measure theory question

**squarerootof2** · May 15th 2008, 08:48 PM

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 set Ω. Prove that A=intersection {A_α}(where α is in I)

**Isomorphism** · May 15th 2008, 09:21 PM

Courtesy Wiki:

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.

I dont know topology but I think proving 2 and 3 is easy. Since they follow directly from set properties...
Which axiom are you having trouble with?

**squarerootof2** · May 15th 2008, 09:41 PM

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..

**Isomorphism** · May 15th 2008, 10:22 PM

For 2) $\forall \alpha \in I, E \in A_{\alpha}\Rightarrow \Omega - E \in A_{\alpha}$ since $A_{\alpha}$ is a sigma-algebra.

And how is (1) trivial? It is easily possible that intersection of a few sets can easily be empty

**frenzy** · May 19th 2008, 11:26 AM

Originally Posted by Isomorphism

For 2) $\forall \alpha \in I, E \in A_{\alpha}\Rightarrow \Omega - E \in A_{\alpha}$ since $A_{\alpha}$ is a sigma-algebra.

And how is (1) trivial? It is easily possible that intersection of a few sets can easily be empty

1 is trivial...

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

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

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

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

Hence

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

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