A measurable function $F$ sends measurable sets to measurable sets $\iff$ it has the Luzin n property. (In $R$)

Question

The forward direction is pretty clear and has been answered on here. My question, however, is on the backward direction. My idea is to use lusin's theorem. Let $A$ be a measurable set that is bounded. Then by lusins there is a compact set $F_\epsilon$, $F$ is continuous on and $m(A-F_\epsilon)<\epsilon$. Then $A=\cup F_{1/n}\cup B$ where $B$ is a set of measure $0$. Now then F clearly sends $A$ to a measurable set as it sends compact sets to compact and $F(B)$ to a measurable set of measure zero. I assumed that A is bounded to make sure $F_\epsilon$ are bounded. How do i prove the statement in full generality?

Answer 1

$A=\bigcup_n A_n$ where $A_n=A\cap (-n,n)$ and $F(A)=\bigcup_n F(A_n)$. You already know that $F(A_n)$ is measurable for each $n$ so the union is measurable.