description/proof of that for set, intersection of \(\sigma\)-algebras is \(\sigma\)-algebra
Topics
About: measure
The table of contents of this article
Starting Context
- The reader knows a definition of \(\sigma\)-algebra of set.
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.
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.
Main Body
1: Structured Description
Here is the rules of Structured Description.
Entities:
\(S\): \(\in \{\text{ the sets }\}\)
\(\{A_j \vert j \in J\}\): \(J \in \{\text{ the possibly uncountable index sets }\}\), \(A_j \in \{\text{ the } \sigma \text{ -algebras of } S\}\)
\(A\): \(= \cap_{j \in J} A_j\)
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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) \(\forall a \in A (S \setminus a \in A)\): \(a \in A_j\) 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) \(\forall s: \mathbb{N} \to A (\cup_{j \in \mathbb{N}} s (j) \in A)\): 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, \(\cup_{k \in \mathbb{N}} s (k) \in \cap_{j \in J} A_j = A\).
