Is this Proof Concerning Measure Theory Valid? So I know it is true that if $\mu(E)\gt0$ (where $\mu$ is Lebesgue measure of $\mathbb{R}$) then there exists $z,\ x,\ y$ such that $x,y,z\in E $, $x\neq y$ and $\frac{x+y}{2}=z$.
I was just curious about the most efficient way to prove it.  I've seen this, but am not 100% confident this works and would appreciate some commentary on ...  
There exist the sequence of functions $\chi_{E_n} \rightarrow \chi_E$ such that $$\int_{\mathbb{R}}\chi_{E_n} = E_n$$ where $\chi_{E_n}$ is a sequence of characteristic functions of intervals.  With Dominated Convergence, $$\lim_{n\to \infty} \int \chi_{E_n} = \int \lim_{n \to \infty} \chi_E = \mu(E).$$
But does it work to define
$$Z_n= \left\{z\mid z=\frac{x+y}{2} ;x,y,z\in E_n\right\}$$
and
$$Z= \left\{z\mid z=\frac{x+y}{2} ;x,y,z\in E\right\}$$ and say $Z_n \to Z$? 
I believe it works but am having some uneasiness with stating it as fact.
 A: Perhaps something like this might work : Consider the function $f:E\times E \to \mathbb{R}$ given by
$$
f(x,y) = \frac{x+y}{2}
$$
Your goal is to prove that $f^{-1}(E)\setminus\Delta(E) \neq \emptyset$ where $\Delta(E) = \{(x,x) : x\in E\}$


*

*Note that $\Delta(E) \subset \mathbb{R}^2$ must have measure zero. Thus, it would suffice to prove that $f^{-1}(E)$ has positive measure.

*First prove this when $E$ is an open set. This should be true because $f$ is continuous, so $f^{-1}(E)$ is open.

*For any general $E$, find an $G_{\delta}$ set $U$ and set $Z$ of measure zero such that
$$
U = E\sqcup Z
$$

*Check that $f^{-1}(Z)$ has measure zero (see https://mathoverflow.net/questions/128477/preimage-of-zero-measure-sets).
Edit: Just realized $U$ is a $G_{\delta}$ set, not open. Perhaps this might not work after all :(
A: Let $\xi$ be a Lebesgue density point of $E$, that is $\frac{\mu(E \cap B)}{\mu(B)} \rightarrow 1$ as $\mu(B) \rightarrow 0$ and $B \ni \xi$. Let $B_0$ an admissible ball such that, for some small $\epsilon$, $\mu(B_0 \cap E) \geq (1 - \epsilon)\mu(B_0)$. Consider $x_0 \in B_0 \cap E$ such that it is true that $\frac{x_0 + y}{2} \notin E$ for any $y \in E$; then the map $f \colon y \mapsto \frac{x_0 + y}{2}$ is injective and has the property that $f(E \cap B_0) \cap E = \emptyset$ and $f(B_0) \subset B_0$. Thus 
$\mu(B_0) \geq \mu((E \cap B_0) \cup f(E \cap B_0)) = \mu((E \cap B_0)) + \mu( f(E \cap B_0)) \geq (1-\epsilon) \mu(B_0) + \frac{1 - \epsilon}{2} \mu(B_0)$
if $\epsilon$ is sufficiently small, the inequality is false. The fact that $\mu(E) > 0$ implies the existence of some Lebesgue density points, hence the argument above applies and gives a contradiction.
