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nfrer
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Let f in AC[0,1] monotonic,Prove that if m(E)=0 then m(f(E))=0
A function f is absolutely continuous on [0,1] if for any given ε > 0, there exists a δ > 0 such that for every finite sequence of pairwise disjoint subintervals [ai,bi] of [0,1] whose total length is less than δ, the sum of the absolute values of the differences between the function values at the endpoints of the subintervals is less than ε.
Absolute continuity is a stronger condition than continuity, and it ensures that the function has a nice degree of smoothness. It is often used in the study of integration and differentiation, and is also useful for proving the fundamental theorem of calculus.
If a function is absolutely continuous on [0,1], then it must be monotonic on that interval. However, the converse is not necessarily true - a function can be monotonic but not absolutely continuous.
One way to prove absolute continuity is to show that the function is differentiable almost everywhere on [0,1] and its derivative is integrable on [0,1]. Another way is to use the fact that a monotonic function on a closed interval is absolutely continuous if and only if it is of bounded variation.
Yes, a function can be absolutely continuous on any interval [a,b] as long as its derivative is integrable on [a,b]. Additionally, if a function is absolutely continuous on [0,1], it is also absolutely continuous on any subinterval of [0,1].