I need help in understanding the proof of the following problem. Both the problem and the proof are given in the book Probability and Measure (by Patrick Billingsley, p. 32-34) and are presented below. There book is old. There can be typos. Also below there is a first question of mine regarding the proof.$\\$
$\textbf{Problem:}$ $\\$ $\\$For any class $\mathscr{H}$ of sets in $\mathscr{\Omega}$ let $\mathscr{H^*}$ consist of the sets in $\mathscr{H}$, the complements of sets in $\mathscr{H}$, and the finite and countable unions of sets in $\mathscr{H}$. Given a class $\mathscr{A}$, put $\mathscr{A_0}$ = $\mathscr{A}$ and define $\mathscr{A_1}$, $\mathscr{A_2}, \cdots$ inductively by
$$\mathscr{A_n}=\mathscr{A_{n-1}^*}. \qquad\textbf{2.27}\qquad$$
That each $\mathscr{A_n}$ is contained in $\sigma$($\mathscr{A}$) follows by induction. One might hope that $\mathscr{A_n}$ = $\sigma$($\mathscr{A}$) for some n, or at least that $\cup_{n=0}^{\infty} \mathscr{A_n}=\sigma(\mathscr{A})$. But this process applied to the class of intervals fails to account for all the Borel sets.
Let $\mathscr{I_0}$ consist of the empty set and the intervals of the form $(a,b]$ in $\Omega = (0,1]$ with rational endpoints, and define $\mathscr{I_n}=\mathscr{I_{n-1}^*}$ for $n = 1, 2, \cdots$ . $\\$$\textbf{Show that }$ $\cup_{n=0}^{\infty} \mathscr{I_n}$ $\textbf{is strictly smaller than }\mathscr{B}$ = $\sigma$($\mathscr{I}_0).$ $\\$
$\textbf{Proof:}$$\\$
If $a_n$ and $b_n$ are rationals decreasing to $a$ and $b$, then$(a,b]=\cup_m\cap_n (a_n,b_n]=\cup_m(\cup_n (a_n,b_n]^c)^c\in\mathscr{I_4}$.The result would therefore not be changed by including in $\mathscr{I_0}$ all the intervals in $(0,1]$. $\\$
To prove $\cup_{n=0}^{\infty} \mathscr{I_n}$ smaller than $\mathscr{B}$, first put
$$\psi (A_1, A_2,\cdots)=A_1^c \cup{A_2}\cup{A_3}\cup{A_4}\cup{\cdots}. \qquad\textbf{2.28}\qquad$$
Since $\mathscr{I_{n-1}}$ contains $\Omega = (0,1]$ and the empty set, every element of $\mathscr{I_n}$ has the form (2.28) for some sequence $A1, A2, …$ of sets in $\mathscr{I_{n-1}}$. Let every positive integer appear exactly once in the square array
$$ \begin{matrix} m_{11} & m_{12} & \cdots \\ m_{21} & m_{22} & \cdots \\ \vdots & \vdots & \\ \end{matrix} $$
Inductively define
$$\Phi_0 (A_1, A_2,\cdots)=A_1,$$ $$\Phi_n (A_1, A_2,\cdots)=\psi (\Phi_{n-1}(A_{m_{11}}, A_{m_{12}},\cdots),\Phi_{n-1}(A_{m_{21}}, A_{m_{22}},\cdots),\cdots), \qquad\textbf{}\qquad \qquad\textbf{}\qquad \qquad\textbf{}\qquad \qquad\textbf{}\qquad \qquad\textbf{}\qquad \qquad\textbf{}\qquad n=1,2,3,\cdots.\qquad\textbf{2.29}\qquad$$
$\bbox[yellow]{\underline{\color{red}{It\ follows\ by\ induction}}}$ that every element of $\mathscr{I}_n$ has the form $\Phi_n (A_1, A_2,\cdots)$ for some sequence of sets in $\mathscr{I}_0$. Finally, put
$$\Phi (A_1, A_2,\cdots)=\Phi_{1}(A_{m_{11}}, A_{m_{12}},\cdots)\cup\Phi_{2}(A_{m_{21}},A_{m_{22}},\cdots)\cup\cdots). \qquad\textbf{2.30}\qquad $$ Then every element of $\cup_{n=0}^{\infty} \mathscr{I_n}$ has the form (2.30) for some sequence $A_1, A_2,\cdots$ of sets in $\mathscr{I}_0$.
