proof of functional monotone class theorem
We start by proving the following version of the monotone class theorem.
Theorem 1.
Let $\mathrm{(}X\mathrm{,}\mathrm{A}\mathrm{)}$ be a measurable space^{} and $\mathrm{S}$ be a $\pi $system (http://planetmath.org/PiSystem) generating the $\sigma $algebra (http://planetmath.org/SigmaAlgebra) $\mathrm{A}$. Suppose that $\mathrm{H}$ be a vector space^{} of realvalued functions on $X$ containing the constant functions and satisfying the following,

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if $f:X\to {\mathbb{R}}_{+}$ is bounded^{} and there is a sequence of nonnegative functions ${f}_{n}\in \mathscr{H}$ increasing pointwise to $f$, then $f\in \mathscr{H}$.

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for every set $A\in \mathcal{S}$ the characteristic function^{} ${1}_{A}$ is in $\mathscr{H}$.
Then, $\mathrm{H}$ contains every bounded and measurable function^{} from $X$ to $\mathrm{R}$.
Let $\mathcal{D}$ consist of the collection^{} of subsets $B$ of $X$ such that the characteristic function ${1}_{B}$ is in $\mathscr{H}$. Then, by the conditions of the theorem, the constant function ${1}_{X}$ is in $V$ so that $X\in \mathcal{D}$, and $\mathcal{S}\subseteq \mathcal{D}$. For any $A\subseteq B$ in $\mathcal{D}$ then ${1}_{B\setminus A}={1}_{B}{1}_{A}\in \mathscr{H}$, as $\mathscr{H}$ is closed under linear combinations^{}, and therefore $B\setminus A$ is in $\mathcal{D}$. If ${A}_{n}\in \mathcal{D}$ is an increasing sequence, then ${1}_{{A}_{n}}\in \mathscr{H}$ increases pointwise to ${1}_{{\bigcup}_{n}{A}_{n}}$, which is therefore in $\mathscr{H}$, and ${\bigcup}_{n}{A}_{n}\in \mathcal{D}$. It follows that $\mathcal{D}$ is a Dynkin system, and Dynkin’s lemma shows that it contains the $\sigma $algebra $\mathcal{A}$.
We have shown that ${1}_{A}\in \mathscr{H}$ for every $A\in \mathcal{A}$. Now consider any bounded and measurable function $f:X\to \mathbb{R}$ taking values in a finite set^{} $S\subseteq \mathbb{R}$. Then,
$$f=\sum _{s\in S}s{1}_{{f}^{1}(\{s\})}$$ 
is in $\mathscr{H}$.
We denote the floor function by $\lfloor \cdot \rfloor $. That is, $\lfloor a\rfloor $ is defined to be the largest integer less than or equal to the real number $a$. Then, for any nonnegative bounded and measurable $f:X\to \mathbb{R}$, the sequence of functions ${f}_{n}(x)={2}^{n}\lfloor {2}^{n}f(x)\rfloor $ each take values in a finite set, so are in $\mathscr{H}$, and increase pointwise to $f$. So, $f\in \mathscr{H}$.
Finally, as every measurable and bounded function $f:X\to \mathbb{R}$ can be written as the difference^{} of its positive and negative parts $f={f}_{+}{f}_{}$, then $f\in \mathscr{H}$.
We now extend this result to prove the following more general form of the theorem.
Theorem 2.
Let $X$ be a set and $\mathrm{K}$ be a collection of bounded and real valued functions on $X$ which is closed under multiplication, so that $f\mathit{}g\mathrm{\in}\mathrm{K}$ for all $f\mathrm{,}g\mathrm{\in}\mathrm{K}$. Let $\mathrm{A}$ be the $\sigma $algebra on $X$ generated by $\mathrm{K}$.
Suppose that $\mathrm{H}$ is a vector space of bounded real valued functions on $X$ containing $\mathrm{K}$ and the constant functions, and satisfying the following

