Intuition for proof of monotone class theorem? Here is the monotone class theorem from my real analysis textbook.

Suppose $\mathcal{A}_0$ is an algebra, $\mathcal{A}$ is the smallest $\sigma$-algebra containing $\mathcal{A}_0$, and $\mathcal{M}$ is the smallest monotone class containing $\mathcal{A}_0$. Then $\mathcal{M} = \mathcal{A}$.

Here is the proof in the book.

A $\sigma$-algebra is clearly a monotone class, so $\mathcal{M} \subset \mathcal{A}$. We must show $\mathcal{A} \subset \mathcal{M}$.
Let $\mathcal{N}_1 = \{A \in \mathcal{M} : A^c \in \mathcal{M}\}$. Note $\mathcal{N}_1$ is contained in $\mathcal{M}$ and contains $\mathcal{A}_0$. If $A_i \uparrow A$ and each $A_i \in \mathcal{N}_1$, then each $A_i^c \in \mathcal{M}$ and $A_i^c \downarrow A^c$. Since $\mathcal{M}$ is a monotone class, $A^c \in \mathcal{M}$, and so $A \in \mathcal{N}_1$. Similarly, if $A_i \downarrow A$ and each $A_i \in \mathcal{N}_1$, then $A \in \mathcal{N}_1$. Therefore $\mathcal{N}_1$ is a monotone class. Hence $\mathcal{N}_1 = \mathcal{M}$, and we conclude $\mathcal{M}$ is closed under the operation of taking complements.
Let $\mathcal{N}_2 = \{A \in \mathcal{M} : A \cap B \in \mathcal{M} \text{ for all }B \in \mathcal{A}_0\}$. Note the following: $\mathcal{N}_2$ is contained in $\mathcal{M}$ and $\mathcal{N}_2$ contains $\mathcal{A}_0$ because $\mathcal{A}_0$ is an algebra. If $A_i \uparrow A$, each $A_i \in \mathcal{N}_2$, and $B \in \mathcal{A}_0$, then $A \cap B = \cup_{i = 1}^\infty (A_i \cap B)$. Because $\mathcal{M}$ is a monotone class, $A \cap B \in \mathcal{M}$, which implies $A \in \mathcal{N}_2$. We use a similar argument when $A_i \downarrow A$. Therefore $\mathcal{N}_2$ is a monotone class, and we conclude $\mathcal{N}_2 = \mathcal{M}$. In other words, if $B \in \mathcal{A}_0$ and $A \in \mathcal{M}$, then $A \cap B \in \mathcal{M}$.
Let $\mathcal{N}_3 = \{A \in \mathcal{M} : A \cap B \in \mathcal{M} \text{ for all }B \in \mathcal{M}\}$. As in the preceding paragraph, $\mathcal{N}_3$ is a monotone class contained in $\mathcal{M}$. By the last sentence of the preceding paragraph, $\mathcal{N}_3$ contains $\mathcal{A}_0$. Hence $\mathcal{N}_3 = \mathcal{M}$.
We thus have that $\mathcal{M}$ is a monotone class closed under the operations of taking complements and taking finite intersections. If $A_1, A_2, \ldots$ are elements of $\mathcal{M}$, then $B_n = A_1 \cap \ldots \cap A_n \in \mathcal{M}$ for each $n$ and $B_n \downarrow \cap_{i = 1}^\infty A_i$. Since $\mathcal{M}$ is a monotone class, we have that $\cap_{i = 1}^\infty A_i \in \mathcal{M}$. If $A_1, A_2, \ldots $ are in $\mathcal{M}$, then $A_1^c, A_2^c, \ldots$ are in $\mathcal{M}$, hence $\cap_{i = 1}^\infty A_i^c \in \mathcal{M}$, and then$$\cup_{i = 1}^\infty A_i = (\cap_{i = 1}^\infty A_i^c)^c \in \mathcal{M}.$$This shows that $\mathcal{M}$ is a $\sigma$-algebra, and so $A \subset \mathcal{M}$.

The proof of this theorem is rather technical. I have a few questions about it.

*

*What is the underlying intuition behind the proof?

*What are the one to three key ideas this proof boils down to?

*What is the geometric significance of this result/how can I visualize it?

Thanks in advance!
 A: The phenomenon of closedness conditions, corresponding closed classes, and closure generations is quite common in mathematics. Often, you have a definition like „a class of some objects is called something if it is closed under the following operations“. For example:


*

*Topology is a collection of sets closed under finite intersections and arbitrary unions.

