Exercise 1.4.5 (Examples of non-atomic Boolean algebras)

We use the fact that the uncountable singleton sets $\{x\}$ for $x\in {ℝ}^d$ are contained in any of these algebras.

(i)

elementary Boolean algebra
Let $B$ be elementary Boolean algebras $E({ℝ}^d)$, and suppose for the sake of contradiction that it has the atoms ${(A_{\alpha })}_{\alpha \in I}$. Note that $$\forall <!--\; FUNCTION\; APPLICATION\; -->x\in {ℝ}^d:\{x\}\in E({ℝ}^d)$$

since any point $\{x\}$ can be represented as a trivial box $\{x\}={\prod <!--\; FUNCTION\; APPLICATION\; -->}_{i=1}^d[x_i,x_i]$. Furthermore, we have

$$\left\{\{x\}\in E:x\in {ℝ}^d\right\}\subseteq \{A_{\alpha }:\alpha \in I\}$$

i.e., any singleton set is an atom, since no other smaller set can generate the singletons other than the singletons themeselves. But using the singletons we can generate sets other than elementary sets - balls ${\bigcup <!--\; FUNCTION\; APPLICATION\; -->}_{x\in B(0,r)}\{x\}$ for instance. This is a contradiction.

(ii)

Jordan Boolean algebra
Here we can construct the set $[0,1]∕\mathbb{Q}$ which is not Jordan-measurable.

(iii)

Lebesgue Boolean algebra
Here we can construct the Vitali set using the singletons.

(iv)

null Boolean algebra
Here we can take enough (uncountably many) singletons to create a set of a non-zero measure.