Can we extend an a.e. Lipschitz map defined on a closed subset of R^N to the whole space so that it is still a.e. Lipschitz? I have the following question. Let $A$ be a metrically oriented $n$-dimensional subset of $\mathbb{R}^N$ and $f$ a continuous map from $A$ to $\mathbb{R}^M$. We know that $\operatorname{Lip} f < +\infty$ holds $L^N$-almost everywhere, where $\operatorname{Lip} f$ is the local Lipschitz constant of $f$. Can we then continuously extend $f$ to the whole of $\mathbb{R}^N$ in a way that $\operatorname{Lip} f < +\infty$ holds $L^N$-almost everywhere?
 A: You only need to assume that $A$ is a closed subset of $\mathbb{R}^N$ and then construct an extension of $f$ so that it is locally Lipschitz outside $A$. Something like what I explained in Can we extend a continuous function with keeping Hausdorff dimension? should work (extending by hand using a Whitney decomposition). Now the extended mapping is locally Lipschitz exactly outside the same exceptional set as the original mapping. One has to be careful with the boundary points: if the original mapping was locally Lipschitz at the boundary, the extension is also (because of the way it is constructed).
Edit: Only now I noticed who was asking the question. You can drop by my office to discuss more, if there are any problems with the extension. :)
A: This is not a solution to your problem as I do not know what "metrically oriented" sets are. 
However, you could try to use Kirszbraun's extension construction and see what it gives in the context of your question: 
Kirszbraun's proves that every Lipschitz function $f: A \to {\mathbb R}$ 
defined on an arbitrary subset of ${\mathbb R}^m$ has a Lipschitz extension to 
${\mathbb R}^m$ with the same Lipschitz constant.
M.D. Kirszbraun, Uber die zusammenziehende und Lipschitzsche Transformationen, Fundamenta Math. 22 (1934), p. 77-108.
If you do not read German, you can find here a generalization of Kirszbraun's theorem. 
