# Verification that the Borel $\sigma$-algebra on $\mathbb{R}$ is not atomic.

Let $$X$$ be a set, and let $$\mathcal{A} = (A_n)_{n=1}^{\infty}$$ be a sequence of disjoint, nonempty subsets whose union is $$X$$. Then the set $$\mathcal{M}$$ of all finite or countable unions of elements of $$\mathcal{A}$$ together with $$\emptyset$$ is a $$\sigma$$-algebra. A $$\sigma$$-algebra of this form is called atomic. Then the Borel $$\sigma$$-algebra $$\mathcal{B}_{\mathbb{R}}$$ on $$\mathbb{R}$$ is not atomic.

$$\text{Proof.}$$

Suppose for sake of contradiction that $$\mathcal{B}_{\mathbb{R}}$$ is the collection of all finite or countable unions of sets in $$\mathcal{A} = (A_n)_{n=1}^{\infty}$$, where the $$A_i$$ are mutually disjoint, non-empty subsets of $$\mathbb{R}$$ whose union is $$\mathbb{R}$$. In particular it follows from this that each set in $$\mathcal{A}$$ is itself a Borel-set in $$\mathbb{R}$$. Now let $$\mathcal{U}= \left\{\left\{p\right\}:0. Then each singleton in $$\mathcal{U}$$ is a Borel-set, being closed with respect to the standard topology on $$\mathbb{R}$$. Hence, for each $$p\in (0,1)$$, it follows that $$\left\{p \right\}$$ is a union of a finite or countable sub-collection of the $$A_i$$. But since the $$A_i$$ are each nonempty, $$\left\{p\right\}$$ cannot be a union of more than one $$A_i$$, since otherwise $$\left\{p\right\}$$ would have more than one element. Thus, for each $$p\in (0,1)$$, we can injectively associate a set $$A_i\in\mathcal{A}$$ with $$\left\{p\right\} = A_i$$. But this is absurd because $$\mathcal{U}$$ is uncountable and $$\mathcal{A}$$ is countable.

I'd appreciate if anyone here could check for the accuracy of the above proof. Thanks.

• Looks correct to me. There indeed are uncountably many disjoint Borel sets in $\mathbb{R}$ and this is the main point. Sep 7, 2020 at 19:48

Suppose such $$\mathcal{A}$$ existed. Consider two cases: either every $$A_i$$ is a singleton, or some $$A_i$$ contains at least two points. The first case is impossible since $$\mathbb{R}$$ is uncountable. In the second case, if $$A_i$$ contains two points $$x,y$$, then $$\{x\}$$ is a Borel set which, as you argued, cannot be written as a union of elements of $$\mathcal{A}$$.