description/proof of that for map between measurable spaces, if codomain \(\sigma\)-algebra is generated by set of subsets and preimage of element of set of subsets is measurable, map is measurable
Topics
About: measurable space
The table of contents of this article
Starting Context
- The reader knows a definition of measurable map between measurable spaces.
- The reader knows a definition of \(\sigma\)-algebra of set generated by set of subsets.
- The reader admits the proposition that the preimage of the codomain minus any codomain subset under any map is the domain minus the preimage of the subset.
- The reader admits the proposition that for any map, the map preimage of any union of sets is the union of the map preimages of the sets.
Target Context
- The reader will have a description and a proof of the proposition that for any map between any measurable spaces, if the codomain \(\sigma\)-algebra is generated by any set of subsets and the preimage of each element of the set of subsets is measurable, the map 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:
\((M_1, A_1)\): \(\in \{\text{ the measurable spaces }\}\)
\((M_2, A_2)\): \(\in \{\text{ the measurable spaces }\}\)
\(f\): \(: M_1 \to M_2\)
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Statements:
\(\exists S \subseteq Pow M_2 \text{ such that } A_2 = \sigma (S) (\forall s \in S (f^{-1} (s) \in A_1))\)
\(\implies\)
\(f \in \{\text{ the measurable maps }\}\)
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2: Proof
Whole Strategy: Step 1: consider the set of the subsets of \(M_2\) the preimage of whose each element is measurable, and see that the set is a \(\sigma\)-algebra that contains \(S\); Step 2: conclude the proposition.
Step 1:
Let \(\widetilde{S}\) be the set of the subsets of \(M_2\) the preimage of whose each element is in \(A_1\).
Let us see that \(\widetilde{S}\) is a \(\sigma\)-algebra of \(M_2\).
1) \(M_2 \in \widetilde{S}\): \(f^{-1} (M_2) = M_1 \in A_1\).
2) for each \(a \in \widetilde{S}\), \(M_2 \setminus a \in \widetilde{S}\): \(f^{-1} (M_2 \setminus a) = M_1 \setminus f^{-1} (a)\), by the proposition that the preimage of the codomain minus any codomain subset under any map is the domain minus the preimage of the subset, where \(f^{-1} (a) \in A_1\), so, \(M_1 \setminus f^{-1} (a) \in A_1\).
3) for each sequence into \(\widetilde{S}\), \(s\), \(\cup_{j \in \mathbb{N}} s (j) \in \widetilde{S}\): \(f^{-1} (\cup_{j \in \mathbb{N}} s (j)) = \cup_{j \in \mathbb{N}} f^{-1} (s (j))\), 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, where \(f^{-1} (s (j)) \in A_1\), so, \(\cup_{j \in \mathbb{N}} f^{-1} (s (j)) \in A_1\).
\(S \subseteq \widetilde{S}\), by the supposition.
Step 2:
As \(\sigma (S)\) is the intersection of all the \(\sigma\)-algebras that contain \(S\), \(A_2 = \sigma (S) \subseteq \widetilde{S}\).
That means that for each element of \(A_2\), its preimage under \(f\) is in \(A_1\), so, \(f\) is measurable.
