description/proof of that for measure space and measure subspace for measurable subset, locally negligible subset of subspace is locally negligible on space
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, any locally negligible subset of the subspace is locally negligible on the space.
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:
\((M', A', \mu')\): \(\in \{\text{ the measure spaces }\}\)
\(M\): \(\in A'\)
\((M, A, \mu)\): \(= \text{ the measure subspace }\)
\(S\): \(\in \{\text{ the locally negligible subsets of } M\}\)
//
Statements:
\(S \in \{\text{ the locally negligible subsets of } M'\}\)
//
2: Proof
Whole Strategy: Step 1: take any \(a' \in A'\) such that \(\mu' (a') \lt \infty\), and see that \(S \cap a'\) is a negligible subset of \(M'\).
Step 1:
Let \(a' \in A'\) be any such that \(\mu' (a') \lt \infty\).
\(S \cap a' = S \cap a' \cap M\), because \(S \subseteq M\) anyway.
\(a' \cap M \in A\), by the definition of measure subspace of measure space for measurable subset.
\(\mu (a' \cap M) = \mu' (a' \cap M) \le \mu' (a') \lt \infty\).
As \(S\) is locally negligible on \(M\), \(S \cap a' \cap M\) is negligible on \(M\), so, there is an \(N \in A\) such that \(S \cap a' \cap M \subseteq N\) and \(\mu (N) = 0\).
\(N = N' \cap M\) for an \(N' \in A'\), by the definition of measure subspace of measure space for measurable subset.
But \(N = N' \cap M \in A'\), because \(M \in A'\), and \(S \cap a' = S \cap a' \cap M \subseteq N\) and \(\mu' (N) = \mu (N) = 0\).
So, \(S \cap a'\) is a negligible subset of \(M'\).
So, \(S\) is locally negligible on \(M'\).
