Given a set Ω with a σ-field F defined on it. Let B be a...

Frederick Kramer

Frederick Kramer

Answered

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

Expert

2022-07-09Added 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

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