Exercise 1.4.17 (Borel measurable sets and Cartesian products)

We first consider the case when $F$ is box.

• Case I: $E$ is Borel measurable, $F=B$ is a box.
  Following the Remark 1.4.15, we verify whether the following property $P$

  $$\forall <!--\; FUNCTION\; APPLICATION\; -->E\in B\left({ℝ}^{d+1}\right):P(E)⟺\left[F\text{ is a box}⟹E\times F\text{ is Borel measurable}\right]$$

  is a $\sigma$-algebraic property.

  (i)

  (empty set) Let $\varnothing \in {ℝ}^{d_1}$ be the empty set in ${ℝ}^{d_1}$. Then $P(\varnothing )$ is true, since for any box $B\subseteq {ℝ}^{d_2}$ we have $\varnothing \times B=\varnothing$ is an empty set in ${ℝ}^{d_1+d_2}$, and thus Borel measurable in ${ℝ}^{d_1+d_2}$ by definition.

  (ii)

  (complement) Let $E$ be Borel measurable such that $P(E)$ holds, i.e., $B\subseteq {ℝ}^{d_2}$ is a box $⟹E\times B$ is Borel measurable in ${ℝ}^{d_1+d_2}$. Then $P({ℝ}^{d_1}∕E)$ must hold also, since $({ℝ}^{d_1}∕E)\times B=({ℝ}^{d_1}\times B)∕(E\times B)$. $E\times B$ is Borel measurable by hypothesis, and ${ℝ}^{d_1}\times B$ is Borel measurable as a countable union of boxes ${\bigcup <!--\; FUNCTION\; APPLICATION\; -->}_{n\in {\mathbb{Z}}^{d_1+d_2}}{\prod }_{i=1}^{d_1}[n_i,n_i+1)\times B$. The set difference of Borel measurable sets is again Borel measurable.

  (iii)

  (countable unions) Let ${\left(E\right)}_{n\in \mathbb{N}}$ be a sequence of subsets of ${ℝ}^{d_1}$ such that for all $n\in \mathbb{N}$ the property $P(E_n)$ holds. We then have $\left({\bigcup <!--\; FUNCTION\; APPLICATION\; -->}_{n=1}^{\infty }{E}_n\right)\times B={\bigcup <!--\; FUNCTION\; APPLICATION\; -->}_{n=1}^{\infty }(E_n\times B)$, and is Borel measurable by axiom 3.

  Now consider for $E$ the set $B\left({ℝ}^{d_1}\right)$ of all boxes in ${ℝ}^{d_1}$, which generates the Borel algebra $⟨B\left({ℝ}^{d_1}\right)⟩=B\left({ℝ}^{d_1}\right)$. The property $P$ holds for all of the sets in the generating set $B\left({ℝ}^{d_1}\right)$ by the first part of Exercise 1.1.4. Thus, it holds for any set in $B\left({ℝ}^{d_1}\right)$, and we are done.

• Case II: $E$ is Borel measurable, $F$ is Borel measurable.now look at the property $P$ given by

  $$\forall <!--\; FUNCTION\; APPLICATION\; -->F\in B\left({ℝ}^{d+2}\right):P(F)⟺\left[E\text{ is Borel measurable}⟹E\times F\text{ is Borel measurable}\right]$$

  1.

  (empty set) Let $\varnothing \in {ℝ}^{d_2}$ be the empty set in ${ℝ}^{d_2}$. Then $P(\varnothing )$ is true, since $E\times \varnothing$ is an empty set in ${ℝ}^{d_1+d_2}$, and thus Borel measurable in ${ℝ}^{d_1+d_2}$ by definition.

  2.

  (complement) Let $F$ be Borel measurable such that $P(E)$ holds, i.e., $F\subseteq {ℝ}^{d_2}$ is Borel measurable $⟹E\times F$ is Borel measurable in ${ℝ}^{d_1+d_2}$. Then $P({ℝ}^{d_2}∕F)$ must hold also, since $E\times ({ℝ}^{d_2}∕F)=(E\times {ℝ}^{d_2})∕(E\times F)$. $E\times F$ is Borel measurable by hypothesis, and $E\times {ℝ}^{d_1}$ is Borel measurable as a countable union of Borel measurable sets ${\bigcup <!--\; FUNCTION\; APPLICATION\; -->}_{n\in {\mathbb{Z}}^{d_1+d_2}}{\prod }_{i=1}^{d_1}E\times [n_i,n_i+1)$ (see Case I). The set difference of Borel measurable sets is again Borel measurable.

  3.

  (countable unions) Let ${(F_n)}_{n\in \mathbb{N}}$ be a sequence of subsets of ${ℝ}^{d_1}$ such that for all $n\in \mathbb{N}$ the property $P(F_n)$ holds. We then have $E\times \left({\bigcup <!--\; FUNCTION\; APPLICATION\; -->}_{n=1}^{\infty }{F}_n\right)={\bigcup <!--\; FUNCTION\; APPLICATION\; -->}_{n=1}^{\infty }(E\times F_n)$, and is Borel measurable by axiom 3.

  For $F$ now consider the elements of the generating set $B\left({ℝ}^{d_1}\right)$. Again, the property $P$ holds for all of the sets in the generating set $B\left({ℝ}^{d_1}\right)$ by the first part of Exercise 1.1.4. Thus, it holds for any set in $B\left({ℝ}^{d_1}\right)$, and we are done.