Prove: Absolute Continuity Let f in AC[0,1] Monotonic

[nfrer]

Let f in AC[0,1] monotonic,Prove that if m(E)=0 then m(f(E))=0

 

[AKG]

What are AC, E, and m? Why wouldn't you bother to define these?

 

[nfrer]

Definitions

Let f in AC[0,1] monotonic,Prove that if m(E)=0 then m(f(E))=0

ie, f is absolutely continuous in [0,1], m denotes the Lebesque measure and E is a subset of [0,1] with meausre 0.

 

[AKG]

Have you tried anything? For every \epsilon > 0, there exists a countable collection of pairwise disjoint open intervals \mathcal{C} such that

E \subseteq \bigcup _{U \in \mathcal{C}} U

and

\sum _{U \in \mathcal{C}} \mbox{vol}(U) < \epsilon

Absolute Continuity

 

[mathwonk]

first try an easier case: let f be lipschitz continuous, i.e. assume there is a constant K such that |f(x)-f(y)| < K|x-y| for all x,y, in domain f.

then prove f preserves measure zero.