Given a set <mi mathvariant="normal">&#x03A9;<!-- Ω --> with a &#x03C3;<!-- σ --> -field

Frederick Kramer

Frederick Kramer

Answered question

2022-07-08

Given a set Ω with a σ-field F defined on it. Let B be a subset of Ω and define
B F = { B F : F F }. Call this F B .
The claim is that this collection of sets is a σ-field and hence (B, F B ) is a measurable space.
I don't see why F B is a σ-field just because it is a subset of one. Any insight appreciated

Answer & Explanation

Brendan Bush

Brendan Bush

Beginner2022-07-09Added 14 answers

Suggested proof. (Axioms are σ-algebra axioms.) Notation different from question.
Let ( ) be
B E := { B E , E E } = E B
To prove the claim it is a σ-algebra on B we need to show:
(i) B E B
(ii) ( B E ) c in E B E E .
(iii) E B satisfies σ-additivity.
Proof of (i): We know that Ω is in E by Axiom (1).
Therefore, by ( ), ( B Ω ) E B .
But B Ω = B, hence B E B .
Proof of (ii): ( B E ) c = ( B c E c ) (de Morgan).
Now B c and E c are in E by Axiom (2) and their union is in E by Axiom (3).
Hence, by ( ) B ( B c E c ) E B .
But ( B c E c ) = ( Ω Ω ) = Ω, and B Ω = B which is in E B by (i).
Proof of (iii): Let E j E .
We want to show that
j N ( B E j ) E B .

But,
j N ( B E j ) = B j N E J j N E J = X ,  say, then  X E
by Axiom (3).
Hence, by ( ) ( B X ) E B

Do you have a similar question?

Recalculate according to your conditions!

Ask your question.
Get an expert answer.

Let our experts help you. Answer in as fast as 15 minutes.

Didn't find what you were looking for?