Dynkin’s lemma
Dynkin’s lemma is a result in measure theory showing that the $\sigma $-algebra (http://planetmath.org/SigmaAlgebra) generated by any given $\pi $-system (http://planetmath.org/PiSystem) on a set $X$ coincides with the Dynkin system generated the $\pi $-system. The result can be used to prove that measures^{} are uniquely determined by their values on $\pi $-systems generating the required $\sigma $-algebra. For example, the Borel $\sigma $-algebra on $\mathbb{R}$ is generated by the $\pi $-system of open intervals $(a,b)$ for $$ and consequently the Lebesgue measure^{} $\mu $ is uniquely determined by the property that $\mu ((a,b))=b-a$.
Note that this lemma generalizes the statement that a Dynkin system which is also a $\pi $-system is a $\sigma $-algebra.
Lemma (Dynkin).
Let $A$ be a $\pi $-system on a set $X$. Then $\mathrm{D}\mathit{}\mathrm{(}A\mathrm{)}\mathrm{=}\sigma \mathit{}\mathrm{(}A\mathrm{)}$. That is, the smallest Dynkin system containing $A$ coincides with the $\sigma $-algebra generated by $A$.
Proof.
As $A$ is a $\pi $-system, the set ${\mathcal{D}}_{1}\equiv \{S\subseteq X:S\cap T\in \mathcal{D}(A)\text{for every}T\in A\}$ contains $A$. We show that ${\mathcal{D}}_{1}$ is also a Dynkin system.
First, for every $T\in A$, $X\cap T=T\in A$ so $X$ is in ${\mathcal{D}}_{1}$. Second, if ${S}_{1}\subseteq {S}_{2}$ are in ${\mathcal{D}}_{1}$ and $T\in A$ then $({S}_{2}\setminus {S}_{1})\cap T=({S}_{2}\cap T)\setminus ({S}_{1}\cap T)$ is in $\mathcal{D}(A)$ showing that ${S}_{2}\setminus {S}_{1}\in {\mathcal{D}}_{1}$. Finally. if ${S}_{n}\in {\mathcal{D}}_{1}$ is a sequence increasing to $S\subseteq X$ and $T\in A$ then ${S}_{n}\cap T$ is a sequence in $\mathcal{D}(A)$ increasing to $S\cap T$. As Dynkin systems are closed under limits of increasing sequences this shows that $S\cap T\in \mathcal{D}(A)$ and therefore $S\in {\mathcal{D}}_{1}$. So ${\mathcal{D}}_{1}$ is indeed a Dynkin system. In particular, $\mathcal{D}(A)\subseteq {\mathcal{D}}_{1}$ and $S\cap T\in \mathcal{D}(A)$ for all $S\in \mathcal{D}(A)$ and $T\in A$.
We now set ${\mathcal{D}}_{2}\equiv \{S\subseteq X:S\cap T\in \mathcal{D}(A)\text{for every}T\in \mathcal{D}(A)\}$ which, as shown above, contains $A$. Also, as in the argument^{} above for ${\mathcal{D}}_{1}$, ${\mathcal{D}}_{2}$ is a Dynkin system. Therefore, $\mathcal{D}(A)$ is contained in ${\mathcal{D}}_{2}$ and it follows that $S\cap T\in \mathcal{D}(A)$ for any $S,T\in \mathcal{D}(A)$. So $\mathcal{D}(A)$ is both a $\pi $-system and a Dynkin system.
We can now show that $\mathcal{D}(A)$ is a $\sigma $-algebra. As it is a Dynkin system, ${S}^{c}=X\setminus S\in \mathcal{D}(A)$ for every $S\in \mathcal{D}(A)$ and, as it is also a $\pi $-system, this shows that $\mathcal{D}(A)$ is an algebra of sets^{} on $X$. Finally, choose any sequence ${A}_{n}\in \mathcal{D}(A)$. Then, ${\bigcup}_{m=1}^{n}{A}_{m}$ is a sequence in $\mathcal{D}(A)$ increasing to ${\bigcup}_{n}{A}_{n}$ which, as $\mathcal{D}(A)$ is Dynkin system, must be in $\mathcal{D}(A)$. So, $\mathcal{D}(A)$ is a $\sigma $-algebra and must contain $\sigma (A)$. Conversely, as $\sigma (A)$ is a Dynkin system (as it is a $\sigma $-algebra) containing $A$, it must also contain $\mathcal{D}(A)$. ∎
References
- 1 David Williams, Probability with martingales, Cambridge Mathematical Textbooks, Cambridge University Press, 1991.
Title | Dynkin’s lemma |
---|---|
Canonical name | DynkinsLemma |
Date of creation | 2013-03-22 18:33:05 |
Last modified on | 2013-03-22 18:33:05 |
Owner | gel (22282) |
Last modified by | gel (22282) |
Numerical id | 11 |
Author | gel (22282) |
Entry type | Theorem |
Classification | msc 28A12 |
Synonym | pi-system d-system lemma |
Related topic | PiSystem |
Related topic | DynkinSystem |
Related topic | UniquenessOfMeasuresExtendedFromAPiSystem |