description/proof of that for map into measurable space and \(\sigma\)-algebra induced on domain, for measurable subset on domain, for point of codomain s.t. \(1\)-point subset is measurable, point preimage is contained in measurable subset or is disjoint from measurable subset
Topics
About: measurable space
The table of contents of this article
Starting Context
- The reader knows a definition of \(\sigma\)-algebra induced on domain of map into measurable space.
- The reader knows a definition of map preimage of subset of codomain.
- The reader admits the proposition that for any map from any set into any measurable space, the smallest \(\sigma\)-algebra of the domain that makes the map measurable is the set of the preimages of the measurable subsets of the codomain.
- The reader admits the proposition that for any map, any subset of the domain, and any point of the codomain, if the intersection of the preimage of the point and the subset is a preimage, the preimage of the point is contained in the subset or is disjoint from the subset.
Target Context
- The reader will have a description and a proof of the proposition that for any map into any measurable space and the \(\sigma\)-algebra induced on the domain, for each measurable subset on the domain, for each point of the codomain such that the \(1\)-point subset is measurable, the point preimage is contained in the measurable subset or is disjoint from the measurable subset.
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:
\(S_1\): \(\in \{\text{ the sets }\}\)
\((M_2, A_2)\): \(\in \{\text{ the measurable spaces }\}\)
\(f\): \(: S_1 \to M_2\)
\(\sigma (f)\): \(= \text{ the } \sigma \text{ -algebra induced on } S_1 \text{ of } f\)
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Statements:
\(\forall a \in \sigma (f) (\forall m_2 \in M_2 \text{ such that } \{m_2\} \in A_2 (f^{-1} (m_2) \subseteq a \lor f^{-1} (m_2) \cap a = \emptyset))\)
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2: Note
In fact, this is a special case of the proposition that for the \(\sigma\)-algebra induced on the domain of any maps into any measurable space, for each measurable subset, each intersection of point preimages is contained in the measurable subset or is disjoint from the measurable subset, in which \(\{m_2\} \in A_2\) is not required.
When \((M_2, A_2) = (\mathbb{R}^d, B (\mathbb{R}^d))\), the Euclidean measurable space, \(\{m_2\} \in A_2\) is guaranteed for each \(m_2 \in M_2\).
3: Proof
Whole Strategy: Step 1: see that \(f^{-1} (m_2) \cap a = f^{-1} (a_2)\) for an \(a_2 \in A_2\), and see that \(f^{-1} (m_2) \subseteq a \lor f^{-1} (m_2) \cap a = \emptyset\).
Step 1:
As \(\{m_2\} \in A_2\), \(f^{-1} (m_2) \in \sigma (f)\), by the proposition that for any map from any set into any measurable space, the smallest \(\sigma\)-algebra of the domain that makes the map measurable is the set of the preimages of the measurable subsets of the codomain, so, \(f^{-1} (m_2) \cap a \in \sigma (f)\).
So, \(f^{-1} (m_2) \cap a = f^{-1} (a_2)\) for an \(a_2 \in A_2\), by the definition of \(\sigma\)-algebra induced on domain of map into measurable space.
So, \(f^{-1} (m_2) \subseteq a\) or \(f^{-1} (m_2) \cap a = \emptyset\), by the proposition that for any map, any subset of the domain, and any point of the codomain, if the intersection of the preimage of the point and the subset is a preimage, the preimage of the point is contained in the subset or is disjoint from the subset.
