914: For Map Between Measurable Spaces, if Preimage of Each Element of Generator of Codomain σ-Algebra Is Measurable, Map Is Measurable

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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

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Topics

About: measure

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The table of contents of this article

• Starting Context
• Target Context
• Orientation
• Main Body
• 1: Structured Description
• 2: Proof

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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.

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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.

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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.

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Main Body

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1: Structured Description

Here is the rules of Structured Description.

Entities:
$(S_1,A_1)$: $\in \{\text{ the measurable spaces }\}$
$(S_2,A_2)$: $\in \{\text{ the measurable spaces }\}$
$f$: $:S_1\to S_2$
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Statements:
(
$\mathrm{\exists }S\subseteq Pow(S_2)\text{ such that }{A}_2=\sigma (S)$
$\wedge$
$\mathrm{\forall }s\in S(f^{-1}(s)\in A_1)$
)
$⟹$
$f\in \{\text{ the measurable maps }\}$
//

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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.

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References

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