Let $|t|<2\alpha$. Both $E$ and its translate $E+t$ lie in an interval of length $1+|t|$. By finite additivity and translation invariance of <Lebesgue measure>,
$$
m(E\cap(E+t))
=m(E)+m(E+t)-m(E\cup(E+t))
\geq2\left(\frac12+\alpha\right)-(1+|t|)
=2\alpha-|t|>0.
$$
Choose $x\in E\cap(E+t)$. Then $x\in E$ and $x-t\in E$, so $t=x-(x-t)\in E-E$. Therefore the <difference-set overlap bound on an interval> gives
$$
\boxed{(-2\alpha,2\alpha)\subseteq E-E.}
$$

Solution

Past exam of the mathematics course of the University of Cambridge / 2018 / ii / Paper 3 / 26J / f / Solution

If a measurable set $E\subseteq[0,L]$ has measure $m(E)=L/2+\alpha$, then for $|t|<2\alpha$ the sets $E$ and $E+t$ lie in an interval of length $L+|t|$, so
$$
m(E\cap(E+t))\geq2m(E)-(L+|t|)>0.
$$
Thus $(-2\alpha,2\alpha)\subseteq E-E$. This quantitative overlap argument is a bounded form of the <Steinhaus theorem>.

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