$A\subset \Bbb{R}$ 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.

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

1 Answer 1


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 $x<x_0$, 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.


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