description/proof of that some para-product maps of measurable maps are measurable
Topics
About: measurable space
The table of contents of this article
- Starting Context
- Target Context
- Orientation
- Main Body
- 1: Structured Description 1
- 2: Proof 1
- 3: Note 1
- 4: Structured Description 2
- 5: Proof 2
Starting Context
- The reader knows a definition of product measurable space.
- The reader knows a definition of measurable map between measurable spaces.
- 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 any finite-product of any \(\sigma\)-algebras is associative.
- The reader admits the proposition that any finite-product map of any measurable maps is measurable.
- The reader admits the proposition that for any measurable maps between any arbitrary subspaces of any measurable spaces, the composition is measurable.
Target Context
- The reader will have a description and a proof of the proposition that some para-product maps of measurable maps are 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 1
Here is the rules of Structured Description.
Entities:
\((M_0, A_0)\): \(\in \{\text{ the measurable spaces }\}\)
\(\{(M_{1, 1}, A_{1, 1}), ..., (M_{n, 1}, A_{n, 1})\}\): \((M_{j, 1}, A_{j, 1}) \in \{\text{ the measurable spaces }\}\)
\(\{(M_{1, 2}, A_{1, 2}), ..., (M_{n, 2}, A_{n, 2})\}\): \((M_{j, 2}, A_{j, 2}) \in \{\text{ the measurable spaces }\}\)
\(\{f_1, ..., f_n\}\): \(f_j: M_0 \times M_{j, 1} \to M_{j, 2} \in \{\text{ the measurable maps }\}\)
\(f\): \(: M_0 \times M_{1, 1} \times ... \times M_{n, 1} \to M_{1, 2} \times ... \times M_{n, 2}, (m_0, m_1, ..., m_n) \mapsto (f_1 (m_0, m_1), ..., f_n (m_0, m_n))\)
//
where each product set has the product \(\sigma\)-algebra.
Statements:
\(f \in \{\text{ the measurable maps }\}\)
//
2: Proof 1
Whole Strategy: Step 1: take \(f': M_0 \times M_{1, 1} \times ... \times M_{n, 1} \to M_0 \times M_{1, 1} \times ... \times M_0 \times M_{n, 1}\) and see that \(f'\) is measurable; Step 2: take \(\widetilde{f}: M_0 \times M_{1, 1} \times ... \times M_0 \times M_{n, 1} \to M_{1, 2} \times ... \times M_{n, 2}\) and see that \(\widetilde{f}\) is measurable; Step 3: see that \(f = \widetilde{f} \circ f'\) and see that \(f\) is measurable.
Step 1:
Note that each product set that will appear hereafter has the product \(\sigma\)-algebra.
Let us take \(f': M_0 \times M_{1, 1} \times ... \times M_{n, 1} \to M_0 \times M_{1, 1} \times ... \times M_0 \times M_{n, 1}, (m_0, m_1, ..., m_n) \mapsto (m_0, m_1, ..., m_0, m_n)\).
Let us see that \(f'\) is measurable.
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 \(M_0 \times M_{1, 1} \times ... \times M_0 \times M_{n, 1}\) is generated by \(\{a_{1, 0} \times a_{1, 1} \times ... \times a_{n, 0} \times a_{n, 1} \vert a_{j, 0} \in A_0 \land a_{j, 1} \in A_{j, 1}\}\), we only need to see that \(f'^{-1} (a_{1, 0} \times a_{1, 1} \times ... \times a_{n, 0} \times a_{n, 1})\) is in the \(\sigma\)-algebra of \(M_0 \times M_{1, 1} \times ... \times M_{n, 1}\).
Let us see that \(f'^{-1} (a_{1, 0} \times a_{1, 1} \times ... \times a_{n, 0} \times a_{n, 1}) = a_{1, 0} \cap ... \cap a_{n, 0} \times a_{1, 1} \times ... \times a_{n, 1}\).
Let \(p = (p_0, p_1, ..., p_n) \in f'^{-1} (a_{1, 0} \times a_{1, 1} \times ... \times a_{n, 0} \times a_{n, 1})\) be any.
\(f' (p) = (p_0, p_1, ..., p_0, p_n) \in a_{1, 0} \times a_{1, 1} \times ... \times a_{n, 0} \times a_{n, 1}\).
So, \(p_0 \in a_{1, 0} \cap ... \cap a_{n, 0}\) and for each \(1 \le j\), \(p_j \in a_{j, 1}\).
