I am reading the proof of the monotone class theorem here and do not understand how Theorem 1 follows from Proposition 1. Let $\Omega$ be any set.
To be clear, we have:
Proposition 1: If $S$ is a $\pi$-system with $\Omega \in S$ and $\mathcal{H}$ is a vector space of real valued functions on $\Omega$ such that
(1) $\mathbb{1}_A \in \mathcal{H} \quad \forall A \in S$
(2) If $0 \leq f_1 \leq f_2 \leq … \leq f_n \quad \forall n \in \mathbb{N}$ and $\lim_{n \rightarrow \infty} f_n \equiv f$ is bounded, then $f \in \mathcal{H}$,
then $\mathcal{H}$ contains all bounded $\sigma(S)$ measurable functions.
Proof:
Set $\mathcal{D} \equiv \{A \subseteq \Omega: 1_A \in \mathcal{H}\}$ and note that $\mathcal{D}$ is a $\lambda$ system containing $S$ so that $1_A \in \mathcal{H} \quad \forall A \in \sigma(S)$ by Dynkin's $\pi/\lambda$ theorem, and for any bounded $f$ that is $\sigma(S)$ measurable, we may write it as an increasing limit of sums of simple functions which are in $\mathcal{H}$ since $\mathcal{H}$ is a vector space, so the theorem follows immediately from (2).
I want to show the following, which is apparently proven in the linked notes package but I don't understand the proof.
Proposition 2: Suppose $\mathcal{M}$ is a set of bounded functions on $\Omega$ such that $f, g \in \mathcal{M} \implies fg \in \mathcal{M}$
Suppose $\mathcal{H}$ is a vector space of real valued functions on $\Omega$ for which (2) above holds and:
(i) $\mathcal{M} \subseteq \mathcal{H}$
(ii) $1_\Omega \in \mathcal{H}$
Then $\mathcal{H}$ contains all bounded $\sigma(\mathcal{M})$ measurable functions, where $\sigma(\mathcal{M}) = \sigma( \{ f^{-1}(B) : f \in \mathcal{M}, B \in \mathscr{B}(\mathbb{R}) \})$
Proof: I understand the first line in the proof in the notes package. It would obviously suffice to show that $1_A \in \mathcal{H}$ for each $A$ in a $\pi$ system that generates $\sigma(\mathcal{M})$. From there, I don't understand what the algebra generated by $\mathcal{M}$ (in their notation $\mathcal{K}$) is.
Please help if you can!
Best Answer
The proof of Theorem 1 takes the $\pi$-system $\mathcal C$ of all finite intersections of sets of the form $\{f > b\}$ and will show that $1_A\in {\mathcal H}$ for every $A\in{\mathcal C}$. To do so it builds up larger and larger classes that contain $\mathcal M$, each of which is found to be within $\mathcal H$. The final class is rich enough to argue that $1_A\in{\mathcal H}$.
First construct ${\mathcal A}_0$, the algebra generated by $\mathcal M$. That is, ${\mathcal A}_0$ is the set containing all $\alpha f + \beta g$ and $fg$ for $f$, $g$ in $\mathcal M$ and $\alpha,\beta\in{\mathbb R}$. Argue that ${\mathcal A}_0\subset {\mathcal H}$.
Next define $\mathcal A$ to be the closure of ${\mathcal A}_0$ under uniform convergence. The Lemma proves that $\mathcal A$ is a subset of $\mathcal H$. Check that $\mathcal A$ remains an algebra (because we need products and sums of functions in $\mathcal A$ to be in $\mathcal A$).
Use Weierstrass' theorem and uniform convergence to prove that $|f|\in {\mathcal A}$ for every $f\in {\mathcal A}$. It follows that $\mathcal A$ is closed under $\max$ and $\min$ and therefore is closed under $f\mapsto f^+$.
Construct $g_n$ as advertised. All the previous work explains why each $g_n$ is a member of $\mathcal A$. Since $\mathcal H$ is closed under nondecreasing limits, $\lim g_n$ is in $\mathcal H$.