This is a graduate-level measure theory problem. I have thought throught it and asked on math.SE but received no satisfying answer.

On P.32 of *[P.Billingsley] Probability and Measure, 3ed, 1993*, the author wrote:

...and there are Borel sets that cannot be arrived at from the intervals by any finite sequence of set-theoretic operations, each operation being finite or countable. It can even be shown that there are Borel sets that cannot be arrived at by any countable sequence of these operations. On the other hand, every Borel set can be arrived at by a

countableordered set of these operations if it is not required that they be performed in asimple sequence.

Then the author referred me to the following exercise which I cannot figure it out:

For any class $\mathcal{H}$ of sets in $\Omega$ let $\mathcal{H}^{*}$ consist of sets in $\mathcal{H}$, the complements of sets in $\mathcal{H}$ and the finite and countable unions of sets in $\mathcal{H}$. Given a class $\mathcal{A}$, put $\mathcal{A}_{0}=\mathcal{A}$ and define $\mathcal{A}_{1},\mathcal{A}_{2},\cdots$ inductively by: $$\mathcal{A}_{n}=\mathcal{A}_{n-1}^{*}$$ That each $\mathcal{A}_{n}$ is contained in $\sigma(\mathcal{A})$ follows by induction.

(1)Prove that if $\Omega=(0,1]$, $\mathcal{A}_{0}=\emptyset$ and the intervals with rational endpoints in $\Omega$,$\cup^{\infty}_{n=0}\mathcal{A}_n$ is strictly smaller than $\sigma(\mathcal{A}_{0})$

(2)Extend the above construction to infinite ordinals $\alpha$ be defining
$$\mathcal{A}_{\alpha}=(\cup_{\beta<\alpha}\mathcal{A}_{\beta})^{*}$$
Show that if $\Omega$ is the first uncountable ordinal, then

$$\cup_{\beta<\Omega}\mathcal{A}_{\beta}=\sigma(\mathcal{A})$$
Moreover, show that if the cardinality of $\mathcal{A}$ does not exceed that of the continuum, then the same is true of $\sigma(\mathcal{A})$, Thus the Borel field has the power of the continuum as its own cardinality.

(i)I cannot figure out how to prove the second problem altough the author provided a hint:

Use the fact that, if $\alpha_{1},\alpha_{2},\cdots$is a sequence of ordinals satisfying $\alpha_{n}<\Omega$, then there exists an ordinal $\alpha$ such that $\alpha<\Omega$ and $\alpha_{n}<\alpha$ for all $n$.

(ii)In particular, I do not quite understand what is the significance of "countable ordered set" here? What role does the "**order**" play in this construction?

(iii)"It can even be shown that there are Borel sets that cannot be arrived at by any countable sequence of these operations." Where can I find a reference for this issue?

Real and Abstract Analysis. $\endgroup$ – Gerald Edgar Sep 10 '15 at 0:55