Past exam of the mathematics course of the University of Cambridge / 2017 / ii / Paper 1 / 26J (Probability and Measure)

The Borel sigma-algebra $\mathcal{B}(\mathbb{R})$ is the smallest sigma-algebra containing all open subsets of $\mathbb{R}$. A real-valued Borel measurable function on $(E,\mathcal{E})$ is a map $f$ with $f^{-1}(B)\in \mathcal{E}$ for every $B\in \mathcal{B}(\mathbb{R})$. It suffices to test the sets $(a,\infty )$ for rational $a$, because these generate the Borel sigma-algebra.

Here “Borel function” describes measurability into the Borel structure of the target; the domain need not itself be a topological space.

Let $f_n$ be Borel measurable and converge pointwise to the finite real value $f$. For every real $a$,If $f(x)>a$, choose a rational $r$ strictly between them; convergence puts $f_n(x)>r$ eventually. Conversely eventual strict inequality $f_n(x)>r>a$ implies $f(x)\ge r>a$. The right side belongs to $\mathcal{E}$ by its countable union/intersection operations. Since these half-lines generate the Borel sigma-algebra, $f$ is a Borel measurable function.

The chosen non-eventually-one binary expansion fixes the ambiguity at dyadic rationals. Its $n$th digit isIn particularThis is a finite union of Borel sets; the zero-digit preimage is its complement in $[0,1)$. Since $R_n$ only takes the values zero and one, every Borel preimage is a union of these two sets. Thus it is Borel measurable. The half-open intervals implement the terminating-zero convention at dyadic endpoints.

Each function $(x,y)\mapsto |R_n(x)-R_n(y)|$ is Borel measurable on the product square, by composition with its coordinate projections and arithmetic of measurable real functions. Its value is one exactly when the $n$th digits disagree. Therefore the partial disagreement countsare nonnegative Borel functions and increase to the extended-real function $f$. For every finite $a$, $\{f>a\}={\bigcup }_N\{f_N>a\}$ is Borel. HenceInfinite disagreement is allowed, so restricting the target to finite real values would lose part of the stated function.

The set of finitely disagreeing pairs is ${\bigcup }_{N\ge 0}{F}_N$, where $F_N$ consists of pairs whose digits agree at every position beyond $N$. For $M>N$, requiring agreement just at positions $N+1,\dots ,M$ defines a union of dyadic rectangles of total area $2^{-(M-N)}$: among the $2^{2M}$ pairs of length-$M$ prefixes, exactly $2^{M+N}$ meet these $M-N$ matching constraints, each rectangle having area $2^{-2M}$.

Since $F_N$ is contained in each of these finite-prefix agreement sets, its measure is at most $2^{-(M-N)}$ for every $M$, hence zero. Countable subadditivity now givesThus almost every pair disagrees in infinitely many digits. Dyadic endpoints, treated with the stipulated expansion convention, do not change any rectangle measure.