definition of \(\sigma\)-algebra induced on domain of maps into measurable space
Topics
About: measurable space
The table of contents of this article
Starting Context
- The reader knows a definition of \(\sigma\)-algebra of set generated by set of subsets.
- The reader knows a definition of measurable map between measurable spaces.
Target Context
- The reader will have a definition of \(\sigma\)-algebra induced on domain of maps into measurable space.
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:
\( S_1\): \(\in \{\text{ the sets }\}\)
\( (M_2, A_2)\): \(\in \{\text{ the measurable spaces }\}\)
\( J\): \(\in \{\text{ the index sets }\}\), such that \(J \neq \emptyset\)
\( \{f_j: S_1 \to M_2 \vert j \in J\}\):
\(*\sigma (\{f_j \vert j \in J\})\): \(= \text{ the smallest } \sigma \text{ -algebra of } S_1 \text{ such that } \forall j \in J (f_j \in \{\text{ the measurable maps }\})\)
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Conditions:
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2: Note
\(\sigma (\{f_j \vert j \in J\})\) is uniquely determined, because it is the intersection of all the \(\sigma\)-algebras of \(S_1\) that make all the \(f_j\) s measurable, while at least \(Pow (S_1)\) is such a \(\sigma\)-algebra: there is the set of the \(\sigma\)-algebras that make all the \(f_j\) s measurable, which contains \(Pow (S_1)\), and take the intersection of the set: the intersection is a \(\sigma\)-algebra, by the proposition that for any set, the intersection of any \(\sigma\)-algebras is a \(\sigma\)-algebra, and the intersection makes all the \(f_j\) s measurable, because for each \(a_2 \in A_2\), \({f_j}^{-1} (a_2)\) is contained in each element of the set, so, \({f_j}^{-1} (a_2)\) is contained in the intersection, and the intersection is the smallest such, because any such \(\sigma\)-algebra is a constituent of the intersection.
When \(\vert J \vert = 1\), \(\sigma (\{f_j \vert j \in J\}) = \sigma (f_{J_1})\) by the definition of \(\sigma\)-algebra induced on domain of map into measurable space, because \(\sigma (f_{J_1})\) is the smallest \(\sigma\)-algebra that makes \(f_{J_1}\) measurable, as is mentioned in Note for the definition of \(\sigma\)-algebra induced on domain of map into measurable space.
\(\sigma (\{f_j \vert j \in J\}) = \sigma (\cup_{j \in J} \sigma (f_j))\), because \(\sigma (\{f_j \vert j \in J\}) \subseteq \sigma (\cup_{j \in J} \sigma (f_j))\), because \(\sigma (\cup_{j \in J} \sigma (f_j))\) makes all the \(f_j\) s measurable, because for each \(a_2 \in A_2\), \({f_j}^{-1} (a_2) \in \sigma (f_j)\), so, \({f_j}^{-1} (a_2) \in \cup_{j \in J} \sigma (f_j) \subseteq \sigma (\cup_{j \in J} \sigma (f_j))\); for each \(j \in J\), \(\sigma (f_j) \subseteq \sigma (\{f_j \vert j \in J\})\), because \(\sigma (\{f_j \vert j \in J\})\) makes \(f_j\) measurable, so, \(\cup_{j \in J} \sigma (f_j) \subseteq \sigma (\{f_j \vert j \in J\})\), so, \(\sigma (\cup_{j \in J} \sigma (f_j)) \subseteq \sigma (\sigma (\{f_j \vert j \in J\})) = \sigma (\{f_j \vert j \in J\})\), by the proposition that for any set and any \(2\) sets of subsets, if the former set of subsets is contained in the latter set of subsets, the \(\sigma\)-algebra generated by the former set is contained in the \(\sigma\)-algebra generated by the latter set and the proposition that for any set and any set of subsets, the \(\sigma\)-algebra generated by the \(\sigma\)-algebra generated by the set of subsets is the \(\sigma\)-algebra generated by the set of subsets.