So, \(p \in a_{1, 0} \cap ... \cap a_{n, 0} \times a_{1, 1} \times ... \times a_{n, 1}\).
Let \(p = (p_0, p_1, ..., p_n) \in a_{1, 0} \cap ... \cap a_{n, 0} \times a_{1, 1} \times ... \times a_{n, 1}\) be any.
\(f' (p) = (p_0, p_1, ..., p_0, p_n) \in a_{1, 0} \times a_{1, 1} \times ... \times a_{n, 0} \times a_{n, 1}\).
So, \(p \in f'^{-1} (a_{1, 0} \times a_{1, 1} \times ... \times a_{n, 0} \times a_{n, 1})\).
So, \(f'^{-1} (a_{1, 0} \times a_{1, 1} \times ... \times a_{n, 0} \times a_{n, 1}) = a_{1, 0} \cap ... \cap a_{n, 0} \times a_{1, 1} \times ... \times a_{n, 1}\).
\(a_{1, 0} \cap ... \cap a_{n, 0} \in A_0\).
As the \(\sigma\)-algebra of \(M_0 \times M_{1, 1} \times ... \times M_{n, 1}\) is generated by \(\{a_0 \times a_{1, 1} \times ... \times a_{n, 1} \subseteq M_0 \times M_{1, 1} \times ... \times M_{n, 1} \vert a_0 \in A_0 \land a_{j, 1} \in A_{j, 1}\}\), \(f'^{-1} (a_{1, 0} \times a_{1, 1} \times ... \times a_{n, 0} \times a_{n, 1})\) 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.
Step 2:
Let us take \(\widetilde{f}: M_0 \times M_{1, 1} \times ... \times M_0 \times M_{n, 1} \to M_{1, 2} \times ... \times M_{n, 2}, (m_{1, 0}, m_1, ..., m_{n, 0}, m_n) \mapsto (f_1 (m_{1, 0}, m_1), ..., f_n (m_{n, 0}, m_n))\).
The domain, \(M_0 \times M_{1, 1} \times ... \times M_0 \times M_{n, 1}\), can be regarded to be \((M_0 \times M_{1, 1}) \times ... \times (M_0 \times M_{n, 1})\) with the \(\sigma\)-algebra as the product of the \(\sigma\)-algebras of \(M_0 \times M_{1, 1}, ..., M_0 \times M_{n, 1}\), by the proposition that any finite-product of any \(\sigma\)-algebras is associative.
In fact, \(\widetilde{f} = f_1 \times ... \times f_n\).
\(\widetilde{f}\) is measurable, by the proposition that any finite-product map of any measurable maps is measurable.
Step 3:
\(f = \widetilde{f} \times f'\), because for each \(p = (p_0, p_1, ..., p_n) \in M_0 \times M_{1, 1} \times ... \times M_{n, 1}\), \(f (p) = (f_1 (p_0, p_1), ..., f_n (p_0, p_n))\) while \(\widetilde{f} \times f' (p) = \widetilde{f} (p_0, p_1, ..., p_0, p_n) = (f_1 (p_0, p_1), ..., f_n (p_0, p_n))\).
So, \(f\) is measurable, by the proposition that for any measurable maps between any arbitrary subspaces of any measurable spaces, the composition is measurable.
3: Note 1
It is called "para-product map" because it is not really a product map, because \(M_0\) is shared by the component maps.
4: Structured Description 2
Here is the rules of Structured Description.
Entities:
\((M_0, A_0)\): \(\in \{\text{ the measurable spaces }\}\)
\(\{(M_{1, 2}, A_{1, 2}), ..., (M_{n, 2}, A_{n, 2})\}\): \((M_{j, 2}, A_{j, 2}) \in \{\text{ the measurable spaces }\}\)
\(\{f_1, ..., f_n\}\): \(f_j: M_0 \to M_{j, 2} \in \{\text{ the measurable maps }\}\)
\(f\): \(: M_0 \to M_{1, 2} \times ... \times M_{n, 2}, m_0 \mapsto (f_1 (m_0), ..., f_n (m_0))\)
//
where each product set has the product \(\sigma\)-algebra.
