Direct proof that f(x)=x sin(1/x) does not satisfy Lusin N condition

Question

Let $f$ be defined as

$$f(x)=\begin{cases} x \sin(\frac{1}{x}) & x\ne 0 \\ 0 & x=0 \end{cases}$$

$f(x)$ is not absolutely continuous so it cannot might not satisfy the Lusin N condition.

Is there a direct proof of that it does not? i.e. I wanted to know how to construct a set of zero measure which does not satisfy Lusin's Condition for this function.

Answer 1

The function sends each set of measure zero to a set of measure zero.

Let $A\subset\mathbb R$ be a null set (i.e. $A$ has measure 0). Then $f(A)=f(A\cap \{0\})\cup f(A\cap(\mathbb{R}\setminus \{0\}))$. The first set in the union has at most one point, so we need only worry about the second. The set $A\cap(\mathbb{R}\setminus \{0\})$ can be expressed as a countable union of sets of the form $A\cap [a,b]$ with $0\lt a$ or $b\lt 0$, and therefore, since the image of a union is the union of the images, $f(A\cap(\mathbb{R}\setminus \{0\}))$ can be expressed as a countable union of sets of the form $f(A\cap [a,b])$ with $0\lt a$ or $b\lt 0$. For each such $a$ and $b$, $f$ is continuously differentiable in a neighborhood of $[a,b]$, and hence the restriction of $f$ to $[a,b]$ is absolutely continuous (e.g. by the fundamental theorem of calculus for $C^1$ functions). Therefore the image of the null set $A\cap[a,b]$ under $f$ is null. Countable unions of null sets are null, so this shows that $f(A)$ is null.