definition of subspace measure over subspace \(\sigma\)-algebra of almost-everywhere 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 almost-everywhere over 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 almost-everywhere 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:
\( (M', A', \mu')\): \(\in \{\text{ the measure spaces }\}\)
\( S\): \(\in \{\text{ the almost-everywhere subsets of } M'\}\)
\( A\): \(= \text{ the subspace } \sigma \text{ -algebra of } S\)
\(*\mu\): \(: A \to [0, + \infty]\), \(\in \{\text{ the measures over } A\}\)
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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.
For each \(a \in A\), let \(a', a'' \in A'\) be any such that \(a = a' \cap S = a'' \cap S\).
\(\mu' (a') = \mu' (a'')\)?
Let \(p \in a'' \setminus a'\) be any.
\(p \notin S\), because if \(p \in S\), \(p \in a'' \cap S = a' \cap S\), which would imply \(p \in a'\), a contradiction.
So, \(a'' \setminus a' \subseteq M' \setminus S\).
But as \(S\) is an almost-everywhere subset of \(M'\), there is an \(n \in A'\) such that \(M' \setminus S \subseteq n\) and \(\mu' (n) = 0\).
So, \(a'' \setminus a' \subseteq M' \setminus S \subseteq n\), and \(\mu' (a'' \setminus a') \le \mu' (n) = 0\), so, \(\mu' (a'' \setminus a') = 0\).
As \(a'' \subseteq a' \cup (a'' \setminus a')\), \(\mu' (a'') \le \mu' (a' \cup (a'' \setminus a')) = \mu' (a') + \mu' (a'' \setminus a') = \mu' (a') + 0 = \mu' (a')\).
By symmetry, \(\mu' (a') \le \mu' (a'')\).
So, yes, \(\mu' (a') = \mu' (a'')\).
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 \(n_1 \neq n_2\), \(s (n_1) \cap s (n_2) = \emptyset\).
\(\mu (\cup_{j \in \mathbb{N}} s (j)) = \sum_{j \in \mathbb{N}} \mu (s (j))\)?
As \(s (j) \in A\), \(s (j) = a'_j \cap S\) for an \(a'_j \in A'\), and \(\cup_{j \in \mathbb{N}} s (j) = \cup_{j \in \mathbb{N}} (a'_j \cap S) = (\cup_{j \in \mathbb{N}} a'_j) \cap S\), by the proposition that for any set, the intersection of the union of any possibly uncountable number of subsets and any subset is the union of the intersections of each of the subsets and the latter subset, while \(\cup_{j \in \mathbb{N}} a'_j \in A'\).
By definition, \(\mu (\cup_{j \in \mathbb{N}} s (j)) = \mu' (\cup_{j \in \mathbb{N}} a'_j)\) and \(\sum_{j \in \mathbb{N}} \mu (s (j)) = \sum_{j \in \mathbb{N}} \mu' (a'_j)\), so, it is about whether \(\mu' (\cup_{j \in \mathbb{N}} a'_j) = \sum_{j \in \mathbb{N}} \mu' (a'_j)\) holds.
Let us define \(b'_j := a'_j \setminus c'_j\) where \(c'_j := a'_j \cap \cup_{l \in \{0, ..., j - 1\}} a'_l\).
For each \(j, m \in \mathbb{N}\) such that \(j \neq m\), \(b'_j \cap b'_m = \emptyset\), because supposing that \(j \lt m\) without loss of generality, for each \(p \in b'_m\), \(p \notin c'_m\), which means that \(p \notin \cup_{l \in \{0, ..., m - 1\}} a'_m\), which means that \(p \notin a'_j\), which means that \(p \notin b'_j\).
