Regarding Proposition 1.23 in Folland's real analysis text

Question

I have read everything in Folland's real analysis book up to Proposition 1.23, but I have a very hard time understanding this proposition and what immediately precedes it using the set theory introduced by the author. I'm paraphrasing now:

Consider a set $X$ and family of sets $\mathcal E\subset\mathcal P(X)$. Let $\mathcal E_1=\mathcal{E}\cup\{E^c:E\in\mathcal E\}$ and then for $j>1$ define $\mathcal E_j$ to be the collection of all sets that are countable unions of sets in $\mathcal E_{j-1}$ or complements of such. Let $\mathcal E_\omega=\bigcup_1^\infty\mathcal E_j$. If $E\in \mathcal E_\omega$, it is in one of the $\mathcal E_j$'s, but then $E^c\in\mathcal E_{j+1}$, so $E^c\in\mathcal E_\omega$. Apparently the set is not closed under countable unions. The claim is that if $E_j\in\mathcal E_j\setminus\mathcal E_{j-1}$ for each $j$, there's no reason for $\bigcup_1^\infty E_j$ to be in $\mathcal E_\omega$.

1. My first question is, why not? Is there an easy-to-understand counterexample for when $\bigcup_1^\infty E_j\notin\mathcal E_\omega$?

The author goes on to say one must start all over again and define $\mathcal E_\alpha$ for each countable ordinal $\alpha$ by transfinite induction: If $\alpha$ has an immediate predecessor $\beta$, $\mathcal E_\alpha$ is the collection of sets that are countable unions of sets in $\mathcal E_\beta$ or complements of such; otherwise, $\mathcal E_\alpha=\bigcup_{\beta<\alpha}\mathcal E_\beta$.

2. The author hasn't really defined what a countable ordinal is nor what a predecessor is (or if there isn't a predecessor). I gather a countable ordinal is an initial segment of $\Omega$, the set of countable ordinals, but I'm not sure what he means by predecessor (or if there isn't one). Moreover, where in defining $\mathcal E_\alpha$ did we use transfinite induction? Transfinite induction has barely been used in the book so far, so I'm sorry if I don't see this immediately.

Finally,

1.23 Proposition. $\mathcal M(\mathcal E)=\bigcup_{\alpha\in\Omega} \mathcal E_\alpha$ where $\mathcal M(\mathcal E)$ is the $\sigma$-algebra generated by $\mathcal E$.

Proof. Transfinite induction shows that $\mathcal E_\alpha\subset \mathcal M(\mathcal E)$ for all $\alpha\in\Omega$, and hence $\bigcup_{\alpha\in\Omega} \mathcal E_\alpha\subset \mathcal M(\mathcal E)$. The reverse inclusion follows from the fact that any sequence in $\Omega$ has a supremum in $\Omega$ (Proposition 0.19): If $E_j\in\mathcal E_{\alpha_j}$ for $j\in\mathbb N$ and $\alpha=\sup\{a_j\}$, then $E_j\in\mathcal E_\alpha$ for all $j$ and hence $\bigcup_1^\infty E_j\in\mathcal E_\beta$ where $\beta$ is the successor of $\alpha$.

3. How does it follow from transfinite induction that $\mathcal E_\alpha\subset \mathcal M(\mathcal E)$ for all $\alpha\in\Omega$? Why is $E_j\in\mathcal E_\alpha$ for all $j$ and why does it follow that $\bigcup_1^\infty E_j\in\mathcal E_\beta$? What's the definition of a successor?

For reference, here's the statement of transfinite induction in the book. Let $X$ be a well ordered set and $x\in X$, then $I_x=\{y\in X:y<x\}$ is an initial segment of $x$.

0.15 The Principle of Transfinite Induction. Let $X$ be a well ordered set. If $A$ is a subset of $X$ such that $x\in A$ whenever $I_x\subset A$, then $A=X$.

Answer 1

For your question 1: Suppose $\bigcup_1^\infty E_j \in \mathcal E_\omega$. Since $\mathcal E_\omega=\bigcup_1^\infty\mathcal E_j$, there should be a $j_0$ such that $\bigcup_1^\infty E_j \in \mathcal E_{j_0}$, but $E_{j_0+1} \in \mathcal E_{j_0+1} \setminus \mathcal E_{j_0}$, it means $E_{j_0+1} \notin \mathcal E_{j_0}$, so, in the general case, we will have $\bigcup_1^\infty E_j \notin \mathcal E_{j_0}$. Contradiction. So, in the general case, $\bigcup_1^\infty E_j \notin \mathcal E_\omega$.

Remark 1: In a special situation, with additional conditions, one could have $\bigcup_1^\infty E_j \in \mathcal E_\omega$, but not in the general case.

Remark 2: My response to question 1 is an explanation of the argument written in Folland's book. Folland does not proivide an actual counterexample to illustrate his argument, although, of course, such counterexamples exist.

