description/proof of that for map between measurable spaces, if preimage of each element of generator of codomain \(\sigma\)-algebra is measurable, map is measurable
Topics
About: measure
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.
- The reader knows a definition of \(\sigma\)-algebra induced on codomain of map from measurable space.
Target Context
- The reader will have a description and a proof of the proposition that for any map between any measurable spaces, if the preimage of each element of any generator of the codomain \(\sigma\)-algebra is measurable, the map is measurable.
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:
\((S_1, A_1)\): \(\in \{\text{ the measurable spaces }\}\)
\((S_2, A_2)\): \(\in \{\text{ the measurable spaces }\}\)
\(f\): \(: S_1 \to S_2\)
//
Statements:
(
\(\exists S \subseteq Pow (S_2) \text{ such that } A_2 = \sigma (S)\)
\(\land\)
\(\forall s \in S (f^{-1} (s) \in A_1)\)
)
\(\implies\)
\(f \in \{\text{ the measurable maps }\}\)
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2: Proof
Whole Strategy: Step 1: take the \(\sigma\)-algebra induced on \(S_2\) by \(f\), \(A\); Step 2: see that \(A_2 \subseteq A\); Step 3: see that the preimage of each element of \(A_2\) is measurable.
Step 1:
Let us take the \(\sigma\)-algebra induced on \(S_2\) by \(f\), \(A\).
Step 2:
By the supposition, \(S \subseteq A\).
As \(A_2 = \sigma (S)\) is the intersection of all the \(\sigma\)-algebras that contain \(S\), \(A_2 \subseteq A\), because \(A\) is a constituent of the intersection.
Step 3:
That means that for each \(a \in A_2\), \(f^{-1} (a) \in A_1\), which means that \(f\) is measurable.
