TITLE: On functions whose derivative equals zero almost everywhere
QUESTION [8 upvotes]: Suppose $f: [0,1] \rightarrow \mathbb{R}$ is continuous everywhere and differentiable almost everywhere in $[0,1]$, and $f'(x)=0$ whenever the derivative exists. Is it true that 
$f(x)$ equals a constant?
It seems like the answer should be yes. However, I'm having trouble finding a version of the fundamental theorem of calculus which would apply to $f$.

REPLY [9 votes]: More dramatically, the Cantor Singular Function $f$ is  increasing, continuous,  $f' = 0$, a.e, $f(0) = 0$ and $f(1) = 1$.
http://en.wikipedia.org/wiki/Cantor_function