description/proof of that for measurable space generated by set of subsets and subset, subspace \(\sigma\)-algebra is generated by set of restricted subsets
Topics
About: measurable space
The table of contents of this article
Starting Context
- The reader knows a definition of measurable subspace.
- The reader knows a definition of \(\sigma\)-algebra of set generated by set of subsets.
- The reader admits the proposition that the intersection of any set minus any set and any set is the intersection of the 1st set and the 3rd set minus the intersection of the 2nd set and the 3rd set.
- The reader admits the proposition that for any set, the intersection of the union of any possibly uncountable number of subsets and any subset is the union of the intersections of each of the subsets and the latter subset.
Target Context
- The reader will have a description and a proof of the proposition that for any measurable space generated by any set of subsets and any subset, the subspace \(\sigma\)-algebra of the subset is generated by the set of the restricted subsets.
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:
\((M', A')\): \(\in \{\text{ the measurable spaces }\}\), where \(A' = \sigma (S')\) for any \(S' \subseteq Pow (M')\)
\((M, A)\): \(\in \{\text{ the measurable subspaces of } (M', A')\}\)
\(S\): \(= \{s' \cap M \in Pow (M) \vert s' \in S'\}\)
\(\sigma (S)\):
//
Statements:
\(A = \sigma (S)\)
//
2: Note
For example, for the extended Euclidean topological space, \(\overline{\mathbb{R}}\), with the topology, \(\overline{O}\), and the Euclidean topological space, \(\mathbb{R}\), \(\mathbb{R}\) has the subspace topology of \(\overline{\mathbb{R}}\), \(O\), and the Borel \(\sigma\)-algebra, \(\sigma (O)\), is the subspace \(\sigma\)-algebra of \(\sigma (\overline{O})\): \(O = \{\overline{o} \cap \mathbb{R} \in Pow (\mathbb{R}) \vert \overline{o} \in \overline{O}\}\).
Likewise, for the topological subspace of \(\overline{\mathbb{R}}\), \([0, \infty]\), with the topology, \(O\), \(\sigma (O)\) is the subspace \(\sigma\)-algebra of \(\sigma (\overline{O})\).
3: Proof
Whole Strategy: Step 1: see that \(\sigma (S) \subseteq A\); Step 2: take \(\widetilde{A'} := \{U \subseteq M' \vert U \cap M \in \sigma (S)\}\), and see that \(A' \subseteq \widetilde{A'}\); Step 3: see that \(A \subseteq \sigma (S)\) using \(A = \{a' \cap M \vert a' \in A'\}\).
Step 1:
Let us see that \(\sigma (S) \subseteq A\).
As \(A = \{a' \cap M \vert a' \in A'\}\) and \(S' \subseteq A'\), \(S \subseteq A\), because for each \(s \in S\), \(s = s' \cap M\) for an \(s' \in S'\), \(s' \in A'\), and \(s' \cap M \in A\).
So, \(A\) is a \(\sigma\)-algebra that contains \(S\).
As \(\sigma (S)\) is the intersection of all the such \(\sigma\)-algebras, \(\sigma (S) \subseteq A\).
Step 2:
Let us take \(\widetilde{A'} := \{U \subseteq M' \vert U \cap M \in \sigma (S)\}\).
Let us see that \(\widetilde{A'}\) is a \(\sigma\)-algebra of \(M'\).
\(M' \in \widetilde{A'}\), because \(M' \cap M = M \in \sigma (S)\).
Let \(U \in \widetilde{A'}\) be any.
\(M' \setminus U \in \widetilde{A'}\), because \((M' \setminus U) \cap M = (M' \cap M) \setminus (U \cap M)\), by the proposition that the intersection of any set minus any set and any set is the intersection of the 1st set and the 3rd set minus the intersection of the 2nd set and the 3rd set, \(= M \setminus (U \cap M) \in \sigma (S)\), because \(U \cap M \in \sigma (S)\).
Let \(s: \mathbb{N} \to \widetilde{A'}\) be any.
\((\cup_{j \in \mathbb{N}} s (j)) \cap M = \cup_{j \in \mathbb{N}} (s (j) \cap M)\), by the proposition that for any set, the intersection of the union of any possibly uncountable number of subsets and any subset is the union of the intersections of each of the subsets and the latter subset, \(\in \sigma (S)\), because \(s (j) \cap M \in \sigma (S)\).
So, \(\widetilde{A'}\) is a \(\sigma\)-algebra of \(M'\).
\(S' \subseteq \widetilde{A'}\), because for each \(s' \in S'\), \(s' \cap M \in S \subseteq \sigma (S)\).
So, \(\widetilde{A'}\) is a \(\sigma\)-algebra of \(M'\) that contains \(S'\).
So, \(A' \subseteq \widetilde{A'}\), because \(A'\) is the intersection of all the such \(\sigma\)-algebras.
Step 3:
\(A = \{a' \cap M \vert a' \in A'\} \subseteq \{a' \cap M \vert a' \in \widetilde{A'}\}\), because \(A' \subseteq \widetilde{A'}\), \(\subseteq \sigma (S)\), by the definition of \(\widetilde{A'}\).
So, \(A = \sigma (S)\).
