# everywhere differentiable function whose derivative is 0 almost everywhere is a constant

There are examples showing that functions with almost everywhere 0 derivative can be increasing. However in those examples, functions are not differentiable everywhere. In fact, invoking theorem 7.21 from Rudin's Real and Complex Analysis, I can deduce that if a function $f$ is differentiable everywhere and its derivative equals $0$ a.e., then $f\equiv constant$. However, I'm wondering if there is some easier proof of such statement, since the proof of theorem 7.21 is quite weird to me. Is there any other theory that I can use to prove the statement?

• @Rahul no, there doesn't. That's equivalent to the question the OP is inquiring about. – Stella Biderman Apr 16 '17 at 22:23