description/proof of that finite-product map of measurable maps 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 product measurable space.
- The reader knows a definition of product map.
- The reader admits 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.
- The reader admits the proposition that the preimage by any product map is the product of the preimages by the component maps.
Target Context
- The reader will have a description and a proof of the proposition that any finite-product map of any measurable maps 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:
\(\{(M_{1, 1}, A_{1, 1}), ..., (M_{n, 1}, A_{n, 1})\}\): \((M_{j, 1}, A_{j, 1}) \in \{\text{ the measurable spaces }\}\)
\(\{(M_{1, 2}, A_{1, 2}), ..., (M_{n, 2}, A_{n, 2})\}\): \((M_{j, 2}, A_{j, 2}) \in \{\text{ the measurable spaces }\}\)
\(\{f_1, ..., f_n\}\): \(f_j: M_{j, 1} \to M_{j, 2} \in \{\text{ the measurable maps }\}\)
\(f\): \(: M_{1, 1} \times ... \times M_{n, 1} \to M_{1, 2} \times ... \times M_{n, 2}, (m_1, ..., m_n) \mapsto (f_1 (m_1), ..., f_n (m_n))\)
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where each product set has the product \(\sigma\)-algebra.
Statements:
\(f \in \{\text{ the measurable maps }\}\)
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2: Proof
Whole Strategy: apply 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; Step 1: see that for each \(a_{1, 2} \times ... \times a_{n, 2} \subseteq M_{1, 2} \times ... \times M_{n, 2}\), \(f^{-1} (a_{1, 2} \times ... \times a_{n, 2}) = {f_1}^{-1} (a_{1, 2}) \times ... \times {f_n}^{-1} (a_{n, 2})\); Step 2: conclude the proposition.
Step 1:
We are going to apply 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.
As the \(\sigma\)-algebra of \(M_{1, 2} \times ... \times M_{n, 2}\) is generated by \(\{a_{1, 2} \times ... \times a_{n, 2} \subseteq M_{1, 2} \times ... \times M_{n, 2} \vert a_j \in A_{j, 2}\}\), we only need to see that \(f^{-1} (a_{1, 2} \times ... \times a_{n, 2})\) is in the \(\sigma\)-algebra of \(M_{1, 1} \times ... \times M_{n, 1}\).
\(f^{-1} (a_{1, 2} \times ... \times a_{n, 2}) = {f_1}^{-1} (a_{1, 2}) \times ... \times {f_n}^{-1} (a_{n, 2})\), by the proposition that the preimage by any product map is the product of the preimages by the component maps.
Step 2:
As \(f_j\) is measurable, \({f_j}^{-1} (a_{j, 2}) \in A_{j, 1}\).
As the \(\sigma\)-algebra of \(M_{1, 1} \times ... \times M_{n, 1}\) is generated by \(\{a_{1, 1} \times ... \times a_{n, 1} \subseteq M_{1, 1} \times ... \times M_{n, 1} \vert a_j \in A_{j, 1}\}\), \({f_1}^{-1} (a_{1, 2}) \times ... \times {f_n}^{-1} (a_{n, 2})\) is in the \(\sigma\)-algebra.
So, \(f\) is measurable, by 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.
