Exercise 1.4.14 (Generators of the Borel $\sigma$-algebra)

For two generating sets $F,F^{\prime }$ to demonstrate that

it is sufficient to demonstrate that

In other words, we show that every $\sigma$-algebra containing $F$ also contains $F^{\prime }$ and vice versa.

• (i)-(ii)
  Let $B$ be an arbitrary $\sigma$-algebra containing all open sets in ${ℝ}^d$. Let $F$ be an arbitrary closed set. Then $F^c$ is open, and $F^c\in B$. Since $B$ is a $\sigma$-algebra, it must contain its complement ${(F^c)}^c=F$.
  The converse statement follows in a similar spirit.
• (ii)-(iii)
  Let $B$ be an arbitrary $\sigma$-algebra containing all closed sets in ${ℝ}^d$. The statement then follows directly by Heine-Borel (cf. Analysis II, Theorem 1.5.7).
  Conversely, let $B$ be an arbitrary $\sigma$-algebra containing all compact sets in ${ℝ}^d$. Let $F$ be an arbitrary closed set. Then, for each $n\in \mathbb{N}$ the set $F\cap \overline{B}(0,n)$ is compact, and is thus contained in $B$. We can say the same about the countable union ${\bigcup <!--\; FUNCTION\; APPLICATION\; -->}_{n=1}^{\infty }F\cap \overline{B}(0,n)=F$.
• (iii)-(iv)
  Let $B$ be an arbitrary $\sigma$-algebra containing all compact sets in ${ℝ}^d$. Let $B$ be an open ball. By Lemma 1.2.11, $B$ can be expressed as a countable union of almost closed cubes. Since closed cubes are compact, and since $B$ is a $\sigma$-algebra, $B$ must be contained in $B$.
  Conversely, let $B$ be an arbitrary $\sigma$-algebra containing all open balls in ${ℝ}^d$. Let $F$ be a compact set. Then the complement $F^c$ of $F$ is open. Every open set can be represented as a countable union of open balls. Therefore, $F^c\in B$, by the properties of $\sigma$-algebra. Hence, ${(F^c)}^c=F\in B$.
• (iv)-(v)
  Let $B$ be an arbitrary $\sigma$-algebra containing all open balls in ${ℝ}^d$. Let $B$ be box in ${ℝ}^d$. Then $B$ can be represented as an intersection of a closed $B^{\prime }$ and an open $B^{\mathit{\prime \prime }}$ boxes. Since $B$ contains all of the open and closed sets, it must contain $B^{\prime }\cap B^{\mathit{\prime \prime }}$.
  Conversely, let $B$ be an arbitrary $\sigma$-algebra containing all boxes in ${ℝ}^d$. Let $B$ be an open ball. By Lemma 1.2.11, $B$ is a countable union of closed boxes. Thus, it must be contained in $B$.
• (v)-(vi)
  Let $B$ be an arbitrary $\sigma$-algebra containing all boxes in ${ℝ}^d$. Let $E$ be an elementary set. Then $E$ is a finite union of boxes and is thus contained in $B$.
  Conversely, let $B$ be an arbitrary $\sigma$-algebra containing all elementary sets in ${ℝ}^d$. Let $B$ be a box. Then $B$ is an elementary set, and we are done.