description/proof of that for probability space, independent indexed set of sub-\(\sigma\)-algebras, and independent indexed set of measurable maps w.r.t. each sub-\(\sigma\)-algebra, indexed set of all maps is independent
Topics
About: measure space
The table of contents of this article
Starting Context
- The reader knows a definition of independent indexed set of sub-\(\sigma\)-algebras of probability space.
- The reader knows a definition of independent indexed set of measurable maps from probability space into same measurable space.
- The reader admits the proposition that for any indexed set of sets and the disjointed union of the indexed set, the intersection by the disjointed union is the intersection by the index after the intersections by the elements of the indexed set.
- The reader admits the proposition that for any finite indexed set of finite sets, the disjointed union of the indexed set, and any commutative ring, the product by the disjointed union is the product by the index after the products by the elements of the indexed set.
Target Context
- The reader will have a description and a proof of the proposition that for any probability space, any independent indexed set of sub-\(\sigma\)-algebras, and any independent indexed set of measurable maps with respect to each of the sub-\(\sigma\)-algebras, the indexed set of all the maps is independent.
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:
\((M, A, \mu)\): \(\in \{\text{ the probability spaces }\}\)
\((M', A')\): \(\in \{\text{ the measurable spaces }\}\)
\(J\): \(\in \{\text{ the possibly uncountable index sets }\}\)
\(\{A_j \in \{\text{ the sub- } \sigma \text{ -algebras of } A\}\}_{j \in J}\): \(\in \{\text{ the independent indexed sets }\}\)
\(\{L_j \in \{\text{ the possibly uncountable index sets }\}\}_{j \in J}\): \(\in \{\text{ the indexed sets }\}\)
\(\{f_{j, l_j}: M \to M' \in \{\text{ the maps measurable with respect to } A_j\}\}_{(j, l_j) \in \cup_{j \in J} \{j\} \times L_j}\): \(\in \{\text{ the indexed sets }\}\)
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Statements:
\(\forall j \in J (\{f_{j, l_j}\}_{l_j \in L_j} \in \{\text{ the independent indexed sets }\})\)
\(\implies\)
\(\{f_{j, l_j}\}_{(j, l_j) \in \cup_{j \in J} \{j\} \times L_j} \in \{\text{ the independent indexed sets }\}\)
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2: Proof
Whole Strategy: Step 1: see that each \(f_{j, l_j}\) is measurable with respect to \(A\); Step 2: take any finite subset of \(\cup_{j \in J} \{j\} \times L_j\), \(\cup_{j^` \in J^`} \{j^`\} \times L^`_{j^`}\) and \(\{{f_{j^`, l^`_{j^`}}}^{- 1} (a'_{j^`, l^`_{j^`}})\}_{(j^`, l^`_{j^`}) \in \cup_{j^` \in J^`} \{j^`\} \times L^`_{j^`}}\); Step 3: see that \(\mu (\cap_{(j^`, l^`_{j^`}) \in \cup_{j^` \in J^`} \{j^`\} \times L^`_{j^`}} {f_{j^`, l^`_{j^`}}}^{- 1} (a'_{j^`, l^`_{j^`}})) = \prod_{j^` \in J^`} \mu (\cap_{l^`_{j^`} \in L^`_{j^`}} {f_{j^`, l^`_{j^`}}}^{- 1} (a'_{j^`, l^`_{j^`}}))\); Step 4: see that \(\mu (\cap_{l^`_{j^`} \in L^`_{j^`}} {f_{j^`, l^`_{j^`}}}^{- 1} (a'_{j^`, l^`_{j^`}})) = \prod_{l^`_{j^`} \in L^`_{j^`}} \mu ({f_{j^`, l^`_{j^`}}}^{- 1} (a'_{j^`, l^`_{j^`}}))\).
Step 1:
For each \((j, l_j) \in \cup_{j \in J} \{j\} \times L_j\), \(f_{j, l_j}\) is measurable with respect to \(A\), because for each \(a' \in A'\), \({f_{j, l_j}}^{- 1} (a') \in A_j\), because \(f_{j, l_j}\) is measurable with respect to \(A_j\), so, \({f_{j, l_j}}^{- 1} (a') \in A_j \subseteq A\).
Step 2:
\(\{f_{j, l_j}\}_{(j, l_j) \in \cup_{j \in J} \{j\} \times L_j}\) is independent if and only if for each \(\{a'_{j, l_j}\}_{(j, l_j) \in \cup_{j \in J} \{j\} \times L_j}\), \(\{{f_{j, l_j}}^{- 1} (a'_{j, l_j})\}_{(j, l_j) \in \cup_{j \in J} \{j\} \times L_j}\) is independent, by the definition of independent indexed set of measurable maps from probability space into same measurable space, if and only if for each finite subset, \(S \subseteq \cup_{j \in J} \{j\} \times L_j\), \(\mu (\cap_{s \in S} {f_s}^{- 1} (a'_s)) = \prod_{s \in S} \mu ({f_s}^{- 1} (a'_s))\), by the definition of independent indexed set of events of probability space.
\(S \subseteq \cup_{j \in J} \{j\} \times L_j\) is a finite subset if and only if \(S = \cup_{j^` \in J^`} \{j^`\} \times L^`_{j^`}\), where \(J^` \subseteq J\) is any finite subset and \(L^`_{j^`} \subseteq L_{j^`}\) is any finite subset.
