Ex3-8

Chapter 3, Question 8

Problem: Show that the smallest $$\sigma$$algebra containing all closed sets of Euclidean space is identical to the Borel $$\sigma$$algebra.

Proof: Let $$\mathcal{A}$$ be the smallest $$\sigma$$algebra containing all the closed sets of Euclidean space, and $$\mathcal{B}$$ the Borel $$\sigma$$algebra, which by definition is the smallest $$\sigma$$algebra containing all the open sets of Euclidean space. We want to show $$\mathcal{A} = \mathcal{B}$$. We do this by showing $$\mathcal{A} \subset \mathcal{B}$$ and $$\mathcal{B} \subset \mathcal{A}$$. We start with $$\mathcal{B} \subset \mathcal{A}$$. Suppose $$U \subset \mathbf{R}^n$$ is an arbitrary open set. We want to show that $$U \subset \mathcal{A}$$. Since $$\mathbf{R}^n \setminus U$$ is closed, we have that $$\mathbf{R}^n \setminus U \in \mathcal{A}$$. Furthermore, $$\mathcal{A}$$ is closed under complements, so $$ U \in \mathcal{A}$$. This shows that $$\mathcal{A}$$ contains all the open sets of Euclidean space. Thus we have $$\mathcal{B} \subset \mathcal{A}$$.

Now we show $$\mathcal{A} \subset \mathcal{B}$$. Let $$C$$ be an arbitrary closed set in Euclidean space. Then by definition $$\mathbf{R}^n \setminus C$$ is open. This means that $$\mathbf{R}^n \setminus C \in \mathcal{B}$$. Again, since $$\mathcal{B}$$ is closed under complements, $$C \in \mathcal{B}$$. This shows that $$\mathcal{B}$$ contains all the closed sets of Euclidean space. Since we assumed $$\mathcal{A}$$ is the smallest $$\sigma$$algebra with such property, we have $$\mathcal{A} \subset \mathcal{B}$$, as desired.