 Solution

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 $\epsilon >0$, some sufficiently small interval $I=(a,b)$ centred at $x$ satisfieswhich is the high-density interval in a positive-measure subset of the real line.

To obtain the final claim, choose $\epsilon <1/2$, put $L=b-a$, and translate $A=E\cap (a,b)$ into $[0,L]$. WritePart (f), after rescaling the interval length, givesThus $E-E$ contains an open interval around zero, which is the Steinhaus theorem.