Confused with the notation used in proof of monotone class theorem and the logic behind the proof We were introduced to the monotone class theorem which stated that (in my own understanding) the minimal sigma field generated by the field $\mathcal{C}$ (denoted $\sigma[\mathcal{C}]$) equals to the minimal monotone class generated by that same field $\mathcal{C}$(denoted $\mathcal{M[C]}$). i.e. $$m[C]=\sigma[C] \ \ \ \text{when } C \text{ is a field}$$
I was introduced to this theorem by the textbook Probability for Statisticians 2nd edition by Galen R.Shorack first. I was able to follow the proof in the textbook by checking all the statements are true, but I didn't quite get the idea behind the proof. I revisited the concept and the proof today using the Youtube source: https://youtu.be/4XhMgu_ekeY?t=606.

I find myself confused with the notation of the proof: I think if $E\subset \mathcal{M[Q]}$ (In this series of video, $ \mathcal{Q}$ is an algebra), then $E$ must be a class of sets. However, in $g(E)=\{F\in\mathcal{M[Q]}:E\backslash F,E\cap F,F\backslash E \in\
\mathcal{M(Q)}\}$ the symbol $E$ seems to represent the element of class, which is set. Also, in the claim, $E\in\mathcal{Q}$ also suggests $E$ is an element in the class.
The lecturer constructs $g(E)$ and show it contains $Q$ and is a monotone class. It seems to me This step means for any given $E\in \mathcal{M(Q)}$, by the function $g$, we will get a corresponding class of sets $g(E)$ which is a monotone class. Does that mean we are mapping a set to a class of set? In fact, in the textbook that I used for studying, the notation is introduced as: For each $A \in \mathcal{M}, \operatorname{let} \mathcal{M}_{A} \equiv\left\{B \in \mathcal{M}: A B, A^{c} B, A B^{c} \in \mathcal{M}\right\} .$ And we were shown that $\mathcal{M}_{A}$ is a monotone class. It didn't occur to me we can think of the construction as some sort of function.
There are many proofs in the textbook involving constructing a set of this kind and prove the constructed sets satisfies the certain property. Whenever I checked the properties of the set as stated, those properties always hold as if by magic. I fancy by thinking those constructions as function may be insightful and illuminating but I lack the mathematical background to make my mind clearer. I apologize for the lengthy explanation of my confusion which may be something very basic and simple. Could someone please explain to me how to think the proof from the perspective of a function and explain the intuition behind the proof? If any fundamental concepts (for example, in topology, functional analysis, set theory) were needed to better understand the idea behind the proof I am more than happy the direction could be pointed.
 A: $\newcommand{\C}{\mathscr{C}}$
$\newcommand{\G}{\mathscr{G}}$
I think your confusion makes perfect sense and I agree that the notation used in the video is at least very loose, if not wrong. Below is a clearer proof.
As you may have already familiarized, to finish the proof, you need to justify two directions:

*

*$\sigma(\C) \subset m(\C).$

*$m(\C) \subset \sigma(\C).$
Since (by definition) $\sigma(\C)$ is the smallest sigma-field that contains $\C$ and $m(\C)$ is the smallest monotone class that contains $\C$, to show $1$, it would be sufficient to show $m(\C)$ is a sigma-field (as we already know $m(\C)$ contains $\C$); similarly, to show 2, it suffices to show $\sigma(\C)$ is a monotone class.
2 is immediate, as a sigma-field is of course a monotone class. In fact, if $A_1, A_2, \ldots \in \sigma(\C)$ and $A_n \uparrow A$, then $A = \cup_{n = 1}^\infty A_n \in \sigma(\C)$, since any sigma-field is closed under the formation of countable unions. Similarly you can check the other condition in the monotone class is also satisfied.
1 is trickier. First note that if $m(\C)$ is a field, then it must be a sigma-field (left you as an exercise, if you have difficulty, I can edit the answer later), thus it suffices to show $m(\C)$ is a field. To this end, first consider the class
\begin{align*}
\G = \{A: A^c \in m(\C)\}.
\end{align*}
Since $m(\C)$ is monotone, so is $\G$. Since $\C$ is a field, $\C \subset \G$, and so $m(\C) \subset \G$ (again, use the minimality). Hence $m(\C)$ is closed under complementation.
Next define the class
\begin{align*}
\G_1 = \{A: A \cup B \in m(\C) \text{ for all } B \in \C\}.
\end{align*}
Then it is straightforward to check (use definition) that $\G_1$ is a monotone class and $\C \subset \G_1$; from the minimality of $m(\C)$ follows $m(\C) \subset
\G_1$. Now define
\begin{align*}
\G_2 = \{B: A \cup B \in m(\C) \text{ for all } A \in m(\C)\}.
\end{align*}
Then $\G_2$ is a monotone class, and from $m(\C) \subset \G_1$ it follows that $A \in m(\C)$ and $B \in \C$ together imply that $A \cup B \in m(\C)$; in other words, $B \in \C$ implies that $B \in \G_2$. Thus $\C \subset \G_2$; by minimality, $m(\C) \subset \G_2$. Therefore, $A, B \in m(\C)$ implies that $A \cup B \in m(\C)$.
This completes showing $m(\C)$ is a field.
To summarize, I hope the route of reaching the conclusion is now clear to you: that is,

*

*Why we would think of defining classes $\G, \G_1, \G_2$ and how they help us to achieve our ultimate goal. I think this corresponds to your statement "Whenever I checked the properties of the set as stated, those properties always hold as if by magic. I fancy by thinking those constructions as function may be insightful and illuminating but I lack the mathematical background to make my mind clearer." At the first glance, the creation of $\G, \G_1, \G_2$ does look like a magic coming out from nowhere, but what if you compare them with the definition of field, which is the destination that we want to arrive? Did you see any similarity? In fact, $\G, \G_1, \G_2$ are so defined that all properties of the field can be checked through them.


*How "minimality" are repeatedly are used in the whole argument.
This trick is extensively used in measure theory, to get more practice of implementing it, you may consider reviewing the proof of the Dynkin's $\pi$-$\lambda$ theorem and trying to prove this exercise by yourself.
