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Archive for February, 2013

## 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 $A$ as 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 for all $x\in S$.

Since $A$ is a nonempty subset of the well-order set $S$, so 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 $\alpha <\omega_1$.

Proof.

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 $\alpha^*\in \Omega$. Observe  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 $\alpha^*$. So it implies 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 $\beta <\alpha$ $\beta <\alpha$

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

Proof.

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 obviously implies that $A\in \Sigma_{\alpha}$. So $\Sigma_{\beta}\subset \Sigma_{\alpha}$ 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 $\alpha_i<\alpha^*<\omega_1$. So it implies that $\alpha_i<\alpha^*<\omega_1$

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. So we can conclude that $\mathfrak{B}$ is the Borel sigma-algebra that we need to understand its structure.

Categories: Real Analysis