# Locally Lipschitz and differentiable almost everywhere

It is said that an absolute continuous function is differentiable almost everywhere; a locally Lipschitz function is absolute continuous and therefore differentiable almost everywhere.

I am confused about the following example: $$V(x_1,x_2)=|x_1|+|x_2|$$ It is easy to check $V$ is locally Lipschitz. However, $V$ is not differentiable at $x_1=0$ or $x_2=0$. The two (infinite-length lines) sets $x_1=0$ and $x_2=0$ are not of measure zero, right? Then does it conflict with that $V$ is differentiable almost everywhere? Where am I mistaken?

They are of measure zero. It's because you're mistaking length with area. In $\mathbb R^2$, the Lebesgue measure is defined using the $\sigma$-algebra generated by rectangles over which the Lebesgue measure is just the area. You can therefore see that the infinite-length zero-area lines you speak of have measure zero.
Note that if you fix say $x_2$, the function $V(x_1) = |x_1| + |x_2|$ is locally Lipschitz and differentiable almost everywhere (i.e. except at $x_1 = 0$).
• It does help. One more question: in $\mathbb{R}^n$ any set in $\mathbb{R}^{m}$ with $m<n$ is of measure zero, right? In $\mathbb{R}$ a point is of measure zero; in $\mathbb{R}^2$ a line is of measure zero; in $\mathbb{R}^3$ a plane is of measure zero, right? – Shiyu Nov 24 '12 at 4:27
• You can't consider subsets of $\mathbb R^m$ as subsets of $\mathbb R^n$ when $n > m$ simply because these sets are disjoint ; but I see your point. A line has measure zero in $\mathbb R^n$ for $n \ge 2$ and a plane has measure zero in $\mathbb R^n$ for $n \ge 3$, yes. – Patrick Da Silva Nov 24 '12 at 4:54