however it is shown that the curvature measure is weakly continuous with respect to uniform convergence of the distance.

Another thing is that you need to assume a priori that your underlying topological space is a 2D surface without boundary (a closed disk with its usual metric is a perfectly legitimate alexandrov space).

The construction goes throught the following process (and actually shows why Gauss-Bonnet should hold). Let $abc$ be a geodesic (maybe convex or small enough, I don't remember the details) triangle in $S$ (considered as an simply connected region in $S$). Consider its three angles $\alpha,\beta,\gamma$, and define the excess of $abc$ to be $\alpha+\beta+\gamma-\pi$. On then sets the curvature of the interior region of $abc$ to be the excess (because one wants local Gauss-Bonnet to hold).

Then the tedious part begins, where one shows that there exists a unique measure on borel subsets of $S$ which gives this result on the interior of geodesic triangles.

For your second question, once all of this has been set up, assume that the curvature measure (call it $\omega$) of $S$ is non-negative and not zero, this implies that $\omega(S)>0$. But Gauss-Bonnet tells you that $\omega(S)=2\pi\chi(S)$. This $S$ is an orientable surface with positive Euler characteristic.

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    $\begingroup$ Yes, GH-continuity is missing in the ref, but it follows easily. $\endgroup$ – Anton Petrunin Jun 11 '15 at 22:52

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