Example of a set $Y$ that has zero Lebesgue measure and a continuous function $f$ such that $f(Y)$ is not a set of zero Lebesgue measure. Could someone give me an example of a set $Y\subset \mathbb{R}$ that has zero Lebesgue measure and a continuous function $f:X\subset \mathbb{R}\to\mathbb{R}$ such that $Y\subset X$ and $f(Y)$ is not a set of zero Lebesgue measure?
Thanks.
 A: The devils staircase on the Cantor set. The Cantor set is a null set and its image is $[0,1]$
The Cantor set can be written as all $x$ of $\mathbb{R}$ such that 
$$x=\sum_{n=1}^\infty \frac{a_n}{3^n}$$
where $a_n\in\{0,2\}$ and your functions maps $x$ to 
$$f(x)=\sum_{n=1}^\infty \frac{a_n}{2^{n+1}}$$
when $x$ is in the Cantor set and the trivial extension on $[0,1]$ (It is constant outside the Cantor set).
A: This is relates very closely to the Luzin N property, which informally states an absolutely continuous function maps measure zero sets to measure zero sets. 
Formally, for $f:X\rightarrow Y$ (let's consider $X,Y\subset \mathbb{R}$ for simplicity), and $\mu(X)=0$, we have $\mu(f(X))=0$. Such a function is said to possess the Luzin N property. As it turns out, a function is absolutely continuous if and only if it possesses this property. Consider the cantor function $f:C\rightarrow [0,1]$, where $C$ denotes the standard ternary Cantor set. Note that $\mu(C)=0$ (measure of the cantor set is $0$) but $f(x)$ maps the Cantor set onto the interval $[0,1]$ and so $\mu(f(C))=1$. The Luzin N property fails, and as can be seen from many other proofs, Cantor's function is not absolutely continuous. 
