# Topological analog of the Lusin-N property

$$A\subset \Bbb{R}$$ 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.

• But the Cantor function will become absolutely continuous if you use a positive measure Cantor set. Jun 4, 2022 at 16:06

Let $$K$$ be a nowhere dense closed subset of $$[0,1]$$ of positive Lebesgue measure $$\delta>0$$. Such a set $$K$$ can be obtained using the standard technique for constructing a nowhere dense closed set of positive measure.
Define a function $$f:\mathbb{R}\rightarrow\mathbb{R}$$ by letting $$f(x)=m(K\cap(-\infty,x))$$ where $$m$$ denotes the Lebesgue measure. Then $$f(0)=0,f(1)=\delta$$. Then the function $$f$$ satisfies $$|f(y)-f(x)|\leq|y-x|$$. Therefore, $$f$$ is Lipschitz continuous and therefore absolutely continuous as well.
I now claim that $$f[K]=[0,\delta]$$. Suppose therefore that $$0\leq y\leq\delta$$. Then let $$x_0$$ be the least element with $$f(x_0)=\delta$$. Therefore, whenever $$x, we have $$f(x)<\delta$$. This implies that $$m((x,x_0)\cap K)>0$$ which means that $$x_0$$ is a limit point of $$K$$. Since $$K$$ is a closed set, we have $$x_0\in K$$ as well. Therefore, $$y=f(x_0)\in K$$. We conclude that $$f[K]=[0,\delta]$$ which has positive measure.
Now, every continuous bijection will automatically map meager sets to meager sets. On the other hand, suppose $$f:[0,1]\rightarrow[0,1]$$ is the Cantor function, and $$I:[0,1]\rightarrow[0,1]$$ is the identity function, then $$f+I:[0,1]\rightarrow[0,2]$$ will be a continuous bijection, so $$f+I$$ maps meager sets to meager sets. The function $$f+I$$ is clearly not absolutely continuous since $$f$$ is not absolutely continuous, so does not satisfy the Luzin N property.