TITLE: evil derivative
QUESTION [10 upvotes]: I'm looking at the following statement in my textbook:

let $u : (0, 1) → (0, 1)$ the devil's staircase function, aka Cantor-Lebesgue function. Then it's derivative is $u' = 0$ pointwise a.e. 

We obtain $$\int_0^1 u \, \varphi_k' \to -1 \, \, (k \to \infty)$$
  if we choose $(\varphi_k)_k \subset C^\infty_c((0,1))$ suitably with $\varphi_k → \chi_{(0,1)}$ $(k → ∞)$
the distributional derivative does not vanish, hence it can't have a weak derivative in $L^1_{\text{loc}}((0,1))$


I get why the the derivative is zero a.e., and why the weak derivative should be zero, if it existed
but I don't understand the reasoning with the $\varphi_k$!

Why does this converge to $-1$?
And what is "suitably"?

Help is much appreciated!
Edit: I made a mistake in my first version; it should be $\varphi_k'$ in the integral with $\varphi_k \to \chi_{(0,1)}$
Now it's updated correctly

REPLY [4 votes]: This question (and @Umberto P.'s good answer) nicely illustrates how distributions/generalized functions explain that the seemingly paradoxical features of the Cantor-Lebesgue staircase function are not paradoxical after all.
Another approach, is via Sobolev spaces and Sobolev imbedding. That is, for a distribution $u$ such that $u\in L^2[0,1]$ and its distributional derivative $u'$ is also in $L^2[0,1]$, we show not only the Sobolev imbedding $u\in C^{0,{1\over 2}}$ (with Lipschitz index ${1\over 2}$), but that, incidentally, that such functions $u$ do satisfy the fundamental theorem of calculus (which the Cantor-Lebesgue function is designed to fail). Thus, for example, the Cantor-Lebesgue function, while in $L^2$, evidently does not have distributional derivative in $L^2$.