<The previous article in this series | The table of contents of this series |
description/proof of that for measure space, subset of locally negligible subset of space is locally negligible
Topics
About:
measure space
The table of contents of this article
Starting Context
Target Context
-
The reader will have a description and a proof of the proposition that for any measure space, any subset of any locally negligible subset of the space is locally negligible.
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:
\((M, A, \mu)\): \(\in \{\text{ the measure spaces }\}\)
\(S'\): \(\in \{\text{ the locally negligible subsets of } M\}\)
\(S\): \(\subseteq S'\)
//
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 \(\mu (a) \lt \infty\).
\(S \cap a \subseteq S' \cap a\).
But as \(S'\) is locally negligible, there is an \(N \in A\) such that \(S' \cap a \subseteq N\) and \(\mu (N) = 0\).
So, \(S \cap a \subseteq S' \cap a \subseteq N\) and \(\mu (N) = 0\).
So, \(S\) is locally negligible.
References
<The previous article in this series | The table of contents of this series |
<The previous article in this series | The table of contents of this series | The next article in this series>
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
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 \(M'\).
References
<The previous article in this series | The table of contents of this series | The next article in this series>
<The previous article in this series | The table of contents of this series | The next article in this series>
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
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 \(M\).
References
<The previous article in this series | The table of contents of this series | The next article in this series>
<The previous article in this series | The table of contents of this series | The next article in this series>
definition of measure subspace of measure space for measurable subset
Topics
About:
measure space
The table of contents of this article
Starting Context
Target Context
-
The reader will have a definition of measure subspace of measure space for measurable subset.
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:
\( (M', A', \mu')\): \(\in \{\text{ the measure spaces }\}\)
\( M\): \(\in A'\)
\( A\): \(= \text{ the subspace } \sigma \text{ -algebra of } M\)
\( \mu\): \(: A \to [0, + \infty]\), \(= \text{ the subspace measure }\)
\(*(M, A, \mu)\):
//
Conditions:
//
References
<The previous article in this series | The table of contents of this series | The next article in this series>
<The previous article in this series | The table of contents of this series | The next article in this series>
description/proof of that for measurable map between measurable spaces, modified map that maps measurable subset to measurable \(1\) point is measurable
Topics
About:
measurable space
The table of contents of this article
Starting Context
Target Context
-
The reader will have a description and a proof of the proposition that for any measurable map between any measurable spaces, the modified map that maps any measurable subset to any measurable \(1\) point is measurable.
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:
\((M_1, A_1)\): \(\in \{\text{ the measurable spaces }\}\)
\((M_2, A_2)\): \(\in \{\text{ the measurable spaces }\}\)
\(f\): \(: M_1 \to M_2\), \(\in \{\text{ the measurable maps }\}\)
\(a_1\): \(\in A_1\)
\(m_2\): \(\in M_2\), such that \(\{m_2\} \in A_2\)
\(f'\): \(: M_1 \to M_2, m \mapsto m_2 \text{ when } m \in a_1; \mapsto f (m) \text{ otherwise }\)
//
Statements:
\(f' \in \{\text{ the measurable maps }\}\)
//
2: Note
This proposition is typically used when \(f\) is somehow invalid for a purpose to modify \(f\) to be valid for the purpose.
For example, when \(M_2 = [- \infty, \infty]\) and \(f'\) is into \(\{- \infty, \infty\}\) only over an \(a_1\) (typically, measure \(0\)), we may want \(f'\) into \(\mathbb{R}\).
3: Proof
Whole Strategy: Step 1: take any \(a_2 \in A_2\), and see that \(f'^{-1} (a_2) = (f^{-1} (a_2 \setminus \{m_2\}) \setminus a_1) \cup (\emptyset \text{ or } a_1 \cup f^{-1} (\{m_2\}))\).
Step 1:
Let \(a_2 \in A_2\) be any.
\(a_2 = (a_2 \setminus \{m_2\}) \cup (a_2 \cap \{m_2\})\), because for each \(p \in a_2\), when \(p \in \{m_2\}\), \(p \in a_2 \cap \{m_2\}\); when \(p \notin \{m_2\}\), \(p \in a_2 \setminus \{m_2\}\).
