description/proof of that for measure space, measure subspace for measurable subset, \(\mathcal{L}^\infty\) for space, and \(\mathcal{L}^\infty\) for subspace, for each element of \(\mathcal{L}^\infty\) for space, its restriction is in \(\mathcal{L}^\infty\) for subspace, and seminorm of restriction is equal to or smaller than seminorm of element of \(\mathcal{L}^\infty\) for space
Topics
About: measure space
The table of contents of this article
Starting Context
- The reader knows a definition of \(\mathcal{L}^p\) over measure space.
- The reader knows a definition of measure subspace of measure space for measurable subset.
- The reader admits the proposition that for any map between sets and its any domain-restriction, the preimage under the domain-restricted map is the intersection of the preimage under the original map and the restricted domain.
- The reader admits 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.
Target Context
- The reader will have a description and a proof of the proposition that for any measure space, the measure subspace for any measurable subset, \(\mathcal{L}^\infty\) for the space, and \(\mathcal{L}^\infty\) for the subspace, for each element of \(\mathcal{L}^\infty\) for the space, its restriction is in \(\mathcal{L}^\infty\) for the subspace, and the seminorm of the restriction is equal to or smaller than the seminorm of the element of \(\mathcal{L}^\infty\) for 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:
\((M', A', \mu')\): \(\in \{\text{ the measure spaces }\}\)
\(M\): \(\in A'\)
\((M, A, \mu)\): \(= \text{ the measure subspace }\)
\(\mathbb{C}\): \(= \text{ the complex Euclidean topological space }\) with the Borel \(\sigma\)-algebra
\(\mathbb{R}\): \(= \text{ the Euclidean topological space }\) with the Borel \(\sigma\)-algebra
\(F\): \(\in \{\mathbb{C}, \mathbb{R}\}\)
\(\mathcal{L}^\infty (M', A', \mu', F)\):
\(\mathcal{L}^\infty (M, A, \mu, F)\):
\(f'\): \(\in \mathcal{L}^\infty (M', A', \mu', F)\)
\(f\): \(= f \vert_{M}\)
//
Statements:
\(f \in \mathcal{L}^\infty (M, A, \mu, F)\)
\(\land\)
\(\Vert f \Vert \le \Vert f' \Vert\)
//
2: Proof
Whole Strategy: Step 1: see that \(f\) is measurable and value-bounded; Step 2: see that \(\Vert f \Vert \le \Vert f' \Vert\) by the definition of seminorm.
Step 1:
Let us see that \(f\) is measurable.
Let \(S \subseteq F\) be any measurable subset of \(F\).
\(f^{-1} (S) = f'^{-1} (S) \cap M\), by the proposition that for any map between sets and its any domain-restriction, the preimage under the domain-restricted map is the intersection of the preimage under the original map and the restricted domain, \(\in A\), because \(f'^{-1} (S) \in A'\).
So, \(f\) is measurable.
\(f\) is value-bounded, because \(f'\) is value-bounded.
So, \(f \in \mathcal{L}^\infty (M, A, \mu, F)\).
Step 2:
Let \(r \in \mathbb{R}\) be any such that \(0 \le r\).
Let us see that \(\{m \in M \vert r \lt \vert f (m) \vert\} = \{m \in M' \vert r \lt \vert f' (m) \vert\} \cap M\).
For each \(p \in \{m \in M \vert r \lt \vert f (m) \vert\}\), \(r \lt \vert f (p) \vert = \vert f' (p) \vert\), so, \(p \in \{m \in M' \vert r \lt \vert f' (m) \vert\} \cap M\).
For each \(p \in \{m \in M' \vert r \lt \vert f' (m) \vert\} \cap M\), \(r \lt \vert f' (m) \vert = \vert f (m) \vert\), so, \(p \in \{m \in M \vert r \lt \vert f (m) \vert\}\).
So, \(\{m \in M \vert r \lt \vert f (m) \vert\} = \{m \in M' \vert r \lt \vert f' (m) \vert\} \cap M\).
For \(r = \Vert f' \Vert\), \(\{m \in M' \vert r \lt \vert f' (m) \vert\}\) is locally negligible on \(M'\).
So, \(\{m \in M' \vert r \lt \vert f' (m) \vert\} \cap M\) is locally negligible on \(M\), by 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, so \(\{m \in M \vert r \lt \vert f (m) \vert\}\) is locally negligible.
As \(\Vert f \Vert\) is the infimum of such \(r\) s, \(\Vert f \Vert \le \Vert f' \Vert\).
