Suppose f:[0,1]->R is absolutely continuous and |f'(x)|<M almost everywhere in [0,1]. Prove that f is Lipschitz. I wrote up the following proof and got significant deductions: By the Mean Value Theorem, for all x,y in [0,1], there exists c between x and y such that |f(x)-f(y)/(x-y)|=|f'(c)|<M. That is, for all x,y in [0,1], |f(x)-f(y)/(x-y)|<M. Then |f(x)-f(y)|<M|x-y|, thus f is Lipschitz. QED What is wrong with my proof? (Since f is AC, I know that f is differentiable almost everywhere in [0,1]. Is that not enough to invoke the mean value theorem? If not, how do I prove this theorem?) Thank you.