description/proof of that for map from set into measurable space, smallest \(\sigma\)-algebra of domain that makes map measurable is set of preimages of measurable subsets
Topics
About: measurable space
The table of contents of this article
Starting Context
- The reader knows a definition of measurable map between measurable spaces.
- The reader admits the proposition that the preimage of the whole codomain of any map is the whole domain.
- The reader admits the proposition that the preimage of the codomain minus any codomain subset under any map is the domain minus the preimage of the subset.
- The reader admits the proposition that for any map, the map preimage of any union of sets is the union of the map preimages of the sets.
Target Context
- The reader will have a description and a proof of the proposition that for any map from any set into any measurable space, the smallest \(\sigma\)-algebra of the domain that makes the map measurable is the set of the preimages of the measurable subsets of the codomain.
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:
\(S_1\): \(\in \{\text{ the sets }\}\)
\((M_2, A_2)\): \(\in \{\text{ the measurable spaces }\}\)
\(f\): \(: S_1 \to M_2\)
\(A_1\): \(= \cap \{A \in \{\text{ the } \sigma \text{ -algebras of } S_1\} \vert f \in \{\text{ the measurable maps from } (S_1, A) \text{ into } (M_2, A_2)\}\}\)
\(Q\): \(= \{f^{-1} (a) \vert a \in A_2\}\)
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Statements:
\(A_1 = Q\)
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2: Proof
Whole Strategy: Step 1: see that \(Q \subseteq A_1\); Step 2: see that \(Q\) is a \(\sigma\)-algebra of \(S_1\); Step 3: conclude the proposition.
Step 1:
Let \(A \in \{A \in \{\text{ the } \sigma \text{ -algebras of } S_1\} \vert f \in \{\text{ the measurable maps from } (S_1, A) \text{ into } (M_2, A_2)\}\}\) be any.
For each \(a \in A_2\), \(f^{-1} (a) \in A\) is a necessary condition for \(f\) to be measurable.
So, \(f^{-1} (a) \in A_1\).
So, \(Q \subseteq A_1\).
Step 2:
Let us see that \(Q\) is a \(\sigma\)-algebra of \(S_1\).
1) \(S_1 \in Q\): \(M_2 \in A_2\), and \(f^{-1} (M_2) = S_1 \in Q\), by the proposition that the preimage of the whole codomain of any map is the whole domain.
2) \(\forall q \in Q (S_1 \setminus q \in Q)\): \(q = f^{-1} (a)\), and \(S_1 \setminus q = S_1 \setminus f^{-1} (a) = f^{-1} (M_2 \setminus a)\), by the proposition that the preimage of the codomain minus any codomain subset under any map is the domain minus the preimage of the subset, but \(M_2 \setminus a \in A_2\).
3) \(\forall s: \mathbb{N} \to Q (\cup_{j \in \mathbb{N}} s (j) \in Q)\): for each \(j \in \mathbb{N}\), \(s (j) = f^{-1} (a_j)\) for an \(a_j \in A_2\), and \(\cup_{j \in \mathbb{N}} s (j) = \cup_{j \in \mathbb{N}} f^{-1} (a_j) = f^{-1} (\cup_{j \in \mathbb{N}} a_j)\), by the proposition that for any map, the map preimage of any union of sets is the union of the map preimages of the sets, but \(\cup_{j \in \mathbb{N}} a_j \in A_2\).
So, \(Q\) is a \(\sigma\)-algebra.
Step 3:
\(f\) is measurable from \((S_1, Q)\) into \((M_2, A_2)\), because for each \(a \in A_2\), \(f^{-1} (a) \in Q\).
So, \(Q\) is an \(A \in \{A \in \{\text{ the } \sigma \text{ -algebras of } S_1\} \vert f \in \{\text{ the measurable maps from } (S_1, A) \text{ into } (M_2, A_2)\}\}\).
So, \(A_1 \subseteq Q\).
Together with \(Q \subseteq A_1\) by Step 1, \(A_1 = Q\).
