If $M\subset \mathbb{R}^n$ is a $m$-dimensional submanifold, does the set with non-zero gaussian curvature always have $m$-dimensional Hausdorff measure greater than zero?
For a manifold homeomorphic to the sphere, is the integral of the curvature always larger than that of the sphere? or equal?
EDIT: Realizing that gaussian curvature is only defined for two dimensional manifolds, I have to rephrase my question: If I have a n-dimensional manifold $M$ embedded in $R^n$, is there always a significant ((n-1)Hausdorff measure>0) subset of the boundary $\partial M$ so that I can only walk away (infinitesimally) from $M$, in directions of the tangent space? Also for two dimensions my question was not really asking what I meant..
For your second question, any space $M$ homeomorphic to the sphere (and thus a closed $2$-manifold) has $\int_M K = 2\pi \chi(M) = 4\pi = \int_{S^2} K$ by the Gauss-Bonnet theorem. The idea that the integral of the curvature is a topological, rather than geometric, invariant (i.e., independent of the metric) holds in higher dimensions, but defining the right notion of curvature requires a bit of nontrivial machinery.