\(\cup_{j \in \mathbb{N}} a'_j = \cup_{j \in \mathbb{N}} b'_j\), because while \(\cup_{j \in \mathbb{N}} b'_j \subseteq \cup_{j \in \mathbb{N}} a'_j\) is obvious, for each \(p \in \cup_{j \in \mathbb{N}} a'_j\), \(p \in a'_j\) for a \(j\), but if \(p \in a'_0\), \(p \in b'_0 = a'_0\), otherwise, if \(p \in a'_1\), \(p \in b'_1 = a'_1 \setminus a'_1 \cap \cup_{l \in \{0\}} a'_l\) because \(p \notin \cup_{l \in \{0\}} a'_l\), otherwise, if \(p \in a'_2\), \(p \in b'_2 = a'_2 \setminus a'_2 \cap \cup_{l \in \{0, 1\}} a'_l\) because \(p \notin \cup_{l \in \{0, 1\}} a'_l\), otherwise, ..., and so on, after all, if it reaches \(a'_j\), that means that \(p \in a'_j\) and \(p \in b'_j = a'_j \setminus a'_j \cap \cup_{l \in \{0, ..., j - 1\}} a'_l\) because \(p \notin \cup_{l \in \{0, ..., j - 1\}} a'_l\) (because otherwise, it would not have reached \(a'_j\)), but if it does not reach \(a'_j\), that means that \(p \in b'_l\) for an \(l \lt j\), so, \(\cup_{j \in \mathbb{N}} a'_j \subseteq \cup_{j \in \mathbb{N}} b'_j\).
So, \(\mu' (\cup_{j \in \mathbb{N}} a'_j) = \mu' (\cup_{j \in \mathbb{N}} b'_j) = \sum_{j \in \mathbb{N}} \mu' (b'_j)\).
So, what we need to see is that \(\mu' (b'_j) = \mu' (a'_j)\).
Obviously, \(\mu' (b'_j) \le \mu' (a'_j)\).
As \(a'_j \subseteq (a'_j \setminus c'_j) \cup c'_j\), \(\mu' (a'_j) \le \mu' (a'_j \setminus c'_j) + \mu' (c'_j)\), so, \(\mu' (a'_j) \le \mu' (b'_j) + \mu' (c'_j)\), so, \(\mu' (a'_j) - \mu' (c'_j) \le \mu' (b'_j)\).
So, \(\mu' (a'_j) - \mu' (c'_j) \le \mu' (b'_j) \le \mu' (a'_j)\).
But \(\mu' (c'_j) = \mu' (a'_j \cap \cup_{l \in \{0, ..., j - 1\}} a'_l) = \mu' (\cup_{l \in \{0, ..., j - 1\}} (a'_j \cap a'_l))\), by the proposition that for any set, the intersection of the union of any possibly uncountable number of subsets and any subset is the union of the intersections of each of the subsets and the latter subset, \(\le \sum_{l \in \{0, ..., j - 1\}} \mu' (a'_j \cap a'_l)\), but \(a'_j \cap a'_l \subseteq M' \setminus S\), because for each \(p \in a'_j \cap a'_l\), if \(p \in S\), \(p \in a'_j \cap S = a_j\) and \(p \in a'_l \cap S = a_l\), which would mean that \(a_j \cap a_l \neq \emptyset\), a contradiction. So, as \(S\) is almost-everywhere, there is an \(n \in A'\) such that \(M' \setminus S \subseteq n\) and \(\mu' (n) = 0\). So, \(\mu' (a'_j \cap a'_l) \le \mu' (n) = 0\). So, \(\mu' (c'_j) = 0\).
So, \(\mu' (a'_j) = \mu' (a'_j) - 0 \le \mu' (b'_j) \le \mu' (a'_j)\), which means that \(\mu' (a'_j) = \mu' (b'_j)\).
So, \(\mu' (\cup_{j \in \mathbb{N}} a'_j) = \sum_{j \in \mathbb{N}} \mu' (a'_j)\).
So, \(\mu (\cup_{j \in \mathbb{N}} s (j)) = \sum_{j \in \mathbb{N}} \mu (s (j))\).
So, \(\mu\) is a measure.
