Continuous functions, null sets and Lebesgue measurable sets so im trying to prove that if i have a continuous function then f transforms null sets in null sets if and only if f transform Lebesgue measurable sets in Lebesgue measurable sets. Anyone has got some advice?? 
I tried using the fact that a Lebesgue measurable set is approximated by an open set from the outside but didn't get very far. 
 A: A continuous function $f$ carries compact sets to compact sets. Since any closed set (I assume you are working on the line) is a countable union of compact sets, $f$ carries closed sets to $F_\sigma$ sets. Thus $f$ carries $F_\sigma$ sets to $F_\sigma$ sets.
A Lebesgue measurable set $E$ can be written as $E = F \cup N$ where $F$ is $F_\sigma$ and $N$ is null. If $f$ carries null sets to null sets then 
$$f(E) = f(F) \cup f(N)$$ is measurable, being the union of an $F_\sigma$ set and a null set.
Conversely, if $f$ carries measurable sets to measurable sets, you can start with a null set $N$. If $f(N)$ is not null it contains a nonmeasurable set $Z$.  But $f^{-1}(Z) \subset N$, which implies $f^{-1}(Z)$ is null. Consequently $f$ carries a measurable set $f^{-1}(Z)$ onto a nonmeasurable set $Z$, contrary to hypothesis. Thus $f(N)$ is null.
A: The continuous image of a null set can fail to be null.
(1). Notational preliminary: 
For $n\in \Bbb N$ let $T_n=\{[(t-1)4^{-n}, t4^{-n}] : 4^n\ge t\in \Bbb N\}$. For $t\in T_n$ let $t'$ be the set of the $4$ members of $T_{n+1}$ that are subsets of $t.$
For $n\in \Bbb N$ let $S_n=\{[(a-1)2^{-n},a2^{-n}]\times [(b-1)2^{-n},b2^{-n}]: a,b \in \Bbb N \land a,b\le 2^{-n}\}.$ 
(2). Consider the Hilbert-Peano space-filling curve $h:[0,1]\to [0,1]\times [0,1],$ which is a continuous surjection with  property that if $t\in T_n$ then  $h(t)\in S_n.$ Let $p_1$ be the projection of $[0,1]\times [0,1]$ to its 1st co-ordinate. Then $f=p_1h$ is a continuous surjection from $[0,1]$ to $[0,1].$
Let $A_1$ be a $2$-member subset of $T_1$ such that $f(\cup A_1)=[0,1].$ Recursively define $A_{n+1}$ from $A_n$ as follows:
$\quad$ For each $a \in A_n$  let $a''$ be a $2$-member subset of $a'$ such that $f(\cup a'')=f(a).$ And let $\quad A_{n+1}=\cup \{a'': a\in A_n\}.$ We have $A_{n+1}\subset A_n$ and $f(\cup A_n)=[0,1]$.
Let $A=\cap_{n\in \Bbb N}(\cup A_n). $
For $y\in [0,1]$ and $n \in \Bbb N$ let $B_n(y)=(\cup A_n) \cap f^{-1}\{y\}.$ Then $B_n(y)$ is a non-empty  closed subset of $[0,1],$ and $B_{n+1}(y)\subset B_n(y).$ 
Therefore $\emptyset \ne \cap_{n\in \Bbb N}B_n(y)=A\cap f^{-1}\{y\}.$
So $f(A)=[0,1].$ But the Lebesgue measure of each $\cup A_n$ is $2^{-n}$ so  $A$ is  Lebesgue-null. 
