description/proof of that finite-product of \(\sigma\)-algebras is associative
Topics
About: measurable space
The table of contents of this article
Starting Context
- The reader knows a definition of product \(\sigma\)-algebra.
- The reader knows a definition of \(\pi\)-system on set.
- The reader knows a definition of \(d\)-system on set.
- The reader admits the proposition that the nested product of any sets are associative in the 'sets - map morphisms' isomorphism sense.
- The reader admits the proposition that for any set and any \(\pi\)-system on the set, the \(\sigma\)-algebra generated by the system is the \(d\)-system generated by the system.
- The reader admits the proposition that the intersection of the same-indices-set products of possibly uncountable number of sets is the product of the intersections of the sets.
- The reader admits the proposition that for any product set, any product subset one of whose components is the subtraction of any 2 subsets is the subtraction of the product subsets whose corresponding components are the 1st subset and the 2nd subset.
- The reader admits the proposition that for any product set, any product subset one of whose components is the union of any subsets is the union of the product subsets whose corresponding components are the subsets.
Target Context
- The reader will have a description and a proof of the proposition that any finite-product of any \(\sigma\)-algebras is associative.
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:
\(\{(M_1, A_1), ..., (M_n, A_n)\}\): \((M_j, A_j) \in \{\text{ the measurable spaces }\}\)
\(M\): \(= M_1 \times ... \times M_n\), \(= \text{ the product set }\)
\(A\): \(= \text{ the product } \sigma \text{ -algebra of } M\)
\(M_{p, p + q}\): \(= M_p \times ... \times M_{p + q}\)
\(A_{p, p + q}\): \(= \text{ the product } \sigma \text{ -algebra of } M_{p, p + q}\)
\(M'\): \(= M_1 \times ... \times M_{p - 1} \times M_{p, p + q} \times M_{p + q + 1} \times ... \times M_n\), \(= \text{ the product set }\)
\(A'\): \(= \text{ the product } \sigma \text{ -algebra of } M'\)
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Statements:
\(A = A'\) with \(M\) canonically identified with \(M'\)
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2: Note
\(M\) and \(M'\) are not exactly the same but are canonically 'sets - map morphisms' isomorphic, by the proposition that the nested product of any sets are associative in the 'sets - map morphisms' isomorphism sense.
\(A = A'\) makes sense only because \(M\) and \(M'\) are identified: any \(\sigma\)-algebras of different sets cannot be equal.
By applying this proposition iteratively, any association of \(M\) is allowed: after \(A = A'\), any association in \(M_{1, p - 1}\) or \(M_{p + q + 1, n}\) can be chosen with the conserved \(\sigma\)-algebra, and so on.
3: Proof
Whole Strategy: use the proposition that for any set and any \(\pi\)-system on the set, the \(\sigma\)-algebra generated by the system is the \(d\)-system generated by the system; Step 1: take \(S_{p, p + q} := \{a_p \times ... \times a_{p + q} \vert a_j \in A_j\}\) and see that \(S_{p, p + q}\) is a \(\pi\)-system on \(M_{p, p + q}\); Step 2: take \(D := \{d \in A_{p, p + q} \vert a_1 \times ... \times a_{p - 1} \times d \times a_{p + q + 1} \times ... \times a_n \in A\}\) and see that \(D\) is a \(d\)-system on \(M_{p, p + q}\) that contains \(S_{p, p + q}\); Step 3: see that \(A_{p, p + q} \subseteq D\) and that \(S' := \{a_1 \times ... \times a_{p - 1} \times a_{p, p + q} \times a_{p + q + 1} \times ... \times a_n \vert a_j \in A_j \land a_{p, p + q} \in A_{p, p + q}\} \subseteq A\); Step 4: see that \(S := \{a_1 \times ... \times a_n \vert a_j \in A_j\} \subseteq S'\); Step 5: conclude the proposition.
Step 1:
Let us define some notations.
\(S := \{a_1 \times ... \times a_n \vert a_j \in A_j\}\).
\(S_{p, p + q} := \{a_p \times ... \times a_{p + q} \vert a_j \in A_j\}\).
\(S' := \{a_1 \times ... \times a_{p - 1} \times a_{p, p + q} \times a_{p + q + 1} \times ... \times a_n \vert a_j \in A_j \land a_{p, p + q} \in A_{p, p + q}\}\).
Note that \(A = \sigma (S)\); \(A_{p, p + q} = \sigma (S_{p, p + q})\); \(A' = \sigma (S')\).
Let us see that \(S_{p, p + q}\) is a \(\pi\)-system on \(M_{p, p + q}\).
Indeed, \(S_{p, p + q} \subseteq Pow (M_{p, p + q})\).
