$A\subset \Bbb{R}$ is meager if $A$ can be expressed as a countable union of nowhere dense sets.
Let $f:[a, b]\to \Bbb{R}$ is absolutely continuous i.e for every $\epsilon>0$ , there exists $\delta>0 $ such that whenever a finite sequence of pairwise disjoint sub-intervals $(a_n, b_n) $ of $[a, b]$ satisfies $\sum_{n} b_n-a_n<\delta$ , then $\sum_{n} |f(b_n) -f(a_n)| <\epsilon$.
Given $f$ is absolutely continuous. Is it always true that $f(M) $ is meager for every $M$ meager set?
Cantor function is an example of a function that maps Cantor set(meager) to whole of $[0, 1]$(non meager). But i found that Cantor function is not absolutely continuous.
One more question , if a function maps meager set to a meager set does it satisfy Lusin-N property? And what about the converse.