# Monotone class theorem

I have some question about the Monotone Class Theorem and its application. First I state the Theorem:

Let $\mathcal{M}:=\{f_\alpha; \alpha \in J\}$ be a set of bounded functions, such that $f_\alpha:N\to \mathbb{R}$ where $N$ is a set. Further we suppose that $\mathcal{M}$ is closed under multiplication and define $\mathcal{C}:=\sigma(\mathcal{M})$. Let $\mathcal{H}$ be a real vector space of bounded real-valued functions on $N$ and assume:

1. $\mathcal{H}$ contains $\mathcal{M}$
2. $\mathcal{H}$ contains the constant function $1$.
3. If $0\le f_{\alpha_1}\le f_{\alpha_2}\le \dots$ is a sequence in $\mathcal{H}$ and $f=\lim_n f_{\alpha_n}$ is bounded, then $f\in \mathcal{H}$

Then $\mathcal{H}$ contains all bounded $\mathcal{C}$ -measurable functions.

My first question: I know the Dynkin lemma which deals with $\sigma$-Algebras and $\pi$-Systems. Which is the stronger one, i.e. does Monotone Class imply Dynkin or the other way around?(or are they equal?) A reference for a proof would also be great!

My second question is about an application. Let $X=(X_t)$ be a right continuous stochastic process with $X_0=0$ a.s. and denote by $F=(F_t)$ a filtration, where $F_t:=\sigma(X_s;s\le t)$. I want to show:

If for all $0\le t_1<\dots<t_n<\infty$ the increments $X_{t_{i}}-X_{t_{i-1}}$ are independent then $X_t-X_s$ is independent of $F_s$ for $t>s$.

The hint in the book is to use Monotone Class Theorem. So $$\mathcal{H}:=\{Y:\Omega\to \mathbb{R} \mbox{ bounded };E[h(X_t-X_s)Y]=E[h(X_t-X_s)]E[Y] \forall h:\mathbb{R}\to\mathbb{R} \mbox{ bounded and Borel-measurable}\}$$

This choice is clear. Now they say $$\mathcal{M}:=\{\prod_{i=1}^n f_i(X_{s_i});0\le s_1\le \dots\le s_n\le s,n\in \mathbb{N},f_i\colon\mathbb{R}\to\mathbb{R} \mbox{ bounded and Borel-measurable}\}$$

1. Why is $\sigma(\mathcal{M})=\mathcal{F}_s$?
2. Why do they define $\mathcal{M}$ like this? (As family of products?)

$F$ is a filtration and $\sigma(\mathcal M)$ is a $\sigma$-algebra/family of measurable maps, right? What do you mean then by $F = \sigma(\mathcal M)$? –  Ilya Mar 13 '12 at 9:24
@ Hulik : I think there is a typo in FMCT item 3, the limit should be in $\mathcal{H}$ not in $\mathcal{M}$. Regards –  TheBridge Mar 14 '12 at 14:38
@ hulik : I nmy opinion $\mathcal{M}$ is not the easiest set of function, you should try with $\mathcal{M}$ built rather with products of functions of increments $X_{t_i}-X_{t_{i-1}}$ not $X_{t_i}$'s in my opinion. Best regards –  TheBridge Mar 14 '12 at 14:45