description/proof of that in nest of measurable subspaces, measurableness of subset of measurable space does not depend on superspace of which space is regarded to be subspace
Topics
About: measurable space
The table of contents of this article
Starting Context
- The reader knows a definition of measurable subspace.
Target Context
- The reader will have a description and a proof of the proposition that in any nest of measurable subspaces, the measurableness of any subset of any measurable space does not depend on the superspace of which the space is regarded to be the subspace.
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:
\((M'', A'')\): \(\in \{\text{ the measurable spaces }\}\)
\((M', A')\): \(\in \{\text{ the measurable subspaces of } (M'', A'')\}\)
\(M\): \(\subseteq M'\)
\((M, A)\): \(\in \{\text{ the measurable subspaces of } (M', A')\}\)
\((M, \widetilde{A})\): \(\in \{\text{ the measurable subspaces of } (M'', A'')\}\)
//
Statements:
\(A = \widetilde{A}\)
//
2: Proof
Whole Strategy: Step 1: take any \(a \in A\), and see that \(a \in \widetilde{A}\); Step 2: take any \(\widetilde{a} \in \widetilde{A}\), and see that \(\widetilde{a} \in A\).
Step 1:
Let \(a \in A\) be any.
\(a = a' \cap M\) for an \(a' \in A'\).
But \(a' = a'' \cap M'\) for an \(a'' \in A''\).
So, \(a = a' \cap M = a'' \cap M' \cap M = a'' \cap M \in \widetilde{A}\).
Step 2:
Let \(\widetilde{a} \in \widetilde{A}\) be any.
\(\widetilde{a} = a'' \cap M\) for an \(a'' \in A''\).
But \(a'' \cap M = a'' \cap M' \cap M\), and \(a'' \cap M' \in A'\).
So, \(\widetilde{a} \in A\).