\(f'^{-1} (a_2) = f'^{-1} ((a_2 \setminus \{m_2\}) \cup (a_2 \cap \{m_2\})) = f'^{-1} (a_2 \setminus \{m_2\}) \cup f'^{-1} (a_2 \cap \{m_2\})\), by the proposition that for any map, the map preimage of any union of sets is the union of the map preimages of the sets.
\(f'^{-1} (a_2 \setminus \{m_2\}) = f^{-1} (a_2 \setminus \{m_2\}) \setminus a_1\), because for each \(p \in f'^{-1} (a_2 \setminus \{m_2\})\), \(f' (p) \in a_2 \setminus \{m_2\}\), but \(p \notin a_1\), so, \(f' (p) = f (p) \in a_2 \setminus \{m_2\}\), so, \(p \in f^{-1} (a_2 \setminus \{m_2\})\), so, \(p \in f^{-1} (a_2 \setminus \{m_2\}) \setminus a_1\); for each \(p \in f^{-1} (a_2 \setminus \{m_2\}) \setminus a_1\), \(f' (p) = f (p) \in a_2 \setminus \{m_2\}\), so, \(p \in f'^{-1} (a_2 \setminus \{m_2\})\).
\(f'^{-1} (a_2 \cap \{m_2\}) = \emptyset \text{ or } a_1 \cup f^{-1} (\{m_2\})\), because when \(a_2 \cap \{m_2\} = \emptyset\), \(f'^{-1} (a_2 \cap \{m_2\}) = \emptyset\); otherwise, \(a_2 \cap \{m_2\} = \{m_2\}\), and \(f'^{-1} (\{m_2\}) = a_1 \cup f^{-1} (\{m_2\})\), because for each \(p \in f'^{-1} (\{m_2\})\), \(p \in a_1\) or \(p \notin a_1\), but when \(p \notin a_1\), \(f' (p) = f (p) = m_2\), so, \(p \in f^{-1} (\{m_2\})\); for each \(p \in a_1 \cup f^{-1} (\{m_2\})\), when \(p \in a_1\), \(f' (p) = m_2\), so, \(p \in f'^{-1} (\{m_2\})\); otherwise, \(f' (p) = f (p) = m_2\), so, \(p \in f'^{-1} (\{m_2\})\).
So, \(f'^{-1} (a_2) = (f^{-1} (a_2 \setminus \{m_2\}) \setminus a_1) \cup (\emptyset \text{ or } a_1 \cup f^{-1} (\{m_2\}))\).
As \(\{m_2\}\) is measurable, \(a_2 \setminus \{m_2\}\) is measurable, and \(f^{-1} (a_2 \setminus \{m_2\})\) is measurable, because \(f\) is measurable, and \(f^{-1} (a_2 \setminus \{m_2\}) \setminus a_1\) is measurable; \(\emptyset\) is measurable, \(f^{-1} (\{m_2\}\) is measurable, and \(a_1 \cup f^{-1} (\{m_2\})\) is measurable.
So, \(f'^{-1} (a_2)\) is measurable.
So, \(f'\) is measurable.
References
<The previous article in this series | The table of contents of this series | The next article in this series>
<The previous article in this series | The table of contents of this series | The next article in this series>
description/proof of that expansion of measurable map on codomain is measurable
Topics
About:
measurable space
The table of contents of this article
Starting Context
Target Context
-
The reader will have a description and a proof of the proposition that any expansion of any measurable map between any measurable spaces on the codomain is measurable.
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:
\((M_1, A_1)\): \(\in \{\text{ the measurable spaces }\}\)
\((M''_2, A''_2)\): \(\in \{\text{ the measurable spaces }\}\)
\((M'_2, A'_2)\): \(\in \{\text{ the measurable subspaces of } (M''_2, A''_2)\}\)
\((M_2, A_2)\): \(\in \{\text{ the measurable subspaces of } (M'_2, A'_2)\}\)
\(f\): \(: M_1 \to M_2\), \(\in \{\text{ the measurable maps }\}\)
\(f'\): \(: M_1 \to M'_2, m \mapsto f (m)\)
//
Statements:
\(f' \in \{\text{ the measurable maps }\}\)
//
2: Note
Whether \((M_2, A_2)\) is regarded to be the measurable subspace of \((M'_2, A'_2)\}\) or the measurable subspace of \((M''_2, A''_2)\}\) does not matter, because they are the same by the proposition that in any nest of measurable subspaces, the measurableness of any subset of any measurable space does not depend on the superspace of which the space is regarded to be the subspace.
