Example of a function that has the Luzin $n$-property and is not absolutely continuous.

The Banach–Zaretsky theorem (page 196) says that a continuous function $f:[a,b]\to\mathbb{R}$ of bounded variation is absolutely continuous if and only if

$$E\subset I \text{ has zero Lebesgue measure }\Rightarrow f(E) \text{ has zero Lebesgue measure }\;\;[\#]$$

I would like see an example of a function that satisfies $[\#]$ but is not absolutely continuous.

Thanks.

Solution 1:

The theorem says that continuity, bounded variation (BV), and # (Property N) imply absolute continuity. Here's how it fails if we have two of the properties:

BV & Property N

Let $f(x) = \operatorname{sign} x$. The range is finite, so $f(E)$ always has zero Lebesgue measure. But $f$ is not absolutely continuous.

Continuity & Property N

Let $f(x) = x\sin (1/x)$, $f(0)=0$. Property $N$ follows from the fact that $f$ is locally Lipschitz in $\mathbb{R} \setminus\{0\}$. But $f$ is not BV, so is not absolutely continuous.

Continuity & BV

Let $f$ be the Cantor function.