Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

Two question about this choice:

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

Thanks in advance!


share|improve this question
$F$ 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 : Regarding your first question I think you should compare Dynkin Lemma with the 'set ' version of the Monotone class Theorem. You can find definitions, axioms and theorems at planetmath.org. But from the proof of Functional MCT that you can find there, Dynkin's lemma is used in the proof of FMCT (theorem 1). Best regards –  TheBridge Mar 13 '12 at 15:16
@ The Bridge: Thank you for your answer! I will check this right now :) –  user20869 Mar 14 '12 at 8:25
@ 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

1 Answer 1

1) Monotone class theorem and $\pi-\lambda$-system of Dynkin are complementary ways to prove that a certain set of subsets contains a $\sigma$- algebra.

One can show that $M(G)$ the smallest monotone class of an algebra $G$ is a $\lambda$- system. Similarly, one can show that $\lambda(P)$ the smallest $\lambda$- system of a $\pi$- system $G$ is a monotone class.

The point is to see which is the simpler criterion.

Q:It is easier to check that $G$ is an algebra or to check that it is a $\pi$- system?

A: It is easier to check that $G$ is a $\pi$ system (every algebra is a $\pi$-system the converse does not follow)

Q:It is easier to check that $M$ is a monotone class or to check that it is a $\lambda$- system?

A: It is easier to check that $M$ is a $\lambda$ system (every algebra is a $\lambda$-system the converse does not follow)

2) To see that $\sigma(M) = \mathcal{F}_s$ note that you can approximate $1_{A_{s_1}}(X_1) 1_{A_{s_2}}(X_2) \ldots 1_{A_{s_k}}(X_k)$ with continuous functions $f_1(X_{s_1})f_2(X_{s_2})\ldots f_k(X_{s_k})$ (see https://en.wikipedia.org/wiki/Urysohn%27s_lemma)

the reason you choose the family of product of continuous functions is that one often deals better with properties for continuous functions.(in fact one may only need a denumerable set of continuous functions , depending on the problem at hand this is very useful)

note that one can know a measure by it's values on measurable sets ($\{\mu(A), A \in \mathcal{F}\}$) But one can also know a measure by it's values on continuous functions $\{\mu(f) = \int f \, d\mu, f \in C(X)\}$ when $X$ is a locally compact Hausdorff space. (see https://en.wikipedia.org/wiki/Riesz%E2%80%93Markov%E2%80%93Kakutani_representation_theorem)

share|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.