Of course, \((M'_2, A'_2)\) can be taken to be \((M''_2, A''_2)\), so, the codomain can be expanded to \(M''_2\) to keep the map measurable.
3: Proof
Whole Strategy: Step 1: take any \(a' \in A'\), and see that \(f'^{-1} (a') = f^{-1} (a' \cap M_2)\).
Step 1:
Let \(a' \in A'\) be any.
Let us see that \(f'^{-1} (a') = f^{-1} (a' \cap M_2)\).
For each \(p \in f'^{-1} (a')\), \(f' (p) \in a'\), but \(f' (p) = f (p) \in M_2\), so, \(f (p) \in a' \cap M_2\), so, \(p \in f^{-1} (a' \cap M_2)\).
For each \(p \in f^{-1} (a' \cap M_2)\), \(f (p) \in a' \cap M_2 \subseteq a'\), but \(f (p) = f' (p)\), so, \(p \in f'^{-1} (a')\).
As \(a' \cap M_2 \in A_2\) and \(f\) is measurable, \(f^{-1} (a' \cap M_2) \in A_1\).
So, \(f'\) is measurable.
References
<The previous article in this series | The table of contents of this series | The next article in this series>
<The previous article in this series | The table of contents of this series | The next article in this series>
description/proof of that in nest of measurable subspaces, measurableness of subset of measurable space does not depend on superspace of which space is regarded to be subspace
Topics
About:
measurable space
The table of contents of this article
Starting Context
Target Context
-
The reader will have a description and a proof of the proposition that in any nest of measurable subspaces, the measurableness of any subset of any measurable space does not depend on the superspace of which the space is regarded to be 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:
\((M'', A'')\): \(\in \{\text{ the measurable spaces }\}\)
\((M', A')\): \(\in \{\text{ the measurable subspaces of } (M'', A'')\}\)
\(M\): \(\subseteq M'\)
\((M, A)\): \(\in \{\text{ the measurable subspaces of } (M', A')\}\)
\((M, \widetilde{A})\): \(\in \{\text{ the measurable subspaces of } (M'', A'')\}\)
//
Statements:
\(A = \widetilde{A}\)
//
2: Proof
Whole Strategy: Step 1: take any \(a \in A\), and see that \(a \in \widetilde{A}\); Step 2: take any \(\widetilde{a} \in \widetilde{A}\), and see that \(\widetilde{a} \in A\).
Step 1:
Let \(a \in A\) be any.
\(a = a' \cap M\) for an \(a' \in A'\).
But \(a' = a'' \cap M'\) for an \(a'' \in A''\).
So, \(a = a' \cap M = a'' \cap M' \cap M = a'' \cap M \in \widetilde{A}\).
Step 2:
Let \(\widetilde{a} \in \widetilde{A}\) be any.
\(\widetilde{a} = a'' \cap M\) for an \(a'' \in A''\).
But \(a'' \cap M = a'' \cap M' \cap M\), and \(a'' \cap M' \in A'\).
So, \(\widetilde{a} \in A\).
References
<The previous article in this series | The table of contents of this series | The next article in this series>
<The previous article in this series | The table of contents of this series | The next article in this series>
description/proof of that for measurable space generated by set of subsets and subset, subspace \(\sigma\)-algebra is generated by set of restricted subsets
Topics
About:
measurable space
The table of contents of this article
Starting Context
Target Context
-
The reader will have a description and a proof of the proposition that for any measurable space generated by any set of subsets and any subset, the subspace \(\sigma\)-algebra of the subset is generated by the set of the restricted subsets.
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:
\((M', A')\): \(\in \{\text{ the measurable spaces }\}\), where \(A' = \sigma (S')\) for any \(S' \subseteq Pow (M')\)
\((M, A)\): \(\in \{\text{ the measurable subspaces of } (M', A')\}\)
\(S\): \(= \{s' \cap M \in Pow (M) \vert s' \in S'\}\)
\(\sigma (S)\):
//
Statements:
\(A = \sigma (S)\)
//
2: Note
For example, for the extended Euclidean topological space, \(\overline{\mathbb{R}}\), with the topology, \(\overline{O}\), and the Euclidean topological space, \(\mathbb{R}\), \(\mathbb{R}\) has the subspace topology of \(\overline{\mathbb{R}}\), \(O\), and the Borel \(\sigma\)-algebra, \(\sigma (O)\), is the subspace \(\sigma\)-algebra of \(\sigma (\overline{O})\): \(O = \{\overline{o} \cap \mathbb{R} \in Pow (\mathbb{R}) \vert \overline{o} \in \overline{O}\}\).
