description/proof of that for measurable maps between arbitrary subspaces of measurable spaces, composition 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 maps between any arbitrary subspaces of any measurable spaces, the composition 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:
\((M'_1, A'_1)\): \(\in \{\text{ the measurable spaces }\}\)
\((M'_2, A'_2)\): \(\in \{\text{ the measurable spaces }\}\)
\((M'_3, A'_3)\): \(\in \{\text{ the measurable spaces }\}\)
\((M_1, A_1)\): \(\in \{\text{ the measurable subspaces of } (M'_1, A'_1)\}\)
\((M_2, A_2)\): \(\in \{\text{ the measurable subspaces of } (M'_2, A'_2)\}\)
\((\widetilde{M_2}, \widetilde{A_2})\): \(\in \{\text{ the measurable subspaces of } (M'_2, A'_2)\}\), such that \(M_2 \subseteq \widetilde{M_2}\)
\((M_3, A_3)\): \(\in \{\text{ the measurable subspaces of } (M'_3, A'_3)\}\)
\(f_1\): \(: M_1 \to M_2\), \(\in \{\text{ the measurable maps }\}\)
\(f_2\): \(: \widetilde{M_2} \to M_3\), \(\in \{\text{ the measurable maps }\}\)
\(f_2 \circ f_1\): \(: M_1 \to M_3\)
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Statements:
\(f_2 \circ f_1 \in \{\text{ the measurable maps }\}\)
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2: Proof
Whole Strategy: Step 1: take any \(a \in A_3\) and see that \((f_2 \circ f_1)^{-1} (a) = f_1^{-1} (f_2^{-1} (a) \cap M_2)\); Step 2: see that \(f_2^{-1} (a) \cap M_2 \in A_2\); Step 3: conclude the proposition.
Step 1:
Let \(a \in A_3\) be any.
\((f_2 \circ f_1)^{-1} (a) = f_1^{-1} (f_2^{-1} (a) \cap M_2)\), by the proposition that for any maps composition, the preimage under the composition is the composition of the map preimages in the reverse order.
Step 2:
\(f_2^{-1} (a) \in \widetilde{A_2}\), because \(f_2\) is measurable.
\(f_2^{-1} (a) = a' \cap \widetilde{M_2}\) for an \(a' \in A'_2\), by the definition of measurable subspace.
\(f_2^{-1} (a) \cap M_2 = a' \cap \widetilde{M_2} \cap M_2 = a' \cap M_2 \in A_2\).
Step 3:
So, \((f_2 \circ f_1)^{-1} (a) = f_1^{-1} (f_2^{-1} (a) \cap M_2) = f_1^{-1} (a' \cap M_2) \in A_1\), because \(f_1\) is measurable.
So, \(f_2 \circ f_1\) is measurable.