*A linear subspace is a set closed under the additive group operations and under multiplication by scalars. More generally, a subalgebra (in the sense of universal algebra) is a subset of an algebra closed under all operations in that algebra.

*An algebraic variaty is a class of algrebras closed under taking subalgebras, products, and homomorphic images.

*An equivalence is a set of pairs closed under diagonal, symetrization and transitivity.

*And here we have set algebras, σ-algebras, and monotone classes, and we consider operations of complement and finite, countably infinite, and monotone countable infinite union and intersection.


For all such propeties it holds that intersection of any collection of closed classes is a closed class, so for any class you may consider the smallest closed class containing the base class – the intersection of all closed classes containing the base class. This may be called the closed class generated by the base class (e.g. linear subspace generated by a set, σ-algebra generated by a collection of sets).
We can view the generation two ways – the “outer” way, which more suitable as a definition – as I said, we consider the intersection of all closed classes containing the base class. We often use the following idea: if a class contains the base class and is closed, then it contains the closed class generated by the base class.
We could call the other way “inner”: you start with the base class and consider the results of the operations with arguments from the base class – you need to add these elements, so you add them. But the resulting class needn't be closed since now we have more arguments and so more potential results, so we have to iterate this construction of adding the required elements. After possibly transfinitely many steps, the construction stabilizes and we have the generated closed class. For example, in vectors spaces, if we consider all finite sums together, the construction stabilizes after one step – a linear combination of linear combinations is just a linear combination. In the case of σ-algebras, we have finitely many operations, but the unions and/or intersections are countably infinite operations, so you may need up to $ω_1$ steps.

Now, it may happen that closing under some condition preserves previous closedness under another condition, which is the case here. We want to show that when starting with an algebra and closing under monotone countable unions and intersections, we already get closedness under all countable unions and intersections and the closedness under complement is preserved.
We observe that since we want to have complement, it is enough to prove just closedness under countable unions or just closedness under countable intersections. Also, since every countable union is a mononote countable union of finite unions (and the same for intersections), it is enough to show that the monotone closure preserves closedness under complements and under finite unions and/or intersections, i.e. that it preserves the property of being an algebra.
To prove that, there are two ways (of course sharing the core property) – the way in the given proof corresponds to the outer way of generation – we want to show that the generated monotone class is closed under complements, so we consider the arguments for which it is true and show that this is in fact he whole class – by showing that it is a monotone class and hence containing the generated class (this is the usual trick).
The other way based on the inner view is to show that the desired closedness is preserved at each step of closing under the other condition(s). Since the operations we want to preserve are finitary, we are done at the limit steps, so it is enough to show: if we have a class closed under complements, then the class of all monotone countable unions and intersetions is also closed under complements.
Both ways boil down to the observation, that you may switch the order of operations: first forming a monotone counable union and then the complements (which may be potentially dangerous) is the same thing as first taking the complements and then forming the countable intersection (which is safe since the base class is closed under complements).
This was quite easy for the unary operation of complement. For the binary operation of intersection (or equivalently union), we need more steps or cases. In the outer way presented, we consider the first arguments such that the property holds for all second arguments from the base class, and then we consider all right arguments such that the property holds for every first argument from the whole generated class. This could be generalized to $n$-ary operation.
Alternatively, we would need to observe that for monotone sequenses $A, B$ we have


*

*$\bigcap_{i = 1}^∞ A_i ∩ \bigcap_{j = 1}^∞ B_i = \bigcap_{i = 1}^∞ (A_i ∩ B_i)$,

*$\bigcap_{i = 1}^∞ A_i ∩ \bigcup_{j = 1}^∞ B_i = \bigcup_{j = 1}^∞ ((\bigcap_{i = 1}^∞ A_i) ∩ B_j) = \bigcup_{j = 1}^∞ \bigcap_{i = 1}^∞ (A_i ∩ B_j)$,

*$\bigcup_{i = 1}^∞ A_i ∩ \bigcup_{j = 1}^∞ B_i = \bigcup_{i = 1}^∞ (A_i ∩ B_i)$.


Again, in both ways the core idea is to switch the order of operations to be safe.

To sum up,


*

* we are in a general situation of generating classes closed under some operations.

* Somethimes we need to observe that closing under some opeartions preserve closedness under some other operations, so generating something more closed may be easier when we are starting with someting already closed to some extent.

* To prove such preservation we need to show that the order of the operations may be switched for all desired operations and situations.