Likewise, for the topological subspace of \(\overline{\mathbb{R}}\), \([0, \infty]\), with the topology, \(O\), \(\sigma (O)\) is the subspace \(\sigma\)-algebra of \(\sigma (\overline{O})\).
3: Proof
Whole Strategy: Step 1: see that \(\sigma (S) \subseteq A\); Step 2: take \(\widetilde{A'} := \{U \subseteq M' \vert U \cap M \in \sigma (S)\}\), and see that \(A' \subseteq \widetilde{A'}\); Step 3: see that \(A \subseteq \sigma (S)\) using \(A = \{a' \cap M \vert a' \in A'\}\).
Step 1:
Let us see that \(\sigma (S) \subseteq A\).
As \(A = \{a' \cap M \vert a' \in A'\}\) and \(S' \subseteq A'\), \(S \subseteq A\), because for each \(s \in S\), \(s = s' \cap M\) for an \(s' \in S'\), \(s' \in A'\), and \(s' \cap M \in A\).
So, \(A\) is a \(\sigma\)-algebra that contains \(S\).
As \(\sigma (S)\) is the intersection of all the such \(\sigma\)-algebras, \(\sigma (S) \subseteq A\).
Step 2:
Let us take \(\widetilde{A'} := \{U \subseteq M' \vert U \cap M \in \sigma (S)\}\).
Let us see that \(\widetilde{A'}\) is a \(\sigma\)-algebra of \(M'\).
\(M' \in \widetilde{A'}\), because \(M' \cap M = M \in \sigma (S)\).
Let \(U \in \widetilde{A'}\) be any.
\(M' \setminus U \in \widetilde{A'}\), because \((M' \setminus U) \cap M = (M' \cap M) \setminus (U \cap M)\), by the proposition that the intersection of any set minus any set and any set is the intersection of the 1st set and the 3rd set minus the intersection of the 2nd set and the 3rd set, \(= M \setminus (U \cap M) \in \sigma (S)\), because \(U \cap M \in \sigma (S)\).
Let \(s: \mathbb{N} \to \widetilde{A'}\) be any.
\((\cup_{j \in \mathbb{N}} s (j)) \cap M = \cup_{j \in \mathbb{N}} (s (j) \cap M)\), 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, \(\in \sigma (S)\), because \(s (j) \cap M \in \sigma (S)\).
So, \(\widetilde{A'}\) is a \(\sigma\)-algebra of \(M'\).
\(S' \subseteq \widetilde{A'}\), because for each \(s' \in S'\), \(s' \cap M \in S \subseteq \sigma (S)\).
So, \(\widetilde{A'}\) is a \(\sigma\)-algebra of \(M'\) that contains \(S'\).
So, \(A' \subseteq \widetilde{A'}\), because \(A'\) is the intersection of all the such \(\sigma\)-algebras.
Step 3:
\(A = \{a' \cap M \vert a' \in A'\} \subseteq \{a' \cap M \vert a' \in \widetilde{A'}\}\), because \(A' \subseteq \widetilde{A'}\), \(\subseteq \sigma (S)\), by the definition of \(\widetilde{A'}\).
So, \(A = \sigma (S)\).
References
<The previous article in this series | The table of contents of this series | The next article in this series>
<The previous article in this series | The table of contents of this series | The next article in this series>
description/proof of that for measurable space with each single-point-subset measurable, map from space into space that replaces countable subset with sequence of points is measurable
Topics
About:
measurable space
The table of contents of this article
Starting Context
Target Context
-
The reader will have a description and a proof of the proposition that for any measurable space with each single-point-subset measurable, any map from the space into the space that replaces any countable subset with any sequence of points is measurable.