If $A_1, A_2,\cdots$ are in $\mathscr{B}$ then (2.28) is in $\mathscr{B}$; it follows by induction that each $\Phi_n (A_1, A_2,\cdots)$ is in $\mathscr{B}$ and therefore that (2.30) is in $\mathscr{B}$. $\\$
With each $\omega$ in $(0,1]$ associate the sequence $(\omega_1, \omega_2,\cdots)$ of positive integers such that $(\omega_1+\cdots+\omega_k)$ is the position of the $k$th $1$ in the nonterminating dyadic expansion of $\omega$ (the smallest $n$ for which $\sum_{i=1}^{n} d_{j}(\omega)=k$). Then $\omega \leftrightarrow (\omega_1, \omega_2, \cdots)$ is a one-to-one correspondence between $(0,1]$ and the set of all sequences of positive integers.
Let $I_1,I_2,\cdots$ be an enumeration of the sets in $\mathscr{I}_0$, put $\varphi(\omega)=\Phi(I_{\omega_1},I_{\omega_2},\cdots)$, and define $B=[\omega:\omega \notin \varphi(\omega)]$. It will be shown that $B$ is a Borel set but is not contained in any of the $\mathscr{I}_n$.
Since $\omega$ lies in $B$ if and only if $\omega$ lies outside $\varphi(\omega)$, $B\ne \varphi(\omega)$ for every $\omega$. But every element of $\cup_{n=0}^{\infty} \mathscr{I_n}$ has the form (2.30) for some sequence in $\mathscr{I_0}$ and hence has the form $\varphi(\omega)$ for some $\omega$. Therefore, $B$ is not a member of $\cup_{n=0}^{\infty} \mathscr{I_n}$.$\\$
It remains to show that $B$ is a Borel set. Let $D_k=[\omega:\omega\in \varphi(I_{\omega_k})].$ $\\$ Since $L_k(n)=[\omega:\omega_1+\cdots+\omega_k=n]=[\omega:\sum_{j=1}^{n-1}d_j(\omega)\lt k=\sum_{j=1}^{n}d_j(\omega)]$ is a Borel set, so are $[\omega:\omega_k=n]=\cup_{m=1}^{\infty}L_{k-1}(m)\cap L_k(m+n)$ and $$D_k=[\omega:\omega\in \varphi(I_{\omega_k})]=\bigcup\limits_n([\omega:\omega_k=n]\cap I_n).$$
Suppose that it is shown that $[\omega:\omega \in\Phi_n (I_{\omega_{u_{1}}},I_{\omega_{u_{2}}},\cdots)]=\Phi_n(D_{u_{1}},D_{u_{2}},\cdots) \qquad\textbf{2.31}\qquad$ for every $n$ and every sequence $u_1, u_2,\cdots$ of positive integers. It will then follow from the definition (2.30) that $\\$ $$B^c=[\omega:\omega\in \varphi(\omega)]=\bigcup\limits_{n=1}^{\infty}[\omega:\omega \in\Phi_n (I_{\omega_{m_{n1}}},I_{\omega_{m_{n2}}},\cdots)]=$$ $\\$ $$=\bigcup\limits_{n=1}^{\infty}\Phi_n(D_{m_{n1}},D_{m_{n1}},\cdots)=\Phi(D_1,D_2,\cdots).$$ But as remarked above, (2.30), is a Borel set if the $A_n$ are. Therefore, (2.31) will imply that $B^c$ and $B$ are Borel sets. $\\$ If $n = 0$, (2.31) holds because it reduces by (2.29) to $[\omega:\omega\in I_{\omega_{u_{1}}}]I=D_{u_{1}}.$ Suppose that (2.31) holds with $n−1$ in place of $n$. Consider the condition $$\omega \in\Phi_{n-1} (I_{\omega_{u_{m_{k1}}}},I_{\omega_{u_{m_{k2}}}},\cdots).\qquad\textbf{2.32}\qquad$$ By (2.28) and (2.29), a necessary and sufficient condition for $\omega \in\Phi_{n} (I_{\omega_{u_{1}}},I_{\omega_{u_{2}}},\cdots)$ is that either (2.32) is false for $k=1$ or else (2.32) is true for some $k$ exceeding $1$. But by the induction hypothesis, (2.32) and its negation can be replaced by $\omega \in\Phi_{n-1} (I_{\omega_{u_{m_{k1}}}},I_{\omega_{u_{m_{k2}}}},\cdots)$ and its negation. Therefore, $\omega \in\Phi_{n} (I_{\omega_{u_{1}}},I_{\omega_{u_{2}}},\cdots)$, if and only if $\omega\in\Phi_n(D_{u_{1}},D_{u_{2}},\cdots).$ $\\$
Thus $\bigcup\limits_{n}^{} I_n \ne B$, and there are Borel sets that cannot be arrived at from the intervals by any finite sequence of set-theoretic operations, each operation being finite or countable. It can even be shown that there are Borel sets that cannot be arrived at by any countable sequence of these operations. On the other hand, every Borel set can be arrived at by a countable ordered set of these operations if it is not required that they be performed in a simple sequence. The proof of this statement - and indeed even a precise explanation of its meaning - depends on the theory of infinite ordinal numbers.
