Prove that if A is a non-(Lebesgue)measurable set and $d(A,B)>0$, show that $A⋃B$ is non-measurable 
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!
 A: 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.

A: If $d(A,B) > 0$, consider the continuous function
$$
f(x) = \frac{d(x,A)}{d(x,A)+d(x,B)}
$$
Then $f$ is well-defined and continuous, $A \subset f^{-1}(\{0\})$ and $B \subset f^{-1}(\{1\})$. We may use $f$ to construct open sets $U,V$ such that $A\subset U, B\subset V$ and $U\cap V = \emptyset$.
Now suppose $A\cup B$ were measurable, then for any $\epsilon > 0, \exists G$ open such that
$$
|G\setminus (A\cup B)|_e < \epsilon
$$
Now check that $H:= G\cap U$ is open and satisfies $H\setminus A\subset G\setminus (A\cup B)$ and so
$$
|H\setminus A|_e <\epsilon
$$
This is true for any $\epsilon > 0$, so $A$ must be measurable - a contradiction.
A: We can use that last theorem you are mentioning. 
Assume that $d(A,B) >0$ and $A\cup B$ is measurable. For any $X$ subset we have 
$$\mu^{\star}(X) = \mu^{\star}(X\cap(A\cup B)) + \mu^{\star}(X\backslash(A\cup B))$$
(in general, we only have the inequality $\le\,$).
However, since $d(X\cap A, X \cap B) >0$ we have 
$$\mu^{\star}(X\cap(A\cup B))= \mu^{\star}(X\cap A) + \mu^{\star}(X\cap B)$$
and so 
$$\mu^{\star}(X)= \mu^{\star}(X\cap A) + \mu^{\star}(X\cap B)  +\mu^{\star}(X\backslash(A\cup B))$$
Now, we also have 
$$\mu^{\star}(X\backslash A) \le \mu^{\star}(X\cap B)  +\mu^{\star}(X\backslash(A\cup B))$$
so 
$$\mu^{\star}(X) \le \mu^{\star}(X\cap A) + \mu^{\star}(X\backslash A) 
\le \mu^{\star}(X\cap A) + \mu^{\star}(X\cap B)  +\mu^{\star}(X\backslash(A\cup B))$$
and since the extreme inequality is an equality, so is the first one. 
We thus see that $A$ is measurable. 
A: Let $\delta\equiv d(A,B)>0$. For each $x\in A$, define $$V(\delta,x)\equiv\{y\in X\,|\,d(y,x)<\delta\}$$ to be the open ball of radius $\delta$ about $x$. Note that $V(\delta,x)\cap B=\varnothing$, for if $y\in B$, then $$d(y,x)=d(x,y)\geq\inf_{(a,b)\in A\times B}d(a,b)=d(A,B)=\delta,$$ so $y\notin V(\delta,x)$. Next, let $$U\equiv\bigcup_{x\in A}V(\delta,x).$$ Then, $U$, being a union of open balls, is an open (and hence Lebesgue-measurable) set containing $A$ and disjoint from $B$. Now, if $A\cup B$ were measurable, then the following set would also be measurable: $$(A\cup B)\cap U=\underbrace{(A\cap U)}_{=A}\cup\underbrace{(B\cap U)}_{\varnothing}=A\cup\varnothing=A,$$ which would contradict the assumption that $A$ is not measurable. It follows that $A$ cannot be Lebesgue-measurable.
