I have the following question: If $A$ is a metrically oriented $n$dimensional subset of $\mathbb{R}^N$ and $f$ is a continuous map from $A$ to $\mathbb{R}^M$ . We know that $\mathrm{Lip} f < +\infty$ $L^N$almost everywhere, can we then continuously extends f to the whole $\mathbb{R}^N$ such that $\mathrm{Lip} f < +\infty$ $L^N$almost everywhere? Here $\mathrm{Lip}$ is the local lipschitz constant of $f$.

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. :) 


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. 77108. If you do not read German, you can find here a generalization of Kirszbraun's theorem. 

