Measurability of the image of a continuous function I am working on this exercise.
Assume that $f$ is continuous on $[a.b]$. Let $m$ be Lebesgue measure, and assume that $f: [a,b] \rightarrow \mathbb{R}$. Prove that $f$ satisfies the condition
\begin{align*}
E \subset[a,b], \qquad m(E) = 0 \Rightarrow \; m(f(E)) = 0
\end{align*}
if and only if
\begin{align*}
\text{ for any measurable } A \subset [a,b], \; f(A) \text{ is measurable}.
\end{align*}
 A: Regularity of $m$ allows us to find, for each $n\in \mathbb N$, sets $K_n\subseteq A\subseteq U_n$ s.t. 
$m(U_n)-m(K_n)<1/n,\ K_n$ are compact and $U_n$ open. 
If we take $G=\bigcap U_n$ and $F=\bigcup K_n$ then $m(G\setminus K)=0$ and 
$A=(A\setminus K)\cup K$ so if we can show that $f(K)$ and $f(A\setminus K)$ are measureable, we are done.
Note that $m(A\setminus K)\leq  m(G\setminus K)=0$ so by assumption $f(A\setminus K)$ is measureable. 
To finish, note that as $K_n$ are compact, $f(K_n)$ are also; hence $f(K_n)$ are closed, which means that $f(K)$ is measureable. 
The other direction is not true: let $f$ be the Cantor function and $C$ the Cantor set on $[0,1]$. Recall that $m(C)=0$. Then $f([0,1] \setminus C)$ is countable, and so has Lebesgue measure zero. This means that $m(f(C))=1$.
A: This might be true if there are additional conditions on the $\sigma$-algebras and measures on the domain and codomain of $f$ but as it stands it is not true.
As a counter example take the identity $id\colon [a,b]\to[a,b]$ and equip the domain with the Lebesgue measure and the codomain with the counting measure. Then $id(A)$ is measurable for all $A\subset [a,b]$ but if $A$ is countable then $m(A)=0$ while clearly $m(f(A))\neq0$
