definition of subspace measure over subspace \(\sigma\)-algebra of measurable subset of measure space
Topics
About: measure space
The table of contents of this article
Starting Context
- The reader knows a definition of measure space.
- The reader knows a definition of subspace \(\sigma\)-algebra of subset of measurable space.
Target Context
- The reader will have a definition of subspace measure over subspace \(\sigma\)-algebra of measurable subset of measure 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:
\( (M', A', \mu')\): \(\in \{\text{ the measure spaces }\}\)
\( S\): \(\in A'\)
\( A\): \(= \text{ the subspace } \sigma \text{ -algebra of } S\)
\(*\mu\): \(: A \to [0, + \infty]\), \(\in \{\text{ the measures over } A\}\)
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Conditions:
\(\forall a = a' \cap S \in A \text{ where } a' \in A' (\mu (a) = \mu' (a))\)
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2: Note
Let us see that \(\mu\) is indeed a measure.
1st, let us see that \(\mu\) is well-defined.
As \(S \in A'\), \(a = a' \cap S \in A'\), so, \(\mu' (a)\) is valid.
So, \(\mu\) is well-defined.
2nd, let us see that \(\mu\) is indeed a measure.
As \(\emptyset = \emptyset \cap S\) where \(\emptyset \in A'\), \(\mu (\emptyset) = \mu' (\emptyset) = 0\).
Let \(s: \mathbb{N} \to A\) be any such that for each \(n_1, n_2 \in \mathbb{N}\) such that \(n_1 \neq n_2\), \(s (n_1) \cap s (n_2) = \emptyset\).
\(\mu (\cup_{j \in \mathbb{N}} s (j)) = \sum_{j \in \mathbb{N}} \mu (s (j))\)?
\(\mu (\cup_{j \in \mathbb{N}} s (j)) = \mu' (\cup_{j \in \mathbb{N}} s (j)) = \sum_{j \in \mathbb{N}} \mu' (s (j)) = \sum_{j \in \mathbb{N}} \mu (s (j))\).
So, \(\mu\) is a measure.