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:
\((M, A)\): \(\in \{\text{ the measurable spaces }\}\), such that \(\forall m \in M (\{m\} \in A)\)
\(J\): \(\in \{\text{ the countable index sets }\}\)
\(S\): \(= \{s_j \in M \vert j \in J\}\)
\(s\): \(: J \to M\)
\(f\): \(: M \to M, m \mapsto s (j) \text{ when } m = s_j \in S; \mapsto m \text{ otherwise }\)
//
Statements:
\(f \in \{\text{ the measurable maps }\}\)
//
2: Note
A motivation for this proposition is to create a measurable map from a measurable map, \(f': M_1 \to M_2\), by tweaking \(f'\) for only some countable image points: take any \(f: M_2 \to M_2\) for this proposition, then, \(f \circ f': M_1 \to M_2\) is measurable, by the proposition that for any measurable maps between any arbitrary subspaces of any measurable spaces, the composition is measurable.
For example, for a measurable \(f': M_1 \to [- \infty, \infty]\), take \(f: [- \infty, \infty] \to [- \infty, \infty]\) with \(S = \{- \infty, \infty\}\) and \(s = \{0, 0\}\), then, \(f \circ f'\) practically becomes a map into \(\mathbb{R}\), measurable.
3: Proof
Whole Strategy: Step 1: take any \(a \in A\) and \(S' := \{s_j \in S \vert s (s_j) \in a\} \subseteq S\), and see that \(f^{-1} (a) = (a \setminus S) \cup S'\); Step 2: conclude the proposition.
Step 1:
Let \(a \in A\) be any.
Let us think of \(S' := \{s_j \in S \vert s (s_j) \in a\} \subseteq S\).
Let us see that \(f^{-1} (a) = (a \setminus S) \cup S'\).
Let \(m \in f^{-1} (a)\) be any.
\(m \in a \cup S\), because if \(m \notin a \cup S\), as \(m \notin S\), \(f (m) = m\), but \(m \notin a\), so, \(f (m) \notin a\), so, \(m \notin f^{-1} (a)\), a contradiction.
\(a \cup S = (a \setminus S) \cup S\).
So, \(m \in a \setminus S\) or \(m \in S\).
When \(m \in S\), \(m \in S'\), because otherwise, \(m \in S \setminus S'\), \(f (m) = s (j) \notin a\) where \(m = s_j\), so, \(m \notin f^{-1} (a)\), a contradiction.
So, \(m \in a \setminus S\) or \(m \in S'\).
So, \(m \in (a \setminus S) \cup S'\).
Let \(m \in (a \setminus S) \cup S'\) be any.
\(m \in a \setminus S\) or \(m \in S'\).
When \(m \in a \setminus S\), \(f (m) = m \in a\), so, \(m \in f^{-1} (a)\).
When \(m \in S'\), \(f (m) = s (j) \in a\) where \(m = s_j\), so, \(m \in f^{-1} (a)\).
So, \(m \in f^{-1} (a)\) anyway.
So, \(f^{-1} (a) = (a \setminus S) \cup S'\).
Step 2:
As each single-point-subset is in \(A\), \(S \in A\), as the union of some countable single-point-subsets.
\(a \setminus S \in A\).
\(S' \in A\), because that is the union of some countable single-point-subsets.
So, \((a \setminus S) \cup S' \in A\).
So, \(f\) is measurable.
References
<The previous article in this series | The table of contents of this series | The next article in this series>
<The previous article in this series | The table of contents of this series | The next article in this series>
description/proof of that for measurable maps between arbitrary subspaces of measurable spaces, composition is measurable
Topics
About:
measurable space
The table of contents of this article
Starting Context
Target Context
-
The reader will have a description and a proof of the proposition that for any measurable maps between any arbitrary subspaces of any measurable spaces, the composition is measurable.