•
if $f:X\to \mathbb{R}$ is bounded and there is a sequence of nonnegative functions ${f}_{n}\in \mathscr{H}$ increasing pointwise to $f$, then $f\in \mathscr{H}$.
Then, $\mathrm{H}$ contains every bounded and real valued $\mathrm{A}$measurable function on $X$.
Let us start by showing that $\mathscr{H}$ is closed under uniform convergence. That is, if ${f}_{n}$ is a sequence in $\mathscr{H}$ and $\parallel {f}_{n}f\parallel \equiv {sup}_{x}{f}_{n}(x)f(x)$ converges^{} to zero, then $f\in \mathscr{H}$. By passing to a subsequence if necessary, we may assume that $\parallel {f}_{n}{f}_{m}\parallel \le {2}^{n}$ for all $m\ge n$. Define ${g}_{n}\equiv {f}_{n}{2}^{1n}+2+\parallel f\parallel $. Then ${g}_{n}\in \mathscr{H}$ since $\mathscr{H}$ is a vector space containing the constant functions. Also, ${g}_{n}$ are nonnegative functions increasing pointwise to $f+2+\parallel f\parallel $ which must therefore be in $\mathscr{H}$, showing that $f\in \mathscr{H}$ as required.
Now let ${\mathscr{H}}_{0}$ consist of linear combinations of constant functions and functions in $\mathcal{K}$ and ${\overline{\mathscr{H}}}_{0}$ be its closure^{} (http://planetmath.org/Closure) under uniform convergence. Then ${\overline{\mathscr{H}}}_{0}\subseteq \mathscr{H}$ since we have just shown that $\mathscr{H}$ is closed under uniform convergence. As $\mathcal{K}$ is already closed under products^{}, ${\mathscr{H}}_{0}$ and ${\overline{\mathscr{H}}}_{0}$ will also be closed under products, so are algebras (http://planetmath.org/Algebra). In particular, $p(f)\in {\overline{\mathscr{H}}}_{0}$ for every $f\in {\overline{\mathscr{H}}}_{0}$ and polynomial^{} $p\in \mathbb{R}[X]$. Then, for any continuous function^{} $p:\mathbb{R}\to \mathbb{R}$, the Weierstrass approximation theorem^{} says that there is a sequence of polynomials ${p}_{n}$ converging uniformly to $p$ on bounded intervals, so ${p}_{n}(f)\to p(f)$ uniformly. It follows that $p(f)\in {\overline{\mathscr{H}}}_{0}$. In particular, the minimum of any two functions $f,g\in {\overline{\mathscr{H}}}_{0}$, $f\wedge g=ffg$ and the maximum $f\vee g=f+gf$ will be in ${\overline{\mathscr{H}}}_{0}$.
We let $\mathcal{S}$ consist of the sets $A\subseteq X$ such that there is a sequence of nonnegative ${f}_{n}\in {\overline{\mathscr{H}}}_{0}$ increasing pointwise to ${1}_{A}$. Once it is shown that this is a $\pi $system generating the $\sigma $algebra $\mathcal{A}$, then the result will follow from theorem 1.
If ${f}_{n},{g}_{n}\in {\overline{\mathscr{H}}}_{0}$ are nonnegative functions increasing pointwise to ${1}_{A},{1}_{B}$ then ${f}_{n}{g}_{n}$ increases pointwise to ${1}_{A\cap B}$, so $A\cap B\in \mathcal{S}$ and $\mathcal{S}$ is a $\pi $system.
Finally, choose any $f\in \mathcal{K}$ and $a\in \mathbb{R}$. Then, ${f}_{n}=((n(fa))\vee 0)\wedge 1$ is a sequence of functions in ${\overline{\mathscr{H}}}_{0}$ increasing pointwise to ${1}_{{f}^{1}((a,\mathrm{\infty}))}$. So, ${f}^{1}((a,\mathrm{\infty}))\in \mathcal{S}$. As intervals of the form $(a,\mathrm{\infty})$ generate the Borel $\sigma $algebra on $\mathbb{R}$, it follows that $\mathcal{S}$ generates the $\sigma $algebra $\mathcal{A}$, as required.
Title  proof of functional monotone class theorem 

Canonical name  ProofOfFunctionalMonotoneClassTheorem 
Date of creation  20130322 18:38:44 
Last modified on  20130322 18:38:44 
Owner  gel (22282) 
Last modified by  gel (22282) 
Numerical id  4 
Author  gel (22282) 
Entry type  Proof 
Classification  msc 28A20 