description/proof of that indexed set of measurable maps from probability space into same measurable space is independent iff indexed set of sub-\(\sigma\)-algebras induced by 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 measurable maps from probability space into same measurable space.
- The reader knows a definition of \(\sigma\)-algebra induced on domain of map into measurable space.
- The reader knows a definition of independent indexed set of sub-\(\sigma\)-algebras of probability space.
Target Context
- The reader will have a description and a proof of the proposition that any indexed set of measurable maps from any probability space into any same measurable space is independent if and only if the indexed set of the sub-\(\sigma\)-algebras induced by 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:
\((M, A, \mu)\): \(\in \{\text{ the probability spaces }\}\)
\((M', A')\): \(\in \{\text{ the measurable spaces }\}\)
\(J\): \(\in \{\text{ the possibly uncountable index sets }\}\)
\(\{f_j: M \to M' \in \{\text{ the measurable maps }\}\}_{j \in J}\): \(\in \{\text{ the indexed sets }\}\)
\(\{\sigma (f_j)\}_{j \in J}\): \(\in \{\text{ the indexed sets }\}\)
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Statements:
\(\{f_j\}_{j \in J} \in \{\text{ the independent indexed sets }\}\)
\(\iff\)
\(\{\sigma (f_j)\}_{j \in J} \in \{\text{ the independent indexed sets }\}\)
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2: Proof
Whole Strategy: Step 1: see that \(\sigma (f_j)\) is a sub-\(\sigma\)-algebra of \(A\); Step 2: suppose that \(\{f_j\}_{j \in J}\) is independent; Step 3: see that \(\{\sigma (f_j)\}_{j \in J}\) is independent; Step 4: suppose that \(\{\sigma (f_j)\}_{j \in J}\) is independent; Step 5: see that \(\{f_j\}_{j \in J}\) is independent.
Step 1:
For each \(j \in J\), \(\sigma (f_j)\) is a sub-\(\sigma\)-algebra of \(A\), because while \(\sigma (f_j) = \{{f_j}^{-1} (a') \vert a' \in A'\}\), as \(f_j\) is measurable, \({f_j}^{-1} (a') \in A\), so, \(\sigma (f_j) \subseteq A\).
So, talking about \(\{\sigma (f_j)\}_{j \in J}\) being independent makes sense.
Step 2:
Let us suppose that \(\{f_j\}_{j \in J}\) is independent.
Step 3:
Let \(\{a_j \in \sigma (f_j)\}_{j \in J}\) be any.
For each \(j \in J\), \(a_j = {f_j}^{- 1} (a'_j)\) for an \(a'_j \in A'\).
\(\{{f_j}^{- 1} (a'_j)\}_{j \in J}\) is independent, so, \(\{a_j\}_{j \in J}\) is independent.
So, \(\{\sigma (f_j)\}_{j \in J}\) is independent.
Step 4:
Let us suppose that \(\{\sigma (f_j)\}_{j \in J}\) is independent.
Step 5:
Let \(\{a'_j \in A'\}_{j \in J}\) be any.
For each \(j \in J\), \({f_j}^{- 1} (a'_j) \in \sigma (f_j)\).
So, \(\{{f_j}^{- 1} (a'_j)\}_{j \in J}\) is independent.
So, \(\{f_j\}_{j \in J}\) is independent.
