description/proof of that for \(\sigma\)-algebra induced on domain of maps into measurable space, for measurable subset, intersection of point preimages is contained in measurable subset or is disjoint from measurable subset
Topics
About: measurable space
The table of contents of this article
Starting Context
Target Context
- The reader will have a description and a proof of the proposition that for the \(\sigma\)-algebra induced on the domain of any maps into any measurable space, for each measurable subset, each intersection of point preimages is contained in the measurable subset or is disjoint from the measurable subset.
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 \(J \neq \emptyset\)
\(\{f_j: S_1 \to M_2 \vert j \in J\}\):
\(\sigma (\{f_j \vert j \in J\})\): \(= \text{ the } \sigma \text{ -algebra induced on } S_1 \text{ by } \{f_j \vert j \in J\}\)
//
Statements:
\(\forall a \in \sigma (\{f_j \vert j \in J\}) (\forall m \in \times_{j \in J} M_2 (\cap_{j \in J} {f_j}^{-1} (m_j) \subseteq a \lor \cap_{j \in J} {f_j}^{-1} (m_j) \cap a = \emptyset))\)
//
2: Note
\(\cap_{j \in J} {f_j}^{-1} (m_j) \subseteq a \lor \cap_{j \in J} {f_j}^{-1} (m_j) \cap a = \emptyset\) is really a natural thing that \(\sigma (\{f_j \vert j \in J\})\) does not need to contain any measurable subset that divides \(\cap_{j \in J} {f_j}^{-1} (m_j)\), which is natural because each preimage under \(f_j\) contains the whole of \({f_j}^{-1} (m_j)\) or is disjoint from \({f_j}^{-1} (m_j)\), and an \(a\) is necessary because it is the result of some set operations on some preimages under \(f_j\) s, but it looks unlikely that the result divides \(\cap_{j \in J} {f_j}^{-1} (m_j)\) (Proof proves that it is really the case).
3: Proof
Whole Strategy: Step 1: define \(A := \{a \in \sigma (\{f_j \vert j \in J\}) \vert \forall m \in \times_{j \in J} M_2 (\cap_{j \in J} {f_j}^{-1} (m_j) \subseteq a \lor \cap_{j \in J} {f_j}^{-1} (m_j) \cap a = \emptyset)\}\) and see that \(A\) is a \(\sigma\)-algebra that makes all the \(f_j\) s measurable.
Step 1:
Let us define \(A := \{a \in \sigma (\{f_j \vert j \in J\}) \vert \forall m \in \times_{j \in J} M_2 (\cap_{j \in J} {f_j}^{-1} (m_j) \subseteq a \lor \cap_{j \in J} {f_j}^{-1} (m_j) \cap a = \emptyset)\}\).
In other words, \(A\) has eliminated from \(\sigma (\{f_j \vert j \in J\})\) the elements that divide a \(\cap_{j \in J} {f_j}^{-1} (m_j)\).
Let us see that \(A\) is a \(\sigma\)-algebra.
1) \(S_1 \in A\): while \(S_1 \in \sigma (\{f_j \vert j \in J\})\), \(\cap_{j \in J} {f_j}^{-1} (m_j) \subseteq S_1\).
2) \(\forall a \in A (S_1 \setminus a \in A)\): as \(a \in \sigma (\{f_j \vert j \in J\})\), \(S_1 \setminus a \in \sigma (\{f_j \vert j \in J\})\), and when \(\cap_{j \in J} {f_j}^{-1} (m_j) \subseteq a\), \(\cap_{j \in J} {f_j}^{-1} (m_j) \cap (S_1 \setminus a) = \emptyset\), and when \(\cap_{j \in J} {f_j}^{-1} (m_j) \cap a = \emptyset\), \(\cap_{j \in J} {f_j}^{-1} (m_j) \subseteq S_1 \setminus a\), so, \(S_1 \setminus a \in A\).
3) \(\forall s: \mathbb{N} \to A (\cup_{n \in \mathbb{N}} s (n) \in A)\): as \(s\) is into \(\sigma (\{f_j \vert j \in J\})\), \(\cup_{n \in \mathbb{N}} s (n) \in \sigma (\{f_j \vert j \in J\})\), and when \(\cap_{j \in J} {f_j}^{-1} (m_j) \subseteq s (n)\) for an \(n \in \mathbb{N}\), \(\cap_{j \in J} {f_j}^{-1} (m_j) \subseteq \cup_{n \in \mathbb{N}} s (n)\), and when \(\cap_{j \in J} {f_j}^{-1} (m_j) \cap s (n) = \emptyset\) for each \(n \in \mathbb{N}\), \(\cap_{j \in J} {f_j}^{-1} (m_j) \cap \cup_{n \in \mathbb{N}} s (n) = \emptyset\).
So, \(A\) is a \(\sigma\)-algebra.
\(A\) makes each \(f_j\) measurable, because for each \(a_2 \in A_2\), \({f_j}^{-1} (a_2) \in \sigma (\{f_j \vert j \in J\})\), because \(\sigma (\{f_j \vert j \in J\})\) makes all the \(f_j\) s measurable, but when \(m_j \in a_2\), \({f_j}^{-1} (m_j) \subseteq {f_j}^{-1} (a_2)\), so, \(\cap_{j \in J} {f_j}^{-1} (m_j) \subseteq {f_j}^{-1} (a_2)\), and when \(m_j \notin a_2\), \({f_j}^{-1} (m_j) \cap {f_j}^{-1} (a_2) = \emptyset\), by the proposition that the preimages of any disjoint subsets under any map are disjoint, so, \(\cap_{j \in J} {f_j}^{-1} (m_j) \cap {f_j}^{-1} (a_2) = \emptyset\), so, \({f_j}^{-1} (a_2) \in A\) anyway.
So, \(A\) is a \(\sigma\)-algebra that makes all the \(f_j\) s measurable.
So, \(\sigma (\{f_j \vert j \in J\}) \subseteq A\).
But as \(A \subseteq \sigma (\{f_j \vert j \in J\})\), \(A = \sigma (\{f_j \vert j \in J\})\), which means that \(A\) did not really eliminate anything.
So, for each \(a \in \sigma (\{f_j \vert j \in J\})\), for each \(m \in \times_{j \in J} M_2\), \(\cap_{j \in J} {f_j}^{-1} (m_j) \subseteq a \lor \cap_{j \in J} {f_j}^{-1} (m_j) \cap a = \emptyset\).
