Okay, so here's a question I was ask forever ago, but to this day, I haven't been able to find a solution the
Let $[a,b] \subset \mathbb{R}$ be compact. Then, if $f \colon F \to \mathbb{R}$ is continuous, show that the following are equivalent, where $E \subset [a,b]$: $m(f(E)) = 0$ whenever $m(E) =0$ iff $f(E)$ is measurable whenever $E$ is measurable.
We begin by showing that $f$ sends $F_{\sigma}$ sets to $F_{\sigma}$ sets. To begin, let $K = \cup_{n=1}^{\infty} F_n$, where $K \subset [a,b]$ and $F_n$ is closed for each $n \geq 1$. Notice that $$ f(K) = f\left( \bigcup_{n=1}^{\infty} F_n \right) = \bigcup_{n=1}^{\infty} f(F_n), $$ and by noting that each $F_n \subset F$ is bounded for each $n \geq 1$, by the continuity of $f$, we see that $f(F_n)$ is compact for each $n$. But then, since each $f(F_n)$ is compact, it follows that each $f(F_n)$ is closed. Thus, we see that $f(K) = \bigcup_{n=1}^{\infty} f(F_n)$ is an $F_{\sigma}$ set.
Now, suppose that $m^*(f(E)) = 0$ whenever $m^*(E) = 0$ for $E \subset [a,b]$. In addition, suppose that $K \subset [a,b]$ is measurable. Then, $K = J \cup W$, where $J, W \subset [a,b]$ are such that $J$ is an $F_{\sigma}-$set and $W$ is a null set. But then, $f(K) =f \left(J \cup W \right) = f(J) \cup f(W)$. But, we show that $f$ sent $F_{\sigma}-$set to $F_{\sigma}-$sets and we assumed that $f$ preserves nullsets, and so $f(W)$ is a null set and $f(J)$ is an $F_{\sigma}-$set. Thus, again, by The Fun Theorem, we see that $f(K)$ is measurable.
I cannot get the other implication. I want to argue by contradiction, however, I'm not sure if this is good.
So suppose there exists a set $E$ such that $m(E) = 0$, and $m(f(E)) \ne 0$. For each $y \in f(E)$, choose one point $x_y \in E$ such that $f(x_y) = x$. Then by replacing $E$ by $\{x_y : y \in f(E)\} \subset E$, we may suppose without loss of generality that $f$ is injective on $E$.
Now if $f(E)$ is not measurable, we are done. Otherwise $m(f(E)) > 0$, so we can find a non-measurable subset $H \subset f(E)$. Then $E' = E \cap f^{-1}(H)$ has measure zero and is hence measurable, but $f(E') = H$ is not measurable.
Let us show how to find $H$. So there exists $n\in \mathbb N$ so that $m(f(E) \cap[-n,n)) > 0$. Let $T_t(x) = ((x+t+n) \bmod 2n) -n$ be the usual circular translation on $[-n,n)$ by $t$. Define an equivalence relation $x \sim y$ if and only if $x-y \in \mathbb Q$, and let $K \subset [-n,n)$ pick one element from each equivalence class of $\sim$. Then $T_q(K) \cap E$ ($q \in \mathbb Q$) is a disjoint countable collection of subsets of $E$. If all of the $T_q(K) \cap E$ are measurable, then ther exists $\tilde q \in \mathbb Q$ such that $T_{\tilde q}(K) \cap E$ has measure greater than $0$. But then $T_q(T_{\tilde q}(K) \cap E)$ ($q \in \mathbb Q$) is a disjoint countable collection of subsets of $[-n,n)$, each having the same positive measure, which is a contradiction. Hence there exists a $q \in \mathbb Q$ such that $T_q(K) \cap E$ is non-measurable.