For your question 2. A countable ordinal corresponds to countable well-ordered set. To say that an ordinal $\alpha$ has an immediate predecessor means that there is an ordinal $\beta$ such that $\alpha=\beta+1$. If an ordinal has an immediate predecessor is call a successor ordinal. There are ordinals that has not an immediate predecessor, they are not successor ordinals. For instance $0$ and $\omega$ (which corresponds to $\Bbb N$ with the usual order). If an ordinal, other than $0$, is not a successor ordinal, it is called a limit ordinal. So $\omega$ is the first limit ordinal.

Transfinite Induction is used in defining that: If $\alpha$ has an immediate predecessor $\beta$, $\mathcal E_\alpha$ is the collection of sets that are countable unions of sets in $\mathcal E_\beta$ or complements of such; otherwise, $\mathcal E_\alpha=\bigcup_{\beta<\alpha}\mathcal E_\beta$. We need Transfinite induction to prove that $\mathcal E_\alpha$ is defined for all ordinals.

Remark: Proving that $\mathcal E_\alpha$ is defined for all ordinals.

1. For $\alpha =0$, $\mathcal E_0$ is defined. In fact, $\mathcal E_0=\mathcal{E}\cup\{E^c:E\in\mathcal E\}$.

2. Given an ordinal $\alpha$, suppose we already know that $\mathcal E_\beta$ is defined, for all $\beta < \alpha$.

2.a. if $\alpha$ is a sucessor ordinal, there is an ordinal $\beta < \alpha$ such that $\alpha = \beta+1$. So, $\mathcal E_\alpha$ is the collection of sets that are countable unions of sets in $\mathcal E_\beta$ or complements of such. So, $\mathcal E_\alpha$ is defined.

2.b. if $\alpha$ is a limit ordinal, $\mathcal E_\alpha=\bigcup_{\beta<\alpha}\mathcal E_\beta$. So, $\mathcal E_\alpha$ is defined.

So, by Transfinite Induction, $\mathcal E_\alpha$ is define for all ordinals.

For your question 3: Proving that $\mathcal E_\alpha\subset \mathcal M(\mathcal E)$ for all $\alpha\in\Omega$ by Transfinite Induction.

1. Note that $\mathcal E_0=\mathcal{E}\cup\{E^c:E\in\mathcal E\}\in M(\mathcal E)$

2. Given an ordinal $\alpha$, suppose we already know that $\mathcal E_\beta\subset \mathcal M(\mathcal E)$ for all $\beta < \alpha$.

2.a. if $\alpha$ is a sucessor ordinal, there is an ordinal $\beta < \alpha$ such that $\alpha = \beta+1$. So, $\mathcal E_\alpha$ is the collection of sets that are countable unions of sets in $\mathcal E_\beta$ or complements of such. Since $\mathcal E_\beta\subset \mathcal M(\mathcal E)$, it follows immediately that $\mathcal E_\alpha\subset \mathcal M(\mathcal E)$.

2.b. if $\alpha$ is a limit ordinal, $\mathcal E_\alpha=\bigcup_{\beta<\alpha}\mathcal E_\beta \subset \mathcal M(\mathcal E)$.

So the proof is complete.

You also asked: why is $E_j\in\mathcal E_\alpha$ for all $j$ and why does it follow that $\bigcup_1^\infty E_j\in\mathcal E_\beta$? The answer to this in in proposition 0.19. Any countable set of countable ordinals has a supremum and it is a countable ordinal.

Here are more details: Using such property, we can prove that $\bigcup_{\alpha\in\Omega}\mathcal E_\alpha$ is a $\sigma$-algebra.

1. Even if $\emptyset \notin \mathcal E_0$, we have that $\emptyset \in \mathcal E_1 \subset \bigcup_{\alpha\in\Omega}\mathcal E_\alpha$.

2. If $E \in \bigcup_{\alpha\in\Omega}\mathcal E_\alpha$, then there is $\alpha_0 \in \Omega$ such that $E \in \mathcal E_{\alpha_0}$. So $E^c \in \mathcal E_{\alpha_0+1} \subset \bigcup_{\alpha\in\Omega}\mathcal E_\alpha$.

3. If $\{E_i\}_{i=1}^\infty$ is a countable family of sets such that, for all $i$, $E_i \in \bigcup_{\alpha\in\Omega}\mathcal E_\alpha$, then there are $\alpha_i \in \Omega$ such that $E_i \in \mathcal E_{\alpha_i}$. Since any countable set of countable ordinals has a supremum and it is a countable ordinal, there is a countable ordinal $\beta$, that is an ordinal $\beta \in \Omega$, such that, for all $i$, $\alpha_i \leqslant \beta$. So, for all $i$, $E_i \in \mathcal E_\beta$. So $\bigcup_1^\infty E_i \in \mathcal E_{\beta+1} \subset \bigcup_{\alpha\in\Omega}\mathcal E_\alpha$.

So $\bigcup_{\alpha\in\Omega}\mathcal E_\alpha$ is $\sigma$-algebra and $\mathcal E \subset \bigcup_{\alpha\in\Omega}\mathcal E_\alpha$. It follows immediately that $\mathcal M(\mathcal E) \subset \bigcup_{\alpha\in\Omega}\mathcal E_\alpha$.