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 real-valued functions on $X$ containing the constant functions and satisfying the following,

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

• for every set $A\in 𝒮$ the characteristic function $1_A$ is in $\mathscr{H}$.

Then, $\mathrm{H}$ contains every bounded and measurable function from $X$ to $\mathrm{R}$.

Let $𝒟$ 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 $X\in 𝒟$, and $𝒮\subseteq 𝒟$. For any $A\subseteq B$ in $𝒟$ 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 $𝒟$.If $A_n\in 𝒟$ 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 ${\bigcup }_n{A}_n\in 𝒟$. It follows that $𝒟$ is a Dynkin system, and Dynkin’s lemma shows that it contains the $\sigma$-algebra $𝒜$.

We have shown that $1_A\in \mathscr{H}$ for every $A\in 𝒜$. Now consider any bounded and measurable function $f:X\to ℝ$ taking values in a finite set $S\subseteq ℝ$. 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 $f:X\to ℝ$, 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 $f:X\to ℝ$ 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 $f:X\to ℝ$ 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 $g_n\equiv f_n-2^{1-n}+2+\parallel f\parallel$. 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 $𝒦$ 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 $𝒦$ 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 $p\in ℝ[X]$.Then, for any continuous function $p:ℝ\to ℝ$, 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 $f,g\in {\overline{\mathscr{H}}}_0$, $f\wedge g=f-|f-g|$ and the maximum $f\vee g=f+|g-f|$ will be in ${\overline{\mathscr{H}}}_0$.

We let $𝒮$ 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 $𝒜$, 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 $A\cap B\in 𝒮$ and $𝒮$ is a $\pi$-system.

Finally, choose any $f\in 𝒦$ and $a\in ℝ$. Then, $f_n=((n(f-a))\vee 0)\wedge 1$ is a sequence of functions in ${\overline{\mathscr{H}}}_0$ increasing pointwise to $1_{f^{-1}((a,\mathrm{\infty }))}$. So, $f^{-1}((a,\mathrm{\infty }))\in 𝒮$. As intervals of the form $(a,\mathrm{\infty })$ generate the Borel $\sigma$-algebra on $ℝ$, it follows that $𝒮$ generates the $\sigma$-algebra $𝒜$, as required.