description/proof of that expansion of measurable map on codomain is measurable
Topics
About: measurable space
The table of contents of this article
Starting Context
- The reader knows a definition of measurable subspace.
- The reader knows a definition of measurable map between measurable spaces.
Target Context
- The reader will have a description and a proof of the proposition that any expansion of any measurable map between any measurable spaces on the codomain is measurable.
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:
\((M_1, A_1)\): \(\in \{\text{ the measurable spaces }\}\)
\((M''_2, A''_2)\): \(\in \{\text{ the measurable spaces }\}\)
\((M'_2, A'_2)\): \(\in \{\text{ the measurable subspaces of } (M''_2, A''_2)\}\)
\((M_2, A_2)\): \(\in \{\text{ the measurable subspaces of } (M'_2, A'_2)\}\)
\(f\): \(: M_1 \to M_2\), \(\in \{\text{ the measurable maps }\}\)
\(f'\): \(: M_1 \to M'_2, m \mapsto f (m)\)
//
Statements:
\(f' \in \{\text{ the measurable maps }\}\)
//
2: Note
Whether \((M_2, A_2)\) is regarded to be the measurable subspace of \((M'_2, A'_2)\}\) or the measurable subspace of \((M''_2, A''_2)\}\) does not matter, because they are the same by 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.
Of course, \((M'_2, A'_2)\) can be taken to be \((M''_2, A''_2)\), so, the codomain can be expanded to \(M''_2\) to keep the map measurable.
3: Proof
Whole Strategy: Step 1: take any \(a' \in A'\), and see that \(f'^{-1} (a') = f^{-1} (a' \cap M_2)\).
Step 1:
Let \(a' \in A'\) be any.
Let us see that \(f'^{-1} (a') = f^{-1} (a' \cap M_2)\).
For each \(p \in f'^{-1} (a')\), \(f' (p) \in a'\), but \(f' (p) = f (p) \in M_2\), so, \(f (p) \in a' \cap M_2\), so, \(p \in f^{-1} (a' \cap M_2)\).
For each \(p \in f^{-1} (a' \cap M_2)\), \(f (p) \in a' \cap M_2 \subseteq a'\), but \(f (p) = f' (p)\), so, \(p \in f'^{-1} (a')\).
As \(a' \cap M_2 \in A_2\) and \(f\) is measurable, \(f^{-1} (a' \cap M_2) \in A_1\).
So, \(f'\) is measurable.
