# Can we obtain the Borel $\sigma$-algebra on $[0;1]$ as a limit of finite algebra?

The Borel sets of $[0;1]$ are defined to be the smallest $\sigma$-algebra which contains all the subintervals $\subseteq [0;1]$.

Is there a construction where we can obtain it as a limit of a countable sequence of finite algebra? Or as a limit of limits of countable sequences.

For instance something similar (in form) to : $$\lim_{n \to \infty} \bigcup_{k = 1}^{n} A_k \quad \text{where A_k is a finite algebra}$$

• You mean something like the following: Let $\mathcal{A}_n$ be the $\sigma-$algebras of the Borel subsets of $\left[0,\frac{n+1}{n}\right]$. Then $\mathcal{B}_n\to\mathcal{A}$ where $\mathcal{A}$ is the $\sigma-$algebra of all the Borel subsets of $[0,1]$? – Βασίλης Μάρκος May 9 '18 at 12:45
• Hum, no, I will edit to be more clear. Edit: done -- I'm looking for a limit of algebra – Julien__ May 9 '18 at 12:49
• Your edit is unreasonable, but transfinite induction could help. – Olivier May 9 '18 at 12:58
• What does limit mean in this context? Are we just asking about $\bigcup_{k=1}^{\infty}A_k$, or does the limit mean something else? (Also, I think you have a mistake in your subscript in $A_n$) – Milo Brandt May 9 '18 at 12:59
• @MiloBrandt subscript fixed thanks. I'm not sure I understand your question. The limit $\lim_{n \to \infty} u_n$ of any sequence of algebra ? – Julien__ May 9 '18 at 13:01

Yes there is a natural and useful construction of $$\sigma$$-algebras as a "limit". But not as the limit of a sequence, rather the limit of an uncountable collection, indexed by the countable ordinals (something logicians might call an $$\omega_1$$-sequence, but which the rest of us don't usually call a "sequence", since it's not countable).

First, given $$A\subset\mathcal P(X)$$, let $$S(A)$$ consist of all countable unions of elements of $$A$$ together with the complements of elements of $$A$$.

Now suppose $$A\subset\mathcal P(X)$$. By "transfinite recursion" there exists a family $$(A_\alpha)_{\alpha<\omega_1}$$ of subsets of $$\mathcal P(X)$$ with $$A_0=A$$and $$A_\alpha=S\left(\bigcup_{\beta<\alpha}A_\beta\right)$$for $$\alpha>0$$. Then it's not hard to show that$$\sigma(A)=\bigcup_{\alpha<\omega_1}A_\alpha$$is precisely the $$\sigma$$-algebra generated by $$A$$. (Of course we haven't defined what we mean by "limit" here, but it's natural to regard this as the limit of $$A_\alpha$$ as $$\alpha\to\omega_1$$.)

Proof. Say $$\sigma(A)$$ is defined as above and let $$\sum(A)$$ be the $$\sigma$$-algebra generated by $$A$$. Note first that if $$F$$ is a $$\sigma$$-algebra and $$B\subset F$$ then $$S(B)\subset F$$ by definition. So since $$A\subset\sum(A)$$ it follows by transfinite induction that $$A_\alpha\subset\sum(A)$$ for every $$\alpha<\omega_1$$, hence $$\sigma(A)\subset\sum(A)$$.

So we need only show that $$\sigma(A)$$ is a $$\sigma$$-algebra. This is clear. If $$E\in\sigma(A)$$ then there exists $$\alpha < \omega_1$$ with $$E\in A_\alpha$$; hence $$X\setminus E\in A_{\alpha+1}$$, so $$X \setminus E \in \sigma(A)$$.

Suppose that $$E_1,E_2 \dots \in \sigma(A)$$. For every $$n$$ there exists $$\alpha_n<\omega_1$$ with $$E_n\in A_{\alpha_n}$$. Since $$\omega_1$$ is uncountable there exists $$\alpha<\omega_1$$ with $$\alpha>\alpha_N$$ for every $$n$$; hence $$\bigcup E_n\in A_\alpha\subset\sigma(A)$$ $$\square.$$

This is useful, for example as far as I know it's the only way to show that the Borel algebra on $$\Bbb R$$ has cardinality $$\mathfrak c$$.

The answer is negative. Let $\mathcal{A_n}$, $n\in\mathbb{N}$ be a collection of finite algebras such that: $$\mathcal{B}=\bigcup_{n=1}^\infty\mathcal{A}_n$$ where $\mathcal{B}$ is the $\sigma-$algebra of all the Borel subsetes of $\mathbb{[0,1]}$. Now, on the one hand we have that $\#\mathcal{B}$ is uncountable, on the other hand $\#\bigcup\limits_{n=1}^\infty\mathcal{A}_n$ is obviously countable - countable union of finite sets.

As for your edit, it is also false that: $$\bigcup_{n=1}^\infty\mathcal{P}(\{1,\dots,n\})=\lim_{k\to\infty}\bigcup_{n=1}^k\mathcal{P}(\{1,\dots,n\})$$ is uncountable, since it is a countable union of finite sets. It is important to see that: $$\bigcup_{n=1}^\infty\mathcal{P}(\{1,\dots,n\})\underset{\neq}{\subset}\mathcal{P}(\mathbb{N})$$ since $\mathbb{N}\in\mathcal{P}(\mathbb{N})$ but $\mathbb{N}\not\in\bigcup\limits_{n=1}^\infty\mathcal{P}(\{1,\dots,n\})$ - obviously, since there is not $n\in\mathbb{N}$ such that $\mathbb{N}\in\mathcal{P}(\{1,2,\dots,n\})$.

Edit/Further Info: What is true is that: $$\bigcup_{n=1}^\infty\mathcal{P}(\{1,\dots,n\})=\{A\mid A\subset\mathbb{N},\ A\text{ is finite}\}.$$