description/proof of that for set and sets of subsets, \(\sigma\)-algebra generated by union of \(\sigma\)-algebras generated by sets of subsets is \(\sigma\)-algebra generated by union of sets of subsets
Topics
About: measurable space
The table of contents of this article
Starting Context
- The reader knows a definition of \(\sigma\)-algebra of set generated by set of subsets.
- The reader admits the proposition that for any set and any \(2\) sets of subsets, if the former set of subsets is contained in the latter set of subsets, the \(\sigma\)-algebra generated by the former set is contained in the \(\sigma\)-algebra generated by the latter set.
- The reader admits the proposition that for any set and any set of subsets, the \(\sigma\)-algebra generated by the \(\sigma\)-algebra generated by the set of subsets is the \(\sigma\)-algebra generated by the set of subsets.
Target Context
- The reader will have a description and a proof of the proposition that for any set and any sets of subsets, the \(\sigma\)-algebra generated by the union of the \(\sigma\)-algebras generated by the sets of subsets is the \(\sigma\)-algebra generated by the union of the sets of subsets.
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:
\(S'\): \(\in \{\text{ the sets }\}\)
\(J\): \(\in \{\text{ the index sets }\}\)
\(\{S_j \subseteq Pow (S') \vert j \in J\}\):
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Statements:
\(\sigma (\cup_{j \in J} \sigma (S_j)) = \sigma (\cup_{j \in J} S_j)\)
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2: Proof
Whole Strategy: Step 1: see that \(\cup_{j \in J} S_j \subseteq \cup_{j \in J} \sigma (S_j)\) and \(\sigma (\cup_{j \in J} S_j) \subseteq \sigma (\cup_{j \in J} \sigma (S_j))\); Step 2: see that \(\cup_{j \in J} \sigma (S_j) \subseteq \sigma (\cup_{j \in J} S_j)\) and \(\sigma (\cup_{j \in J} \sigma (S_j)) \subseteq \sigma (\sigma (\cup_{j \in J} S_j)) = \sigma (\cup_{j \in J} S_j)\); Step 3: conclude the proposition.
Step 1:
For each \(j \in J\), \(S_j \subseteq \sigma (S_j)\), by the definition of \(\sigma\)-algebra of set generated by set of subsets.
So, \(\cup_{j \in J} S_j \subseteq \cup_{j \in J} \sigma (S_j)\).
So, \(\sigma (\cup_{j \in J} S_j) \subseteq \sigma (\cup_{j \in J} \sigma (S_j))\), by the proposition that for any set and any \(2\) sets of subsets, if the former set of subsets is contained in the latter set of subsets, the \(\sigma\)-algebra generated by the former set is contained in the \(\sigma\)-algebra generated by the latter set.
Step 2:
Let \(S^` \in \cup_{j \in J} \sigma (S_j)\) be any.
\(S^` \in \sigma (S_j)\) for a \(j \in J\).
As \(S_j \subseteq \cup_{j \in J} S_j\), \(\sigma (S_j) \subseteq \sigma (\cup_{j \in J} S_j)\), by the proposition that for any set and any \(2\) sets of subsets, if the former set of subsets is contained in the latter set of subsets, the \(\sigma\)-algebra generated by the former set is contained in the \(\sigma\)-algebra generated by the latter set.
So, \(S^` \in \sigma (S_j) \subseteq \sigma (\cup_{j \in J} S_j)\).
So, \(\cup_{j \in J} \sigma (S_j) \subseteq \sigma (\cup_{j \in J} S_j)\).
\(\sigma (\cup_{j \in J} \sigma (S_j)) \subseteq \sigma (\sigma (\cup_{j \in J} S_j))\), by the proposition that for any set and any \(2\) sets of subsets, if the former set of subsets is contained in the latter set of subsets, the \(\sigma\)-algebra generated by the former set is contained in the \(\sigma\)-algebra generated by the latter set.
But \(\sigma (\sigma (\cup_{j \in J} S_j)) = \sigma (\cup_{j \in J} S_j)\), by the proposition that for any set and any set of subsets, the \(\sigma\)-algebra generated by the \(\sigma\)-algebra generated by the set of subsets is the \(\sigma\)-algebra generated by the set of subsets.
So, \(\sigma (\cup_{j \in J} \sigma (S_j)) \subseteq \sigma (\cup_{j \in J} S_j)\).
Step 3:
So, \(\sigma (\cup_{j \in J} \sigma (S_j)) = \sigma (\cup_{j \in J} S_j)\).
