Ex3-8

Chapter 3, Question 8
Problem: Show that the smallest \sigmaalgebra containing all closed sets of Euclidean space is identical to the Borel \sigmaalgebra.
Proof: Let \mathcal{A} be the smallest \sigmaalgebra containing all the closed sets of Euclidean space, and \mathcal{B} the Borel \sigmaalgebra, which by definition is the smallest \sigmaalgebra 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 \sigmaalgebra with such property, we have \mathcal{A} \subset \mathcal{B}, as desired.