Monotone Class Theorem has two types. One is Monotone class theorem for sets and the other for functions. I have no doubt for sets.
Here is a reference of definition of Monotone Class Theorem for functions from Probability: Theory and Examples, Durrett Rick, 4.1-edition.
Theorem 6.1.3. Monotone class theorem. Let $\mathcal {A}$ be a $\pi$-system that contains $\Omega$ and let $\mathcal {H}$ be a collection of real-valued functions that satisfies:
(i) If $A ∈ \mathcal {A}$, then $1_A ∈ \mathcal {H}$.
(ii) If $f, g ∈ \mathcal {H}$, then $f + g$, and $cf ∈ \mathcal {H}$ for any real number $c$.
(iii) If $f_n ∈ \mathcal {H}$ are nonnegative and increase to a bounded function $f$, then $f ∈ \mathcal {H}$.
Then $\mathcal {H}$ contains all bounded functions measurable with respect to $\sigma (\mathcal {A})$.
Proof. The assumption $\Omega \in \mathcal{A}$, (ii), and (iii) imply that $\mathcal {G} = \{A : 1_A ∈ \mathcal {H}\}$ is a $\lambda$-system so by (i) and the $\pi − \lambda$ theorem, Theorem 2.1.2, $\mathcal {G} ⊃ \sigma(A)$. (ii) implies $\mathcal{H}$ contains all simple functions, and (iii) implies that $\mathcal{H}$ contains all bounded measurable functions.
I can't understand one sentence from the proof, "(ii) implies $\mathcal{H}$ contains all simple functions". Why?
P.S.
I think "measurable with respect to $\sigma(\mathcal{A})$" means $\sigma(\mathcal {A})$-measurable. I know $\mathcal{F}$-measurable mapping that is suppose $(\Omega, \mathcal{F})$ and $(E, \mathcal{E})$ are two measurable spaces, $f$ is a mapping from $\Omega \to E$. $f$ is a $\mathcal{F}$-measurable mapping if $f^{-1}(A) \in \mathcal{F}$ for $\forall A \in \mathcal{E}$. So $\sigma(\mathcal {A})$-measurable probably means those $f: (\Omega, \sigma (\mathcal{A})) \to ([0, +\infty), \mathcal{B}{[0, +\infty)})$ where $\mathcal{B}{[0, +\infty)}$ denotes Borel $\sigma$-algebra on $[0, +\infty)$.
Update:
The author didn't make the sentence clear. So I try to make it a little bit more complete. Let $\forall f$ is $\sigma(\mathcal{A})$-measurable. I suppose $g_n =\sum_{k=0}^{n2^n -1} \frac{k}{2^n} I_{\{\frac{k}{2^n} \le f^+ ＜ \frac{k+1}{2^n}\}} + n I_{f^+ \ge n}$ where $f^+ = f ∨ 0$. Then $g_n \in \mathcal{H}$ coz $f$ is $\sigma(\mathcal{A})$-measurable that is $\{x: \frac{k}{2^n} \le f^+ ＜ \frac{k+1}{2^n} \} \in \sigma(\mathcal{A})$ and $\{x: f^+ \ge n\} \in \sigma(\mathcal{A})$. Then $g_n$ increasingly converges to $f^+$ and so does $f^-$. $f = f^+ – f^- \in \mathcal{H}$. Please note that I can only get simple functions defined on $\{x: \frac{k}{2^n} \le f^+ ＜ \frac{k+1}{2^n} \}$ and $\{x: f^+ \ge n\}$ are in $\mathcal{H}$. I can't prove $\mathcal{H}$ will contains all simple functions. Am I right?
Best Answer
A simple function is of the form $$f(x) = \sum_{i = 1}^N a_i 1_{A_i}$$ a finite linear combination of functions $1_{A_i}$ for $A_i \in \sigma(A)$.
We would like to prove that those functions are in $\mathcal{H}$.
So we consider $\mathcal{C} = \{B \in \sigma(A)\mid 1_B \in \mathcal{H}\}$
So $\mathcal{A} \subset \mathcal{C}$ by $(i)$.
Since $\Omega \in \mathcal{A}$, $\Omega \in \mathcal{C}$
Moreover, if $A \subset B \in \mathcal{C}$ then $ B \setminus A \in \mathcal{C}$ by $(ii)$
Now take $A_1, A_2 , \ldots A_n, \ldots$ an increasing sequence $A_i \subset A_{i+1}$. Then $A = \cup_i A_i$ is in $\mathcal{C}$ by $(iii)$
This proves that $\mathcal{C}$ is a $\lambda$-system containing the $\pi$-system $\mathcal{A}$. By Dynkin $\pi-\lambda$ theorem (https://en.wikipedia.org/wiki/Dynkin_system) $\mathcal{C} = \sigma(\mathcal{A})$
So now we know that $1_{A_1}, \ldots, 1_{A_N}$ are in $\mathcal{H}$ for any finite choice of $A_i \in \sigma(\mathcal{A})$, by $(ii)$ $a_1 1_{A_1} + \ldots + a_N 1_{A_N}$ is in $\mathcal{H}$.
This is the claim of the author.
If you would like to see that any $\sigma(\mathcal{A})$-measurable function is in $\mathcal{H}$ It suffices to consider $f = f^+ - f^-$ and to prove that $f^+,f^-$ are in $\mathcal{H}$.
The argument is essentially the one you presented.
Let $f$ be $\sigma(\mathcal{A})$-measurable. define $$g_n =\sum_{k=0}^{n2^n -1} \frac{k}{2^n} 1_{\{\frac{k}{2^n} < f^+ \le \frac{k+1}{2^n}\}} + n 1_{f^+ > n}$$ where $f^+ = f \vee 0$. Then $g_n \in \mathcal{H}$ once $\{x: \frac{k}{2^n} < f^+ \le \frac{k+1}{2^n} \} \in \sigma(\mathcal{A})$ and $\{x: f^+ > n\} \in \sigma(\mathcal{A})$ (because $f$ is $\sigma(\mathcal{A})$-measurable). Then $g_n$ increasingly converges to $f^+$. Since $f$ is bounded so is $f^+$ and therefore by $(iii)$ $f^+ \in \mathcal{H}$. The argument for $f^{-}$ is analogous.
this concludes the proof.