909: For Set, Intersection of σ-Algebras Is σ-Algebra

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description/proof of that for set, intersection of $\sigma$-algebras is $\sigma$-algebra

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Topics

About: measure

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The table of contents of this article

• Starting Context
• Target Context
• Orientation
• Main Body
• 1: Structured Description
• 2: Proof

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Starting Context

• The reader knows a definition of $\sigma$-algebra of set.

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Target Context

• The reader will have a description and a proof of the proposition that for any set, the intersection of any $\sigma$-algebras is a $\sigma$-algebra.

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Orientation

There is a list of definitions discussed so far in this site.

There is a list of propositions discussed so far in this site.

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Main Body

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1: Structured Description

Here is the rules of Structured Description.

Entities:
$S$: $\in \{\text{ the sets }\}$
$\{A_j|j\in J\}$: $J\in \{\text{ the possibly uncountable index sets }\}$, $A_j\in \{\text{ the }\sigma \text{ -algebras of }S\}$
$A$: $={\cap }_{j\in J}{A}_j$
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Statements:
$A\in \{\text{ the }\sigma \text{ -algebras of }S\}$
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2: Proof

Whole Strategy: Step 1: see that $A$ satisfies the requirements to be a $\sigma$-algebra.

Step 1:

1) $S\in A$: for each $j\in J$, $S\in A_j$, so, $S\in {\cap }_{j\in J}{A}_j=A$.

2) $\mathrm{\forall }a\in A(S\setminus a\in A)$: $a\in A_j$ for each $j\in J$, so, $S\setminus a\in A_j$ for each $j$, so, $S\setminus a\in {\cap }_{j\in J}{A}_j=A$.

3) $\mathrm{\forall }s:\mathbb{N}\to A({\cup }_{j\in \mathbb{N}}s(j)\in A)$: as $s(k)\in A$, $s(k)\in A_j$ for each $j\in J$, so, $s$ is $:\mathbb{N}\to A_j$ for each $j$, so, ${\cup }_{k\in \mathbb{N}}s(k)\in A_j$ for each $j$, so, ${\cup }_{k\in \mathbb{N}}s(k)\in {\cap }_{j\in J}{A}_j=A$.

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References

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