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:
\((M'_1, A'_1)\): \(\in \{\text{ the measurable spaces }\}\)
\((M'_2, A'_2)\): \(\in \{\text{ the measurable spaces }\}\)
\((M'_3, A'_3)\): \(\in \{\text{ the measurable spaces }\}\)
\((M_1, A_1)\): \(\in \{\text{ the measurable subspaces of } (M'_1, A'_1)\}\)
\((M_2, A_2)\): \(\in \{\text{ the measurable subspaces of } (M'_2, A'_2)\}\)
\((\widetilde{M_2}, \widetilde{A_2})\): \(\in \{\text{ the measurable subspaces of } (M'_2, A'_2)\}\), such that \(M_2 \subseteq \widetilde{M_2}\)
\((M_3, A_3)\): \(\in \{\text{ the measurable subspaces of } (M'_3, A'_3)\}\)
\(f_1\): \(: M_1 \to M_2\), \(\in \{\text{ the measurable maps }\}\)
\(f_2\): \(: \widetilde{M_2} \to M_3\), \(\in \{\text{ the measurable maps }\}\)
\(f_2 \circ f_1\): \(: M_1 \to M_3\)
//
Statements:
\(f_2 \circ f_1 \in \{\text{ the measurable maps }\}\)
//
2: Proof
Whole Strategy: Step 1: take any \(a \in A_3\) and see that \((f_2 \circ f_1)^{-1} (a) = f_1^{-1} (f_2^{-1} (a) \cap M_2)\); Step 2: see that \(f_2^{-1} (a) \cap M_2 \in A_2\); Step 3: conclude the proposition.
Step 1:
Let \(a \in A_3\) be any.
\((f_2 \circ f_1)^{-1} (a) = f_1^{-1} (f_2^{-1} (a) \cap M_2)\), by the proposition that for any maps composition, the preimage under the composition is the composition of the map preimages in the reverse order.
Step 2:
\(f_2^{-1} (a) \in \widetilde{A_2}\), because \(f_2\) is measurable.
\(f_2^{-1} (a) = a' \cap \widetilde{M_2}\) for an \(a' \in A'_2\), by the definition of measurable subspace.
\(f_2^{-1} (a) \cap M_2 = a' \cap \widetilde{M_2} \cap M_2 = a' \cap M_2 \in A_2\).
Step 3:
So, \((f_2 \circ f_1)^{-1} (a) = f_1^{-1} (f_2^{-1} (a) \cap M_2) = f_1^{-1} (a' \cap M_2) \in A_1\), because \(f_1\) is measurable.
So, \(f_2 \circ f_1\) is measurable.
References
<The previous article in this series | The table of contents of this series | The next article in this series>
<The previous article in this series | The table of contents of this series | The next article in this series>
description/proof of that for measurable map between measurable spaces, restriction on domain subspace and codomain subspace is measurable
Topics
About:
measurable space
The table of contents of this article
Starting Context
Target Context
-
The reader will have a description and a proof of the proposition that for any measurable map between any measurable spaces, the restriction on any domain subspace and any codomain subspace is measurable.
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:
\((M'_1, A'_1)\): \(\in \{\text{ the measurable spaces }\}\)
\((M'_2, A'_2)\): \(\in \{\text{ the measurable spaces }\}\)
\(f'\): \(: M'_1 \to M'_2\), \(\in \{\text{ the measurable maps }\}\)
\(M_1\): \(\subseteq M'_1\)
\((M_1, A_1)\): \(= \text{ the measurable subspace of } (M'_1, A'_1)\)
\(M_2\): \(\subseteq M'_2\) such that \(f' (M_1) \subseteq M_2\)
\((M_2, A_2)\): \(= \text{ the measurable subspace of } (M'_2, A'_2)\)
\(f\): \(: M_1 \to M_2, p \mapsto f' (p)\)
//
Statements:
\(f \in \{\text{ the measurable maps }\}\)
//
2: Proof
Whole Strategy: Step 1: take any \(a \in A_2\) and see that \(a = a' \cap M_2\) where \(a' \in A'_2\); Step 2: see that \(f^{-1} (a) = f'^{-1} (a') \cap M_1\), and conclude the proposition.
Step 1:
Let \(a \in A_2\) be any.
\(a = a' \cap M_2\) where \(a' \in A'_2\), by the definition of measurable subspace.
Step 2:
\(f^{-1} (a) = (f' \vert_{M_1})^{-1} (a) = f'^{-1} (a) \cap M_1\), by the proposition that for any map and its any codomain-restriction, the preimage of any subset of the original codomain under the codomain-restricted map is the preimage of the subset under the original map and 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.
\(= f'^{-1} (a' \cap M_2) \cap M_1 = f'^{-1} (a') \cap f'^{-1} (M_2) \cap M_1\), by the proposition that for any map, the map preimage of any intersection of sets is the intersection of the map preimages of the sets.
