description/proof of that for measurable map between measurable spaces, modified map that maps measurable subset to measurable \(1\) point is measurable
Topics
About: measurable space
The table of contents of this article
Starting Context
Target Context
- The reader will have a description and a proof of the proposition that for any measurable map between any measurable spaces, the modified map that maps any measurable subset to any measurable \(1\) point 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:
\((M_1, A_1)\): \(\in \{\text{ the measurable spaces }\}\)
\((M_2, A_2)\): \(\in \{\text{ the measurable spaces }\}\)
\(f\): \(: M_1 \to M_2\), \(\in \{\text{ the measurable maps }\}\)
\(a_1\): \(\in A_1\)
\(m_2\): \(\in M_2\), such that \(\{m_2\} \in A_2\)
\(f'\): \(: M_1 \to M_2, m \mapsto m_2 \text{ when } m \in a_1; \mapsto f (m) \text{ otherwise }\)
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Statements:
\(f' \in \{\text{ the measurable maps }\}\)
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2: Note
This proposition is typically used when \(f\) is somehow invalid for a purpose to modify \(f\) to be valid for the purpose.
For example, when \(M_2 = [- \infty, \infty]\) and \(f'\) is into \(\{- \infty, \infty\}\) only over an \(a_1\) (typically, measure \(0\)), we may want \(f'\) into \(\mathbb{R}\).
3: Proof
Whole Strategy: Step 1: take any \(a_2 \in A_2\), and see that \(f'^{-1} (a_2) = (f^{-1} (a_2 \setminus \{m_2\}) \setminus a_1) \cup (\emptyset \text{ or } a_1 \cup f^{-1} (\{m_2\}))\).
Step 1:
Let \(a_2 \in A_2\) be any.
\(a_2 = (a_2 \setminus \{m_2\}) \cup (a_2 \cap \{m_2\})\), because for each \(p \in a_2\), when \(p \in \{m_2\}\), \(p \in a_2 \cap \{m_2\}\); when \(p \notin \{m_2\}\), \(p \in a_2 \setminus \{m_2\}\).
\(f'^{-1} (a_2) = f'^{-1} ((a_2 \setminus \{m_2\}) \cup (a_2 \cap \{m_2\})) = f'^{-1} (a_2 \setminus \{m_2\}) \cup f'^{-1} (a_2 \cap \{m_2\})\), by the proposition that for any map, the map preimage of any union of sets is the union of the map preimages of the sets.
\(f'^{-1} (a_2 \setminus \{m_2\}) = f^{-1} (a_2 \setminus \{m_2\}) \setminus a_1\), because for each \(p \in f'^{-1} (a_2 \setminus \{m_2\})\), \(f' (p) \in a_2 \setminus \{m_2\}\), but \(p \notin a_1\), so, \(f' (p) = f (p) \in a_2 \setminus \{m_2\}\), so, \(p \in f^{-1} (a_2 \setminus \{m_2\})\), so, \(p \in f^{-1} (a_2 \setminus \{m_2\}) \setminus a_1\); for each \(p \in f^{-1} (a_2 \setminus \{m_2\}) \setminus a_1\), \(f' (p) = f (p) \in a_2 \setminus \{m_2\}\), so, \(p \in f'^{-1} (a_2 \setminus \{m_2\})\).
\(f'^{-1} (a_2 \cap \{m_2\}) = \emptyset \text{ or } a_1 \cup f^{-1} (\{m_2\})\), because when \(a_2 \cap \{m_2\} = \emptyset\), \(f'^{-1} (a_2 \cap \{m_2\}) = \emptyset\); otherwise, \(a_2 \cap \{m_2\} = \{m_2\}\), and \(f'^{-1} (\{m_2\}) = a_1 \cup f^{-1} (\{m_2\})\), because for each \(p \in f'^{-1} (\{m_2\})\), \(p \in a_1\) or \(p \notin a_1\), but when \(p \notin a_1\), \(f' (p) = f (p) = m_2\), so, \(p \in f^{-1} (\{m_2\})\); for each \(p \in a_1 \cup f^{-1} (\{m_2\})\), when \(p \in a_1\), \(f' (p) = m_2\), so, \(p \in f'^{-1} (\{m_2\})\); otherwise, \(f' (p) = f (p) = m_2\), so, \(p \in f'^{-1} (\{m_2\})\).
So, \(f'^{-1} (a_2) = (f^{-1} (a_2 \setminus \{m_2\}) \setminus a_1) \cup (\emptyset \text{ or } a_1 \cup f^{-1} (\{m_2\}))\).
As \(\{m_2\}\) is measurable, \(a_2 \setminus \{m_2\}\) is measurable, and \(f^{-1} (a_2 \setminus \{m_2\})\) is measurable, because \(f\) is measurable, and \(f^{-1} (a_2 \setminus \{m_2\}) \setminus a_1\) is measurable; \(\emptyset\) is measurable, \(f^{-1} (\{m_2\}\) is measurable, and \(a_1 \cup f^{-1} (\{m_2\})\) is measurable.
So, \(f'^{-1} (a_2)\) is measurable.
So, \(f'\) is measurable.
