Past exam of the mathematics course of the University of Cambridge / 2018 / ii / Paper 3 / 26J (Probability and Measure)

A subset of $\mathbb{R}$ is a Borel set when it belongs to the Borel sigma-algebra, the smallest sigma-algebra containing every open set of $\mathbb{R}$.

A set $E\subseteq \mathbb{R}$ is Lebesgue measurable if it satisfies the Caratheodory criterionfor every $A\subseteq \mathbb{R}$, where $m^{\ast }$ is Lebesgue outer measure. Equivalently, $E$ belongs to the completion of the Borel sigma-algebra with respect to Lebesgue measure.

The rational numbers are countable, sois a countable union of closed sets and hence a Borel set; its Lebesgue measure is zero. Its complement is also Borel and has infinite measure:

The intervals are nested in the relevant directions, and the strongest lower and upper restrictions both occur at $n=2$:Thus this set is closed, hence Borel, and

A function $f:\mathbb{R}\to \mathbb{R}$ is Borel measurable whenis a Borel set for every Borel set $B\subseteq \mathbb{R}$. It is enough to require this for every open set, or for every interval of the form $(-\infty ,a)$.

Yes. In fact, measurability of $f$ is unnecessary. Define the oscillationFor every $\epsilon >0$, the set $\{x:{\omega }_f(x)<\epsilon \}$ is open, and $f$ is continuous at $x$ exactly when ${\omega }_f(x)=0$. Hence the continuity set of a function isa $G_{\delta }$ set and therefore a Borel set.

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,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

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.