Statements:
\(f \in \{\text{ the measurable maps }\}\)
//
5: Proof 2
Whole Strategy: Step 1: take \(\widetilde{f_j}: M_0 \times \{0\} \to M_{j, 2}\) and see that \(\widetilde{f_j}\) is measurable; Step 2: take \(\widetilde{f}: M_0 \times \{0\} \times ... \times \{0\} \to M_{1, 2} \times ... \times M_{n, 2}\) and see that \(\widetilde{f}\) is measurable; Step 3: take \(f': M_0 \to M_0 \times \{0\} \times ... \times \{0\}\) and see that \(f'\) is measurable; Step 4: see that \(f = \widetilde{f} \circ f'\) and \(f\) is measurable.
Step 1:
Note that each product set that will appear hereafter has the product \(\sigma\)-algebra.
Let \(\{0\}\) be the measurable space with the inevitable \(\sigma\)-algebra.
Let us take \(\widetilde{f_j}: M_0 \times \{0\} \to M_{j, 2}, (m_0, 0) \mapsto f_j (m_0)\).
Let us see that \(\widetilde{f_j}\) is measurable.
Let \(a_{j, 2} \in A_{j, 2}\) be any.
\({\widetilde{f_j}}^{-1} (a_{j, 2}) = {f_j}^{-1} (a_{j, 2}) \times \{0\}\), because for each \(p = (p_0, 0) \in {\widetilde{f_j}}^{-1} (a_{j, 2})\), \(\widetilde{f_j} (p) = f_j (p_0) \in a_{j, 2}\), so, \(p_0 \in {f_j}^{-1} (a_{j, 2})\), so, \(p \in {f_j}^{-1} (a_{j, 2}) \times \{0\}\); for each \(p = (p_0, 0) \in {f_j}^{-1} (a_{j, 2}) \times \{0\}\), \(\widetilde{f_j} (p) = f_j (p_0) \in a_{j, 2}\), so, \(p \in {\widetilde{f_j}}^{-1} (a_{j, 2})\).
As \({f_j}^{-1} (a_{j, 2}) \in A_0\) and \(\{0\}\) is measurable, \({f_j}^{-1} (a_{j, 2}) \times \{0\}\) is measurable in the product \(\sigma\)-algebra.
Step 2:
Let us take \(\widetilde{f}: M_0 \times \{0\} \times ... \times \{0\} \to M_{1, 2} \times ... \times M_{n, 2}, (m_0, 0, ..., 0) \mapsto (f_1 (m_0), ..., f_n (m_0))\).
In fact, \(\widetilde{f} (m_0, 0, ..., 0) = (\widetilde{f_1} (m_0, 0), ..., \widetilde{f_n} (m_0, 0))\).
So, by Description 1, \(\widetilde{f}\) is measurable.
Step 3:
Let us take \(f': M_0 \to M_0 \times \{0\} \times ... \times \{0\}, m_0 \mapsto (m_0, 0, ..., 0)\).
Let us see that \(f'\) is measurable.
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 \(M_0 \times \{0\} \times ... \times \{0\}\) is generated by \(\{a_0 \times a_{1, 0} \times ... \times a_{n, 0} \subseteq M_0 \times \{0\} \times ... \times \{0\} \vert a_0 \in A_0 \land a_{j, 0} \in \{\emptyset, \{0\}\}\}\), we only need to see that \({f'}^{-1} (a_0 \times a_{1, 0} \times ... \times a_{n, 0}) \in A_0\).
When \(a_{j, 0} = \emptyset\) for a \(j\), \(a_0 \times a_{1, 0} \times ... \times a_{n, 0} = \emptyset\), and \({f'}^{-1} (\emptyset) = \emptyset \in A_0\).
When \(a_{j, 0} = \{0\}\) for each \(j\), \({f'}^{-1} (a_0 \times \{0\} \times ... \times \{0\}) = a_0 \in A_0\), because for each \(p \in {f'}^{-1} (a_0 \times \{0\} \times ... \times \{0\})\), \(f' (p) = (p, 0, ..., 0) \in a_0 \times \{0\} \times ... \times \{0\}\), so, \(p \in a_0\); for each \(p \in a_0\), \(f' (p) = (p, 0, ..., 0) \in a_0 \times \{0\} \times ... \times \{0\}\).
So, \(f'\) is measurable.
Step 4:
\(f = \widetilde{f} \circ f'\), because for each \(p \in M_0\), \(f (p) = (f_1 (p), ..., f_n (p))\) while \(\widetilde{f} \circ f' (p) = \widetilde{f} (p, 0, ..., 0) = (f_1 (p), ..., f_n (p))\).
So, \(f\) is measurable, by the proposition that for any measurable maps between any arbitrary subspaces of any measurable spaces, the composition is measurable.
