description/proof of that for measure space and measure subspace for measurable subset, intersection of locally negligible subset of space and subspace is locally negligible on subspace
Topics
About: measure space
The table of contents of this article
Starting Context
- The reader knows a definition of measure subspace of measure space for measurable subset.
- The reader knows a definition of locally negligible subset of measure space.
Target Context
- The reader will have a description and a proof of the proposition that for any measure space and the measure subspace for any measurable subset, the intersection of any locally negligible subset of the space and the subspace is locally negligible on the subspace.
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:
\((M', A', \mu')\): \(\in \{\text{ the measure spaces }\}\)
\(M\): \(\in A'\)
\((M, A, \mu)\): \(= \text{ the measure subspace }\)
\(S'\): \(\in \{\text{ the locally negligible subsets of } A'\}\)
\(S\): \(= S' \cap M\)
//
Statements:
\(S \in \{\text{ the locally negligible subsets of } A\}\)
//
2: Proof
Whole Strategy: Step 1: take each \(a \in A\) such that \(\mu (a) \lt \infty\) and see that \(S \cap a\) is negligible.
Step 1:
Let \(a \in A\) be any such that \(\mu (a) \lt \infty\).
We need to see that \(S \cap a\) is negligible on \(M\).
\(a = a' \cap M\) for an \(a' \in A'\), but \(a' \cap M \in A'\), because \(M \in A'\), so, \(a \in A'\).
\(S \cap a = S' \cap M \cap a\), but \(M \cap a = M \cap a' \cap M = M \cap a' \in A'\) while \(\mu' (M \cap a) = \mu (a) \lt \infty\), so, \(S' \cap M \cap a\) is negligible, and there is an \(N' \in A'\) such that \(S' \cap M \cap a \subseteq N'\) and \(\mu' (N') = 0\).
\(S' \cap M \cap a \cap M \subseteq N' \cap M\), but the left hand side is \(S \cap a\), and the right hand side is in \(A\) and \(\mu (N' \cap M) = \mu' (N' \cap M) = 0\).
So, \(S \cap a\) is negligible on \(M\).
So, \(S\) is locally negligible on \(M\).
