$A\subset \Bbb{R}$ 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 an absolutely continuous $f$, is it always true that $f(M) $ is meager for every meager set $M$?

The Cantor function is an example of a function that maps the Cantor set (meager) to all of $[0, 1]$ (non meager). But I found that the Cantor function is not absolutely continuous.

One more question, if a function maps meager sets to meager sets, does it satisfy the Lusin-N property? And what about the converse?

I asked this question on MSE. https://math.stackexchange.com/q/4463264/977780

This is my first question on MO.