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?

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.

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.