First intersect $E$ with a sufficiently large bounded interval to obtain a finite positive-measure set. By the <Lebesgue density theorem>, this set has a density-one point $x$. Hence, for every $\varepsilon>0$, some sufficiently small interval $I=(a,b)$ centred at $x$ satisfies
$$
\boxed{m(E\cap(a,b))>(1-\varepsilon)(b-a),}
$$
which is the <high-density interval in a positive-measure subset of the real line>.

To obtain the final claim, choose $\varepsilon<1/2$, put $L=b-a$, and translate $A=E\cap(a,b)$ into $[0,L]$. Write
$$
m(A)=\frac L2+\alpha,
\qquad
\alpha>\left(\frac12-\varepsilon\right)L>0.
$$
Part (f), after rescaling the interval length, gives
$$
(-2\alpha,2\alpha)\subseteq A-A\subseteq E-E.
$$
Thus \b[$E-E$ contains an open interval around zero], which is the <Steinhaus theorem>.

Solution

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

If a measurable set $E\subseteq\mathbb R$ has positive measure, then for every $\varepsilon>0$ there is a bounded interval $I$ such that
$$
m(E\cap I)>(1-\varepsilon)m(I).
$$
Choose a density-one point of a finite positive-measure portion of $E$ using the <Lebesgue density theorem>, and take a sufficiently small interval centred there.

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