Gromov-Hausdroff convergence for Alexandrov spaces

Let $\{X_n\}_{n=1}^\infty$ be a sequence of compact Alexandrov spaces (with curvature $\geq k$) converging to (in the sense of Gromov-Hausdroff convergence) an Alexandrov spaces $X$, and $f_n:X_n\mapsto X$ is the corresponding $\nu-$ approximation. The set of regular points in $X_n$ is denoted by $R_n$.

Question: (1) When $n$ sufficiently large, can we prove that $f_n$ is Lipschitz continuous on $R_n$ ? Or a weaker conclusion: $f_n$ is Lipschitz continuous on some $Y_n\subset R_n$ with $\mathrm{vol}(R_n-Y_n)=o(1)$ ,as $n\to\infty$ ?

(2) If question (1) is not true, can we construct a $g_n:X_n\mapsto X$ (maybe related to f_n), such that $g_n$ is Lipschitz continuous on some $Y_n\subset R_n$ with $\mathrm{vol}(R_n-Y_n)=o(1)$ ,as $n\to\infty$ ?

The approximations $f_n$ (by def.) need not to be continuous, but you can approximate Lipschitz ones. I will give a hint how to prove such thing, as far as I know the statement is not written anywhere.
If $X$ is Riemannian the statement follows from Yamaguchi theorem and nearly the same proof works if all the tangent spaces of $X$ are close to Euclidean.
Now assume $X$ has an isolated strongly singular point $p$. Then the above argument gives an approximation which is Lipschitz approximations outside a fixed neighborhood of $p$. Contract all this neighborhood into $p$ and compose the resulting map with the approximation $X_n\to X$. This way you get a Lipschitz approximation.