Real Analysis

Construction of Borel sigma-algebra

Let (S, <) be an uncountable well-ordered set, and let 1 stand for the minimum element of S. We denote by A the set of all elements \alpha\in S such that the set of its predecessors \text{Pre}_{\alpha}=\{x\in S \ | \ x<\alpha\} is uncountable. If the set A is empty, we will replace S by S\cup\{s^*\}, where s^*\notin S, and set x<s^* for all x\in S.

Since A is a nonempty subset of the well-order set S, there exists a unique minimum element \omega_1 called first uncountable ordinal. Therefore, if x<\omega_1 then \text{Pre}_x is countable.

Exercise. Give an example for an uncountable set with a well-order relation.

Theorem 1. (Transfinite induction) Let P(\alpha) be a property defined for all \alpha \in S. Suppose that:

1. P(1) is true,

2. for each \alpha<\omega_1, if P(\beta) is true for all \beta < \alpha, then P(\alpha) is also true.

Therefore, P(\alpha) is true for all \alpha <\omega_1.


Let \Omega be the set of \alpha <\omega_1 such that P(\alpha) is false. Suppose that \Omega \neq \emptyset. Since \Omega is nonempty subset of the well-ordered set S, there exists an unique minimum element \alpha^*\in \Omega. Note that \alpha^*\neq 1 and the set of its predecessors \text{Pre}_{\alpha^*} is countable. Hence, the property P is true for 1 and all  predecessors of \alpha^*. It follows that P(\alpha^*) is also true. This inconsistency completes our proof.

Theorem 2. For any sequence \alpha_1,\alpha_2 .... in S, \alpha_i<\omega_1, there exists \alpha^*<\omega_1 such that \alpha_i<\alpha^*<\omega_1 .

Proof. Note that T =\bigcup_{i=1}^{\infty}\left(\{\alpha_i\}\cup \text{Pre}_{\alpha_i}\right) is countable. Therefore, we can choose \alpha^* as an element in the uncountable set \text{Pre}_{\omega_1}\setminus T .

Now let \Sigma_1 stand for the family of all open sets, and \Pi_1 be the family of all closed sets (considered in a topology space). Suppose that \Sigma_{\beta}, \Pi_{\beta} are defined for all \beta <\alpha,  then we can define
\displaystyle \Sigma_{\alpha}=\left\{ \bigcup_{i=1}^{\infty} A_i\ | \ A_i\in \Pi_{\beta_i}, \beta_i<\alpha \right\}
\displaystyle \Pi_{\alpha}=\left\{\bigcap_{i=1}^{\infty} A_i\ | \ A_i\in \Sigma_{\beta_i}, \beta_i<\alpha \right\}

By the principle of transfinite induction, \Sigma_{\alpha},\Pi_{\alpha} are well-defined for every \alpha<\omega_1.

Lemma 3. 1. \Pi_{\alpha}=\{A^c \ | \ A\in\Sigma_{\alpha}\}, \Sigma_{\alpha}=\{A^c \ | \ A\in\Pi_{\alpha}\} for 1\le \alpha <\omega_1

and for 1\le \beta<\alpha<\omega_1 then

2.  \Sigma_{\beta}\subset \Pi_{\alpha}, \Pi_{\beta}\subset \Sigma_{\alpha};

3. \Sigma_{\beta}\subset \Sigma_{\alpha}, \Pi_{\beta}\subset \Pi_{\alpha}.


1. The fact is true for \alpha=1. Assume that it is true for all \beta<\alpha, then

for each A\in \Pi_{\alpha}, we have A=\bigcap_{i=1}^{\infty} A_{i}, where A_i\in \Sigma_{\beta_i}, \beta_i<\alpha. Moreover,  A_{i}^c \in \Pi_{\beta_i}. It allows that A^c=\bigcup_{i=1}^{\infty} A_{i}^c \subset \Sigma_{\alpha}. Therefore, by the principle of transfinite induction, we conclude that \Pi_{\alpha}=\{A^c \ | \ A\in\Sigma_{\alpha}\}, and similarly, \Sigma_{\alpha}=\{A^c \ | \ A\in\Pi_{\alpha}\} for 1\le \alpha <\omega_1.

2. This fact is obvious.

3. If A\in \Sigma_{\beta}, then  A=\bigcup_{i=1}^{\infty} A_{i}, A_i\in \Pi_{\gamma_i} where \gamma_i<\beta < \alpha. This immediately follows that A\in \Sigma_{\alpha}. Thus \Sigma_{\beta}\subset \Sigma_{\alpha} and similarly, \Pi_{\beta}\subset \Pi_{\alpha}.

Theorem 4. The Borel sigma-algebra, i.e. smallest sigma-algebra that consists all open (closed) sets, is given by \mathfrak{B}=\bigcup_{\alpha<\omega_1}\Sigma_{\alpha}=\bigcup_{\alpha<\omega_1}\Pi_{\alpha}

Proof.  Lemma 3.2 help us to prove the identity \bigcup_{\alpha<\omega_1}\Sigma_{\alpha}=\bigcup_{\alpha<\omega_1}\Pi_{\alpha}.

It is easy to show that \mathfrak{B} is closed under complement by applying Lemma 3.1.

If A_i\in \mathfrak{B}, i=1,2,... then A_i \in \Pi_{\alpha_i}, \alpha_i<\omega_1.

Applying Theorem 2, there exists \alpha^* such that \alpha_i<\alpha^*<\omega_1. It follows that

\displaystyle \bigcup_{i=1}^{\infty} A_i\in \Sigma_{\alpha^*}\subset \mathfrak{B}

Moreover, one can show by the transfinite induction that \Sigma_{\alpha}, \Pi_{\alpha} are subsets of any sigma algebra that contains all closed (open) sets.  Hence, we can conclude that \mathfrak{B} is the Borel sigma-algebra that we want to understand its structure.