Approximating Lipschitz Functions by $C^1$ functions According to Evans-Gariepy as a corollary of the Whitney's Extension Theorem we have the following
Theorem (Approximating Lipschitz Functions)
Suppose $f: \mathbb R^n \to \mathbb R$ is Lipschitz continuous. Then for each $\epsilon > 0$, there exists a $C^1$ function $\bar f: \mathbb R^n \to \mathbb R$ such that $\mathcal L^n(\{x|\bar f(x)\neq f(x) \hspace 0,5cm or \hspace 0,5cm D\bar f(x) \neq Df(x)\}) \leq \epsilon$.
Questions: 
a) This is telling us something like we can approximate $f$ with $C^1$ functions on a subset as big as we please. But in fact these approximating functions are exactly equal to our function on a set as big as we need. Is it true than that a Lipschitz function is  $C^1$ almost everywhere?
b) Are there any good examples or applications of this remarkable result?
Thanks in advance!
 A: Lipschitz functions are almost everywhere differentiable (see the second bullet of the wikipedia article). You ask if they're continuously differentiable almost everywhere. It seems that the easy answer would be 
$$
\mu \left( \left\{ x \in \mathbb R^n \;\middle|\; \tilde f(x) = f(x) \text{ or } D \tilde f(x) = D f(x) \right\} \right) \leq \varepsilon \implies \mu \left( \left\{ x \in \mathbb R^n \;\middle|\; D \tilde f(x) = D f(x) \right\} \right) \leq \varepsilon
$$
since the latter set is a subset of the former. (Here, $\mu$ is the Lebesgue measure on $\mathbb R^n$; I assume this is what you mean by $\mathcal{L}^n$.) Now, it seems implicit in writing this down that $Df(x)$ exists for all $x$ in the latter subset, i.e., off a set of measure at most $\varepsilon$. So I'd say the answer to your question is probably "yes", although I don't have the proof of the theorem you cited in mind, so I'm not positive.
A: a) "Lipschitz function is $C^1$ almost everywhere" is not a welldefined statments as noted in the comments. If I am not wrong, you are thinking in lines of why not take a union of sets on each of which f agrees with a $C^1$ function. Since on the overlap the functions coincide we obtain one ultimate function that agrees with f except on a set of zero measure. So, do we get that f agrees with a $C^1$ function outside a set of measure zero? No! A limit of $C^1$ functions is not necessarily a $C^1$ function. For the same reason that limit of continuous functions is not always a continuous function. As an example, let f be the absolute value function on R. It is Lipschitz, but there is no $C^1$ function on R that agrees with f outside a set of measure zero. (Proof needed!) The derivatives of the sequence of the approximating functions from above converge to the discontinuous function that is -1 on negatives and +1 on positive real numbers.
b) There are (or can be) absolutely many applications of this. Informally, in light of this lemma "everything" that holds for $C^1$ maps also holds for Lipschitz maps. For example, area and coarea formulas. Hajlasz and Zimmerman have also used this approximation result to obtain an implicit function theorem for Lipschitz maps (also when they map into $\ell^\infty$!)
Finally, there is a less known proof of this theorem that does not use Whitney's extension theorem.
