If $E$ has positive measure, is it necessarily true that $x = \frac 12(y + z)$ for distinct $y, z\in E$ Working in $\Bbb R$ and assuming that $E$ has positive Lebesgue measure, does there necessarily exist $x \in E$ s.t. $x = \frac 12(y + z)$ for distinct $y, z \in E$? I think a sufficient condition that could be provable is the following: if $f(x) = m(\frac E2 \cap (x - \frac E2)) > 0$ for some $x \in E$, then there are two distinct $y_1, y_2 \in E/2$ s.t. $y_1 +z_1 = x$ and $y_2 + z_2 = x$ for $z_1, z_2 \in E/2$. Because the $y_i$'s are distinct, so are the $z_i$'s and we are done. It remains to show that $f(x) \neq 0$ on all of $E$. Noting that
$$\chi_{x - E/2}(y) = 1 \Leftrightarrow x = y +z/2, z \in E \Leftrightarrow \chi_{y+E/2}(x)$$
we express $f(x) = \int \chi_{E/2}(y) \chi_{x - E/2}(y)\text d y= \int \chi_{E/2}(y) \chi_{y + E/2}(x)\text d y$ and apply Fubini's Theorem to compute
\begin{align}
\int_E f(x) \text d x &= \int_E \int_\Bbb R \chi_{E/2}(y)\chi_{y+E/2}(x)\text d y \text d x
\\
&=\int_\Bbb R\chi_{E/2}(y)\int_\Bbb R \chi_{E \cap (y+E/2)}(x)\text dx \text d y
\\
&= \int_{E/2}m(E \cap (y+E/2))\text dy
\end{align}
It feels to me like this last integral should be bounded below by $m(E)$, but I have no idea how to show this. Am I on the right track or is there a simpler solution?
 A: There is an easier solution I found in this question which I will summarize for my own future reference. By the Lebesgue Density Theorem, we know there exists some $x \in E$ with density equal to $1$. In other words, there is $\epsilon > 0$ s.t.
$$ \frac{m(E \cap (x - \epsilon, x + \epsilon))}{2\epsilon} > 1/2 \Rightarrow m(E \cap (x - \epsilon, x + \epsilon)) > \epsilon $$
Expressing the measure as an integral yields:
\begin{align}
m(E \cap (x - \epsilon, x + \epsilon)) &= \int_\Bbb R \chi_{E \cap (x - \epsilon, x + \epsilon)}(t)\text d t =  \int_{x-\epsilon}^{x + \epsilon} \chi_{E}(t)\text d t = \int_{-\epsilon}^{\epsilon} \chi_{E}(x + t)\text d t
\\
&= \int_{-\epsilon}^0 \chi_E(x + t)\text d t + \int_0^\epsilon\chi_E(x + t)\text d t
\\
&= \int_0^\epsilon (\chi_E(x - t) + \chi_E(x + t))\text d t 
\end{align}
If both of $\chi_E(x \pm t)$ are equal to $1$, then $x = \frac 12(x+t + x - t)$ and we would be done, so assume the integrand is bounded above by $1$. This would imply $m(E \cap (x- \epsilon, x + \epsilon))\le \epsilon$, contradicting the bound furnished by the density.
