TITLE: Sard's theorem for absolutely continuous function
QUESTION [5 upvotes]: Can anyone help me proving Sard's theorem where $ f $ is a real valued absolutely continuous function on $ [a,b] $ that is to prove $ f(A) $ is measure zero where $$ A = \{x\in [a,b]\ |\ f'(x) = 0 \}$$  I need to use the definition of absolute continuity of real functions using estimates on sum and possibly properties like $f$ being differentiable a.e , $ f' \in L^1 $ and validity of fundamental theorem of calculus. I can't figure out ways to replicate or generalize the proof for continuously differentiable functions.

REPLY [4 votes]: Well its kind of weird to answer my own question but this is due to the visitors as pointed out by Willie Wong.  We can see here that for $ x\in A$ there exists $h_x$ such that $ |f(y)-f(x)| \leq \epsilon |y-x| $ for all $y \in [x,x+h_x] $. We can show all such $ [x,x+h_x] $ form a Vitali covering of $A$. So we have finitely many disjoint intervals $ I_1,...,I_k $ whose combined measure has very small difference with $A$ . So I estimate $$\lambda (f(A)) \leq \lambda(f(\cup_k I_k)) + \lambda (f(A\setminus \cup_k I_k )) $$ We can estimate both seperately where first one is obvious and second one by containing the set in a open set with small measure and using fundamental theorem of calculus and uniform integrability as for AC functions $f'\in L^1 $.