So, what we need to see is that \(\mu (\cap_{(j^`, l^`_{j^`}) \in \cup_{j^` \in J^`} \{j^`\} \times L^`_{j^`}} {f_{j^`, l^`_{j^`}}}^{- 1} (a'_{j^`, l^`_{j^`}})) = \prod_{(j^`, l^`_{j^`}) \in \cup_{j^` \in J^`} \{j^`\} \times L^`_{j^`}} \mu ({f_{j^`, l^`_{j^`}}}^{- 1} (a'_{j^`, l^`_{j^`}}))\) for each \(J^` \in \{\text{ the finite subsets of } J\}\) and each \(\{L^`_{j^`} \in \{\text{ the finite subsets of } L_{j^`} \}\}_{j^` \in J^`}\).
Step 3:
Let us think of \(\mu (\cap_{(j^`, l^`_{j^`}) \in \cup_{j^` \in J^`} \{j^`\} \times L^`_{j^`}} {f_{j^`, l^`_{j^`}}}^{- 1} (a'_{j^`, l^`_{j^`}}))\).
\(\cap_{(j^`, l^`_{j^`}) \in \cup_{j^` \in J^`} \{j^`\} \times L^`_{j^`}} {f_{j^`, l^`_{j^`}}}^{- 1} (a'_{j^`, l^`_{j^`}}) = \cap_{j^` \in J^`} \cap_{l^`_{j^`} \in L^`_{j^`}} {f_{j^`, l^`_{j^`}}}^{- 1} (a'_{j^`, l^`_{j^`}})\), by the proposition that for any indexed set of sets and the disjointed union of the indexed set, the intersection by the disjointed union is the intersection by the index after the intersections by the elements of the indexed set.
But \(\cap_{l^`_{j^`} \in L^`_{j^`}} {f_{j^`, l^`_{j^`}}}^{- 1} (a'_{j^`, l^`_{j^`}}) \in A_{j^`}\), because \({f_{j^`, l^`_{j^`}}}^{- 1} (a'_{j^`, l^`_{j^`}}) \in A_{j^`}\) and it is a finite intersection.
As \(\{A_j\}_{j \in J}\) is independent, \(\{A_{j^`}\}_{j^` \in J^`}\) is independent, so, \(\{\cap_{l^`_{j^`} \in L^`_{j^`}} {f_{j^`, l^`_{j^`}}}^{- 1} (a'_{j^`, l^`_{j^`}})\}_{j^` \in J^`}\) is independent, by the definition of independent indexed set of sub-\(\sigma\)-algebras of probability space.
So, \(\mu (\cap_{j^` \in J^`} \cap_{l^`_{j^`} \in L^`_{j^`}} {f_{j^`, l^`_{j^`}}}^{- 1} (a'_{j^`, l^`_{j^`}})) = \prod_{j^` \in J^`} \mu (\cap_{l^`_{j^`} \in L^`_{j^`}} {f_{j^`, l^`_{j^`}}}^{- 1} (a'_{j^`, l^`_{j^`}}))\), by the definition of independent indexed set of events of probability space.
Step 4:
But for each \(j^` \in J^`\), \(\mu (\cap_{l^`_{j^`} \in L^`_{j^`}} {f_{j^`, l^`_{j^`}}}^{- 1} (a'_{j^`, l^`_{j^`}})) = \prod_{l^`_{j^`} \in L^`_{j^`}} \mu ({f_{j^`, l^`_{j^`}}}^{- 1} (a'_{j^`, l^`_{j^`}}))\), because \(\{f_{j^`, l_{j^`}}\}_{l_{j^`} \in L_{j^`}}\) is independent for each \(j^` \in J^`\).
So, \(\mu (\cap_{j^` \in J^`} \cap_{l^`_{j^`} \in L^`_{j^`}} {f_{j^`, l^`_{j^`}}}^{- 1} (a'_{j^`, l^`_{j^`}})) = \prod_{j^` \in J^`} \prod_{l^`_{j^`} \in L^`_{j^`}} \mu ({f_{j^`, l^`_{j^`}}}^{- 1} (a'_{j^`, l^`_{j^`}}))\).
\(= \prod_{(j^`, l^`_{j^`}) \in \cup_{j^` \in J^`} L^`_{j^`}} \mu ({f_{j^`, l^`_{j^`}}}^{- 1} (a'_{j^`, l^`_{j^`}}))\), by the proposition that for any finite indexed set of finite sets, the disjointed union of the indexed set, and any commutative ring, the product by the disjointed union is the product by the index after the products by the elements of the indexed set.
Step 5:
So, \(\mu (\cap_{(j^`, l^`_{j^`}) \in \cup_{j^` \in J^`} \{j^`\} \times L^`_{j^`}} {f_{j^`, l^`_{j^`}}}^{- 1} (a'_{j^`, l^`_{j^`}})) = \prod_{(j^`, l^`_{j^`}) \in \cup_{j^` \in J^`} L^`_{j^`}} \mu ({f_{j^`, l^`_{j^`}}}^{- 1} (a'_{j^`, l^`_{j^`}}))\).
So, \(\{f_{j, l_j}\}_{(j, l_j) \in \cup_{j \in J} \{j\} \times L_j}\) is independent.
