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
- The reader knows a definition of measurable map between measurable spaces.
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 \(\forall m \in M (\{m\} \in A)\)
\(J\): \(\in \{\text{ the countable index sets }\}\)
\(S\): \(= \{s_j \in M \vert j \in J\}\)
\(s\): \(: J \to M\)
\(f\): \(: M \to M, m \mapsto s (j) \text{ when } m = s_j \in S; \mapsto m \text{ otherwise }\)
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Statements:
\(f \in \{\text{ the measurable maps }\}\)
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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.
