Must the (continuous) image of a null set be null?

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?

• What is a null set? – Matemáticos Chibchas Aug 30 '13 at 20:05
• Depends on $f$. If $f$ is Lipschitz (absolutely continuous should suffice), then $f(E)$ is a null set. In general, it need not be one. – Daniel Fischer Aug 30 '13 at 20:06
• With null set you mean empty, right? Because Wikipedia thinks differently (from what I thought). – JMCF125 Aug 30 '13 at 20:11
• In the future, please use more descriptive titles. I've edited this one. – Nate Eldredge Sep 2 '13 at 2:54

If we do not put any conditions on $f$, the answer is "not necessarily."
For example the Cantor set is a null set, but there is a one-to-one onto mapping $f$ of the Cantor set to the unit interval.
• And there is a (uniformly) continuous, at most $2$-$1$ mapping $f$ of the Cantor set onto $[0,1]$ as well, so even uniform continuity isn’t enough. – Brian M. Scott Aug 30 '13 at 20:17
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.
• Incidentally, $C^1$ functions on $[0,1]$ are Lipschitz, and Lipschitz functions on $[0,1]$ are absolutely continuous. To see that absolute continuity is strictly stronger than Luzin N, note that if a continuous function on $[0,1]$ is absolutely continuous on each interval $[\varepsilon,1]$ with $\varepsilon>0$, then it has the Luzin N property on $[0,1]$, but need not be absolutely continuous. – Jonas Meyer Sep 4 '13 at 0:20