\(S_{p, p + q}\) is not empty, because \(M_p \times ... \times M_{p + q} \in S_{p, p + q}\).
For each \(a_p \times ... \times a_{p + q}, a'_p \times ... \times a'_{p + q} \in S_{p, p + q}\), \((a_p \times ... \times a_{p + q}) \cap (a'_p \times ... \times a'_{p + q}) = (a_p \cap a'_p) \times ... \times (a_{p + q} \cap a'_{p + q})\), by the proposition that the intersection of the same-indices-set products of possibly uncountable number of sets is the product of the intersections of the sets, \(\in S_{p, p + q}\), because \(a_j \cap a'_j \in A_j\).
So, \(S_{p, p + q}\) is a \(\pi\)-system on \(M_{p, p + q}\).
Step 2:
Let us take \(D := \{d \in A_{p, p + q} \vert a_1 \times ... \times a_{p - 1} \times d \times a_{p + q + 1} \times ... \times a_n \in A\}\).
\(S_{p, p + q} \subseteq D\), because for each \(a_p \times ... \times a_{p + q} \in S_{p, p + q}\), \(a_p \times ... \times a_{p + q} \in A_{p, p + q}\) and \(a_1 \times ... \times a_{p - 1} \times a_p \times ... \times a_{p + q} \times a_{p + q + 1} \times ... \times a_n \in A\).
Let us see that \(D\) is a \(d\)-system on \(M_{p, p + q}\).
Indeed, \(D \subseteq A_{p, p + q} \subseteq Pow (M_{p, p + q})\).
\(M_{p, p + q} \in D\), because \(M_{p, p + q} \in A_{p, p + q}\) and \(a_1 \times ... \times a_{p - 1} \times M_{p, p + q} \times a_{p + q + 1} \times ... \times a_n \in A\).
For each \(d_1, d_2 \in D\) such that \(d_1 \subseteq d_2\), \(d_2 \setminus d_1 \in D\), because \(d_2 \setminus d_1 \in A_{p, p + q}\), and \(a_1 \times ... \times a_{p - 1} \times (d_2 \setminus d_1) \times a_{p + q + 1} \times ... \times a_n = (a_1 \times ... \times a_{p - 1} \times d_2 \times a_{p + q + 1} \times ... \times a_n) \setminus (a_1 \times ... \times a_{p - 1} \times d_1 \times a_{p + q + 1} \times ... \times a_n)\), by the proposition that for any product set, any product subset one of whose components is the subtraction of any 2 subsets is the subtraction of the product subsets whose corresponding components are the 1st subset and the 2nd subset, \(\in A\).
For each \(s: \mathbb{N} \to D\) such that for each \(n \in \mathbb{N}\), \(s (n) \subseteq s (n + 1)\), \(\cup_{n \in \mathbb{N}} s (n) \in D\), because \(\cup_{n \in \mathbb{N}} s (n) \in A_{p, p + q}\), and \(a_1 \times ... \times a_{p - 1} \times \cup_{n \in \mathbb{N}} s (n) \times a_{p + q + 1} \times ... \times a_n =\cup_{n \in \mathbb{N}} (a_1 \times ... \times a_{p - 1} \times s (n) \times a_{p + q + 1} \times ... \times a_n)\), by the proposition that for any product set, any product subset one of whose components is the union of any subsets is the union of the product subsets whose corresponding components are the subsets, \(\in A\).
So, \(D\) is a \(d\)-system on \(M_{p, p + q}\).
Step 3:
\(A_{p, p + q} \subseteq D\), by the proposition that for any set and any \(\pi\)-system on the set, the \(\sigma\)-algebra generated by the system is the \(d\)-system generated by the system, because \(A_{p, p + q} = \sigma (S_{p, p + q})\) and \(D\) is a \(d\)-system that contains the \(d\)-system generated by \(S_{p, p + q}\), which is the intersection of the \(d\)-systems that contain \(S_{p, p + q}\).
So, \(S' \subseteq A\), because for each \(a_1 \times ... \times a_{p - 1} \times a_{p, p + q} \times a_{p + q + 1} \times ... \times a_n \in S'\), \(a_{p + q + 1} \in A_{p, p + q} \subseteq D\), which means that \(a_1 \times ... \times a_{p - 1} \times a_{p + q + 1} \times a_{p + q + 1} \times ... \times a_n \in A\).
So, \(A' = \sigma (S') \subseteq A\).
Step 4:
\(S \subseteq S'\), because \(a_p \times ... \times a_{p + q} \in A_{p, p + q}\).
So, \(A = \sigma (S) \subseteq \sigma (S') = A'\).
Step 5:
So, \(A = A'\).