$\textbf{Question 1:}$
I will probably have several questions to this proof. But my 1st question is about proving the induction that I underlined and put in red/yellow color. Lets start with the base of the induction, which is $n=1$. Every set in $\mathscr{I}_1$ can be presented in the form $$\psi (A_{\omega_1}, A_{\omega_2},\cdots)=A_{\omega_1}^c \cup{A_{\omega_2}}\cup{A_{\omega_3}}\cup{A_{\omega_4}}\cup{\cdots} $$ for some countable sequence of sets $A_{\omega_1}, A_{\omega_2},\cdots$ in $\mathscr{I}_0$.$\\$
I will write sets in $\mathscr{I}_n$ in the form ${I^n}_{\alpha}$. $\\$
If a set in $\mathscr{I}_1$ is a set or a countable union of sets in $\mathscr{I}_0$ (for example $A_{\omega_1} \cup{A_{\omega_2}}\cup{A_{\omega_3}}\cup{A_{\omega_4}}\cup{\cdots}$ for some integers $\omega_1,\omega_2,\omega_3,\omega_4,\cdots$ such that non of the $A_{\omega_k}$ is equal to $\Omega$ and $\omega_k\ne\omega_j$ if $k\ne j$) we can write it in the form ${I^1}_{\alpha}=A_{\omega_1} \cup{A_{\omega_2}}\cup{A_{\omega_3}}\cup{A_{\omega_4}}\cup{\cdots}=\psi(\Omega, A_{\omega_1},A_{\omega_2},A_{\omega_3},A_{\omega_4},\cdots)=\psi(\Phi_0(\Omega),\Phi_0(A_{\omega_1}),\Phi_0(A_{\omega_2}),\Phi_0(A_{\omega_3}),\Phi_0(A_{\omega_4}),\cdots).$ $\\$
If a set in $\mathscr{I}_1$ is a conjugate of a set in $\mathscr{I}_0$ (for example it is equal to $A_{\omega_1}^c$), then it can be written in the form: ${I^1}_{\alpha}=A_{\omega_1}^c=\psi(A_{\omega_1})=\psi(\Phi_0(A_{\omega_1}))$. Here $A_{\omega_1}$ could be equal to empty set, thus ${I^1}_{\alpha}$ could be equal to $\Omega$. So the base of induction is proven. $\\$
Now lets prove that the induction property holds for $n$ if it holds for $n-1.$ $\\$
Here I am not sure what to do next. What is $\Phi_{n}(\Omega)$ equal to? Is this correct?: $\Phi_1(\Omega)=\psi(\Phi_0(\Omega))=\psi(\Omega)=\emptyset; \Phi_n(\Omega)=\psi(\Phi_{n-1}(\Omega))= \emptyset$ (if n is odd) or $\Omega$ (if n is even). $\\$
If a set in $\mathscr{I}_n$ is also a set or a countable union of sets in $\mathscr{I}_{n-1}$ then it has the form: ${I^n}_{\alpha}$=${I^{n-1}}_{\omega_1}\cup{I^{n-1}}_{\omega_2}\cup{I^{n-1}}_{\omega_3}\cdots=\psi(\Omega,{I^{n-1}}_{\omega_1},{I^{n-1}}_{\omega_2},{I^{n-1}}_{\omega_3},\cdots)=\psi(\Omega,\Phi_{n-1}(A_{m_{11}},A_{m_{12}},\cdots),\Phi_{n-1}(\cdots,but\ here, \ e.g. \ A_{m_{12}}could \ be \ present?),\cdots).$ $\\$
But it was stated that all $m_{kj}$ appear only once in the array
$$ \begin{matrix} m_{11} & m_{12} & \cdots \\ m_{21} & m_{22} & \cdots \\ \vdots & \vdots & \\ \end{matrix} $$ Please help me to prove this induction.
