for the lower bound take infimum of f'(x).
I guess by m you mean the Lebesgue measure function?
Well, it is intuitively clear that a function cannot turn a set of Lebesgure-measure zero to a set which has Lebesgue-measure non-zero.
Your example doesn't work because the sine function is differentiable on whole of [0,1].
Think about it: Is the set that you chose where sin is differentiable really a set of measure zero? And think about what you integrate over.
That's not true. The Cantor function on takes a set of Lebesgue measure zero to a set of Lebesgue measure 1. It doesn't matter that is differentiable everywhere. E merely needs to be a subset of of places where it is differentiable. That's besides the point, anyway. I merely was questioning the validity of the first inequality posted by InvisibleMan.
Okay, you are right that the Cantor function does that.
I mentioned the fact that sin is differentiable everywhere because the set you chose for the sin-function isn't a set of Lebesgue-measure zero (interval [0,pi]).
As for the inequality that InvisibleMan posted:
I think there might be some absolute value signs missing.