TITLE: Must the (continuous) image of a null set be null?
QUESTION [12 upvotes]: Say $E \subset [0,1]$ is a null set. Let $f: [0,1] \rightarrow [0,1] $. Do you think $f(E)$ is a null set or not? Just being curious.
(DEF): A set $A$ is null if given any $\epsilon > 0$, there exists a sequence of intervals $\{I_n\}_{n\geq1}$ such that
$$ A \subseteq \bigcup _{n=1}^{\infty}I_n$$ and $$ \sum |I_n| < \epsilon $$
if $f$ is continuous, is $f(E)$ nullset or not?

REPLY [15 votes]: To have that property it is not enough to be just continuous as many above have pointed out. The Devil's ski slope homeomorphism from $[0,1]$ to $[0,2]$ maps the standard cantor set to a fat cantor set (of measure $1$). We can map $[0,2]$ back to $[0,1]$ by composing with the linear transformation $\frac{x}{2}$ (which maps the measure $1$ cantor set in $[0,2]$ to a set of measure $\frac{1}{2}$ in $[0,1]$). So even being a homeomorphism is not enough. What you seek is for your function to have the Luzin N property (http://en.wikipedia.org/wiki/Luzin_N_property). Absolutely continuous functions, Lipschitz functions, and $C^{1}$ diffeomorphisms are types of functions with this property.