description/proof of that for map between measurable spaces and countable cover of domain by elements of \(\sigma\)-algebra, if each domain restriction 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 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 for any map between any measurable spaces and any countable cover of the domain by any elements of the \(\sigma\)-algebra, if each domain restriction 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:
\((M_1, A_1)\): \(\in \{\text{ the measurable spaces }\}\)
\((M_2, A_2)\): \(\in \{\text{ the measurable spaces }\}\)
\(f\): \(: M_1 \to M_2\), \(\in \{\text{ the maps }\}\)
\(J\): \(\in \{\text{ the countable index sets }\}\)
\(\{M_{1, j} \in A_1 \vert j \in J\}\): such that \(\cup_{j \in J} M_{1, j} = M_1\)
\(\{(M_{1, j}, A_{1, j}) \vert j \in J\}\): \((M_{1, j}, A_{1, j}) = \text{ the measurable subspace }\)
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Statements:
\(\forall j \in J (f \vert_{M_{1, j}}: M_{1, j} \to M_2 \in \{\text{ the measurable maps }\})\)
\(\implies\)
\(f \in \{\text{ the measurable maps }\}\)
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2: Proof
Whole Strategy: Step 1: take any \(a \in A_2\), see that \(f^{-1} (a) = \cup_{j \in J} {f \vert_{M_{1, j}}}^{-1} (a)\), and see that \(f^{-1} (a) \in A_1\).
Step 1:
Let \(a \in A_2\) be any.
Let us see that \(f^{-1} (a) = \cup_{j \in J} {f \vert_{M_{1, j}}}^{-1} (a)\).
For each \(p \in f^{-1} (a)\), \(p \in M_{1, j}\) for a \(j\), \(f (p) = f \vert_{M_{1, j}} (p) \in a\), so, \(p \in {f \vert_{M_{1, j}}}^{-1} (a)\), so, \(p \in \cup_{j \in J} {f \vert_{M_{1, j}}}^{-1} (a)\). For each \(p \in \cup_{j \in J} {f \vert_{M_{1, j}}}^{-1} (a)\), \(p \in {f \vert_{M_{1, j}}}^{-1} (a)\) for a \(j\), \(f \vert_{M_{1, j}} (p) = f (p) \in a\), so, \(p \in f^{-1} (a)\).
As \(f \vert_{M_{1, j}}\) is measurable, \({f \vert_{M_{1, j}}}^{-1} (a) \in A_{1, j}\), so, \(= a_j \cap M_{1, j}\) where \(a_j \in A_1\), but as \(M_{1, j} \in A_1\), \(a_j \cap M_{1, j} \in A_1\), and \(\cup_{j \in J} {f \vert_{M_{1, j}}}^{-1} (a) \in A_1\).
