TITLE: Prove that if A is a non-(Lebesgue)measurable set and $d(A,B)>0$, show that $A⋃B$ is non-measurable
QUESTION [7 upvotes]: Prove that if A is a non-(Lebesgue)measurable set and $d(A,B)>0$, show that $A⋃B$ is non-measurable.

$d(A,B)$ is the inf of distance $d(x,y)$ between two points $x\in A, y \in B$. I have tried to prove by contradiction using definition of measurable sets but was not successful. There are two versions of definitions in my textbook.
These are two definitions on my textbook: a set $M$ is measurable iff. $|M⋂A|_e+|M^C⋂A|_e=|A|_e$ for any $A⊆\Bbb{R}$, or $∀ϵ>0$ there exists an open set $M⊆G$ st. $|G−M|_e<ϵ$.
There is a theorem in my textbook saying that if $d(Ω_1,Ω_2 )>0$ then $|Ω_1⋃Ω_2 |_e=|Ω_1 |_e+|Ω_2 |_e$, where $|* |_e$ is outer measure. I don't know if this is useful.
Thank you!

REPLY [7 votes]: Using slightly heavier machinery.

$d(A,B)=d(\operatorname{cl}(A),\operatorname{cl}(B))$. So both are positive, in particular $\operatorname{cl}(A)$ and $\operatorname{cl}(B)$ are disjoint.
Two disjoint closed sets can be separated by disjoint open sets, since $\Bbb R$ is normal. Fix such $U,V$.
If $X=A\cup B$ is measurable, then $X\cap U$ and $X\cap V$ are measurable.
Contradiction! The proof is complete! Huzzah! Have a cold beer to celebrate.