description/proof of that for measurable map between measurable spaces, restriction on domain subspace and codomain subspace 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.
- The reader admits the proposition that for any map and its any codomain-restriction, the preimage of any subset of the original codomain under the codomain-restricted map is the preimage of the subset under the original map.
- The reader admits the proposition that for any map between sets and its any domain-restriction, the preimage under the domain-restricted map is the intersection of the preimage under the original map and the restricted domain.
- The reader admits the proposition that for any map, the map preimage of any intersection of sets is the intersection of the map preimages of the sets.
Target Context
- The reader will have a description and a proof of the proposition that for any measurable map between any measurable spaces, the restriction on any domain subspace and any codomain subspace 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:
\((M'_1, A'_1)\): \(\in \{\text{ the measurable spaces }\}\)
\((M'_2, A'_2)\): \(\in \{\text{ the measurable spaces }\}\)
\(f'\): \(: M'_1 \to M'_2\), \(\in \{\text{ the measurable maps }\}\)
\(M_1\): \(\subseteq M'_1\)
\((M_1, A_1)\): \(= \text{ the measurable subspace of } (M'_1, A'_1)\)
\(M_2\): \(\subseteq M'_2\) such that \(f' (M_1) \subseteq M_2\)
\((M_2, A_2)\): \(= \text{ the measurable subspace of } (M'_2, A'_2)\)
\(f\): \(: M_1 \to M_2, p \mapsto f' (p)\)
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Statements:
\(f \in \{\text{ the measurable maps }\}\)
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2: Proof
Whole Strategy: Step 1: take any \(a \in A_2\) and see that \(a = a' \cap M_2\) where \(a' \in A'_2\); Step 2: see that \(f^{-1} (a) = f'^{-1} (a') \cap M_1\), and conclude the proposition.
Step 1:
Let \(a \in A_2\) be any.
\(a = a' \cap M_2\) where \(a' \in A'_2\), by the definition of measurable subspace.
Step 2:
\(f^{-1} (a) = (f' \vert_{M_1})^{-1} (a) = f'^{-1} (a) \cap M_1\), by the proposition that for any map and its any codomain-restriction, the preimage of any subset of the original codomain under the codomain-restricted map is the preimage of the subset under the original map and the proposition that for any map between sets and its any domain-restriction, the preimage under the domain-restricted map is the intersection of the preimage under the original map and the restricted domain.
\(= f'^{-1} (a' \cap M_2) \cap M_1 = f'^{-1} (a') \cap f'^{-1} (M_2) \cap M_1\), by the proposition that for any map, the map preimage of any intersection of sets is the intersection of the map preimages of the sets.
But \(f'^{-1} (M_2) \cap M_1 = M_1\), because for each \(p \in M_1\), \(f' (p) \in M_2\), so, \(p \in f'^{-1} (M_2)\), so, \(= f'^{-1} (a') \cap M_1\).
But as \(f'\) is measurable, \(f'^{-1} (a') \in A'_1\).
So, \(f^{-1} (a) \in A_1\), by the definition of measurable subspace.
So, \(f\) is measurable.
