1914: For Set and Set of Subsets, \(\sigma\)-Algebra Generated by \(\sigma\)-Algebra Generated by Set of Subsets Is \(\sigma\)-Algebra Generated by Set 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.
Target Context
- The reader will have a description and a proof of 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.
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:
\(S'\): \(\in \{\text{ the sets }\}\)
\(S\): \(\subseteq Pow (S')\)
//
Statements:
\(\sigma (\sigma (S)) = \sigma (S)\)
//
2: Proof
Whole Strategy: Step 1: see that \(\sigma (S) \subseteq \sigma (\sigma (S))\), and \(\sigma (S) \subseteq \sigma (S)\) and \(\sigma (\sigma (S)) \subseteq \sigma (S)\).
Step 1:
\(\sigma (S) \subseteq \sigma (\sigma (S))\), by the definition of \(\sigma\)-algebra generate by set of subsets.
\(\sigma (S) \subseteq \sigma (S)\), so, \(\sigma (S)\) is a \(\sigma\)-algebra such that \(\sigma (S) \subseteq \sigma (S)\), so, \(\sigma (\sigma (S)) \subseteq \sigma (S)\), by the definition of \(\sigma\)-algebra generate by set of subsets.
So, \(\sigma (\sigma (S)) = \sigma (S)\).
