Differentiability of the composition of maps and Differentiability a.e. Let $ f: \mathbb{R} \rightarrow \mathbb{R} $ be non constant Lipschitzian function and $g : \mathbb{R} \rightarrow \mathbb{R}$ be differentiable a.e.in $ \mathbb{R} $. Then it is easy to show that there exists $ A \subset \mathbb{R} $ with positive Lebesgue measure and such that $ g \circ f $ is differentiable at each point of A. I suspect that it is possible to find $ f, g $ in such way that $ g \circ f $ is not differentiable on a set with positive Lebesgue measure. Is it true? If it is so can anyone show an example?
PS. Of course if g is also Lipschitzian $ g \circ f $ is Lipschitzian too!!
 A: You did not assume $g$ continuous; to make the problem more interesting I make this additional assumption. Still, there is such an example. I present the construction in three parts.
1. A function whose set of non-differentiability is  a given Cantor-type set
Let $C$ be a  Cantor-type set, whether of zero measure or positive measure. Follow the process of carving $C$ from a line segment, but instead of just removing a piece of some interval $I$, bend it in a parabolic arc that is as tall as $I$ is long. Like this: 

Let $h$ be the function   whose graph we obtain in this way. To see that it is not differentiable on $C$, note that  for every $x\in C$ and for every scale $\delta>0$, zooming in at that scale uncovers a picture similar to the one above. The graph is substantially non-linear   on every scale. 
2. Lipschitz map squeezing a fat Cantor set into thin one
Let $C_{1/3}$ be the standard middle-third Cantor set; it has zero measure. Let $C$ be the Cantor-type set obtained from $[0,1]$ by making gaps of size $ 3^{-n}/2$   instead of  $3^{-n}$. The measure of $C$ is $1/2$. There is a natural $2$-Lipschitz map $f$ of $[0,1]$ onto itself that sends $C$ onto $C_{1/3}$: namely, stretch the gaps in $C$ by the factor of $2$ and place them where they belong in $C_{1/3}$. This map $f$ is a strictly increasing function, hence invertible.  
3. Conclusion
Let $g = h\circ f^{-1}$. Then 


*

*$g$ is differentiable on the complement of $C_{1/3}$, hence a.e.

*$g\circ f = h$ is not differentiable on $C$, which has positive measure.

*$f$ is Lipschitz, as noted above.

