prove that an arbitrary intersection of σ-algebras is a σ-algebra, i.e. let {A_α} where α is in I (indexing set) be a collection of σ-algebras on a given setΩ. Prove that A=intersection {A_α}(where α is in I)

Printable View

- May 15th 2008, 08:48 PMsquarerootof2measure theory questionprove that an arbitrary intersection of σ-algebras is a σ-algebra, i.e. let {A_α} where α is in I (indexing set) be a collection of σ-algebras on a given set
**Ω**. Prove that A=intersection {A_α}(where α is in I)

- May 15th 2008, 09:21 PMIsomorphism
Courtesy Wiki:

Quote:

A subset Σ of the power set of a set*X*is a σ-algebra if and only if it has the following properties:- Σ is nonempty
- If
*E*is in Σ then so is the complement (*X*\*E*) of*E*. - The union of countably many sets in Σ is also in Σ.

Which axiom are you having trouble with? - May 15th 2008, 09:41 PMsquarerootof2
i think the first condition is pretty trivial, but i'm having trouble with 2nd and 3rd ones. maybe i'm not grasping the definition of sigma algebra that well..

- May 15th 2008, 10:22 PMIsomorphism
For 2) $\displaystyle \forall \alpha \in I, E \in A_{\alpha}\Rightarrow \Omega - E \in A_{\alpha}$ since $\displaystyle A_{\alpha}$ is a sigma-algebra.

And how is (1) trivial? It is easily possible that intersection of a few sets can easily be empty :( - May 19th 2008, 11:26 AMfrenzy
1 is trivial...

for any $\displaystyle \alpha \in I$ we know that $\displaystyle A_{\alpha}$ is nonempty since it is a sigma algebra. Thus $\displaystyle E \in A_{\alpha}$

and therefore $\displaystyle E^c \in A_{\alpha}$

also $\displaystyle \Omega=E\cup E^c \in A_{\alpha}$

thus $\displaystyle \Omega\in A_{\alpha}$ for all $\displaystyle \alpha \in I$

Hence

$\displaystyle \Omega\in \displaystyle\bigcap_{\alpha \in I}{A_{\alpha}}$