But \(f'^{-1} (M_2) \cap M_1 = M_1\), because for each \(p \in M_1\), \(f' (p) \in M_2\), so, \(p \in f'^{-1} (M_2)\), so, \(= f'^{-1} (a') \cap M_1\).
But as \(f'\) is measurable, \(f'^{-1} (a') \in A'_1\).
So, \(f^{-1} (a) \in A_1\), by the definition of measurable subspace.
So, \(f\) is measurable.
References
<The previous article in this series | The table of contents of this series | The next article in this series>
<The previous article in this series | The table of contents of this series | The next article in this series>
description/proof of that preimage under codomain-restricted map is preimage under original map
Topics
About:
set
The table of contents of this article
Starting Context
Target Context
-
The reader will have a description and a proof of the proposition that for any map and its any codomain-restriction, the preimage of any subset of the original codomain under the codomain-restricted map is the preimage of the subset under the original map.
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:
\(S_1\): \(\in \{\text{ the sets }\}\)
\(S'_2\): \(\in \{\text{ the sets }\}\)
\(f'\): \(: S_1 \to S'_2\)
\(S_2\): \(\subseteq S'_2\) such that \(f' (S_1) \subseteq S_2\)
\(f\): \(: S_1 \to S_2, s \mapsto f' (s)\)
\(S\): \(\subseteq S'_2\)
//
Statements:
\(f^{-1} (S) = f'^{-1} (S)\)
//
2: Proof
Whole Strategy: Step 1: for each \(s \in f^{-1} (S)\), see that \(s \in f'^{-1} (S)\); Step 2: for each \(s \in f'^{-1} (S)\), see that \(s \in f^{-1} (S)\).
Step 1:
Let \(s \in f^{-1} (S)\) be any.
\(f (s) \in S\).
But \(f' (s) = f (s) \in S\).
So, \(s \in f'^{-1} (S)\).
Step 2:
Let \(s \in f'^{-1} (S)\) be any.
\(f' (s) \in S\).
But \(f (s) = f' (s) \in S\).
So, \(s \in f^{-1} (S)\).
References
<The previous article in this series | The table of contents of this series | The next article in this series>
<The previous article in this series | The table of contents of this series | The next article in this series>
description/proof of that for measure space, union of sequence of locally negligible subsets is locally negligible
Topics
About:
measure space
The table of contents of this article
Starting Context
Target Context
-
The reader will have a description and a proof of the proposition that for any measure space, the union of any sequence of any locally negligible subsets is locally negligible.
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:
\((M, A, \mu)\): \(\in \{\text{ the measure spaces }\}\)
\(s\): \(: \mathbb{N} \to \{\text{ the locally negligible subsets of } M\}\)
//
Statements:
\(\cup_{j \in \mathbb{N}} s (j) \in \{\text{ the locally negligible subsets of } M\}\)
//
2: Proof
Whole Strategy: Step 1: take any \(a \in A\) such that \(\mu (a) \lt \infty\), and see that \((\cup_{j \in \mathbb{N}} s (j)) \cap a\) is negligible.
Step 1:
Let \(a \in A\) be any such that \(\mu (a) \lt \infty\).
Let us see that \((\cup_{j \in \mathbb{N}} s (j)) \cap a\) is negligible.
\((\cup_{j \in \mathbb{N}} s (j)) \cap a = \cup_{j \in \mathbb{N}} (s (j) \cap a)\), 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.
As \(s (j)\) is locally negligible, \(s (j) \cap a \subseteq N_j\) for an \(N_j \in A\) such that \(\mu (N_j) = 0\).
So, \((\cup_{j \in \mathbb{N}} s (j)) \cap a \subseteq \cup_{j \in \mathbb{N}} N_j\).
\(\cup_{j \in \mathbb{N}} N_j \in A\), and \(\mu (\cup_{j \in \mathbb{N}} N_j) \le \sum_{j \in \mathbb{N}} \mu (N_j) = \sum_{j \in \mathbb{N}} 0 = 0\).
So, \((\cup_{j \in \mathbb{N}} s (j)) \cap a\) is negligible.
So, \(\cup_{j \in \mathbb{N}} s (j)\) is locally negligible.
References
<The previous article in this series | The table of contents of this series | The next article in this series>
