Curvature for $C^{1,\alpha}$-metrics

Question

According to Gromov's Metric Structures for Riemannian and Non-Riemannian Spaces every limit $V_0$ of sequences in the class of manifolds $V$ with $|K(V)| \leq 1$ and $\mathrm{InjRad}(V) \geq \rho > 0$ is a Riemannian $C^{1,1}$-manifold, which according to Gromov's ad hoc definition satisfies the following (among other things, see p. 387):

i. There exists a $C^{1,1}$-atlas (differentiable w/ Lipschitz derivative), and the metric tensor is Lipschitz in these coordinates (using harmonic coordinates I think one should obtain that $g$ is $C^{1,1}$ as well).

ii. The squared distance function is locally $C^{1,1}$.

iii. There exists a bounded measurable quadratic form $B_S$ (one for each hyperplane in $T_v V_0$) satisfying the following tube formula (see p. 372):

For every hypersurface $W\subset V_0$ through $v$ with shape operator $A_0$ and $T_v W = S$, $$ \frac{d}{dt} A^\ast_t \bigg\vert_{t=0} = - (A_0^\ast)^2 + B_S,$$ where $A_t^\ast$ denotes the pullback of the shape operator $A_t$ of the hypersurface $W_t$ which is obtained from $W$ by equidistant translation.

The interesting thing about $B_S$ is that it can be used to define sectional curvature by setting $$ K(\tau_1 \land \tau_2) = - g(B_S(\tau_1),\tau_2),$$ where $\tau_1 \bot S$ and $\tau_2 \in S$.

Question 1: How is B_S obtained? How does one see that it is measurable? I'm assuming that using harmonic coordinates one sees that $A$ is Lipschitz and thus differentiable almost everywhere and then obtains $B_S$ in terms of the derivative of $A$ using the tube formula as definition for $B_S$...

Question 2: Can one obtain a similar definition for sectional curvature if the metric tensor is only $C^{1,\alpha}$ (differentiable with H\"older continuous derivative)?

Answer 1

The following sort of answers the question in case of both sided curvature bounds:

In Convergence of Riemannian Manifold Peters cites a result due to Nikolaev asserting that if a manifold $M$ with a metric tensor $g$ with bounded curvature in the Alexandrov sense admits a domain on which harmonic coordinates are defined, then the metric components $g_{ij}$ are actually of Sobolev class $H^2(\Omega)$. Since harmonic coordinates on approximating manifolds "converge" to harmonic coordnates on the limit manifold, one obtains that the limit metric is of class $H^2(M)$. In particular, its second derivatives and hence the Riemannian curvature tensor and defined outside a Riemannian zero-set.

Since the form $B_S$ (which is, at least in classical surface theory, referred to as Weingarten map) may be defined using $g$ and its first and second derivatives (see Gromov's book or article On Curvature and Sign) one sees how to obtain $B_S$ in the limits space.

As I pointed out, this sort of answers the question. I'm sure that Gromov had a more-geometric-less-analytic method of obtaining the quadratic form. So the question of what Gromov's proof looks like is still open. In fact, I seriously doubt that Gromov has ever had this written down.