description/proof of that Borel \(\sigma\)-algebra of \(d\)-dimensional Euclidean topological space is product \(\sigma\)-algebra of Borel \(\sigma\)-algebras of \(1\)-dimensional Euclidean topological space
Topics
About: measurable space
The table of contents of this article
Starting Context
- The reader knows a definition of Euclidean topological space.
- The reader knows a definition of Borel \(\sigma\)-algebra of topological space.
- The reader knows a definition of product \(\sigma\)-algebra.
- The reader admits the proposition that the \(d\)-dimensional Euclidean topological space is homeomorphic to the product of any combination of some lower-dimensional Euclidean spaces whose (the product's) dimension equals \(d\).
- The reader admits the proposition that for any product topological space, any projection is continuous.
- The reader admits the proposition that any continuous map between any topological spaces with the Borel \(\sigma\)-algebras is measurable.
- The reader admits the proposition that any product subset of any product set is the intersection of the product subsets each of which takes a component subset and the other whole sets.
Target Context
- The reader will have a description and a proof of the proposition that the Borel \(\sigma\)-algebra of the \(d\)-dimensional Euclidean topological space is the product \(\sigma\)-algebra of the Borel \(\sigma\)-algebras of the \(1\)-dimensional Euclidean topological 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:
\(d\): \(\in \mathbb{N} \setminus \{0\}\)
\(\mathbb{R}^d\): \(= \text{ the Euclidean topological space }\)
\(B (\mathbb{R}^d)\): \(= \text{ the Borel } \sigma \text{ -algebra }\)
\(\mathbb{R}\): \(= \text{ the Euclidean topological space }\)
\(B (\mathbb{R})\): \(= \text{ the Borel } \sigma \text{ -algebra }\)
\(A\): \(= \text{ the } d \text{ -product } \sigma \text{ -algebra for } \mathbb{R}^d\)
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Statements:
\(B (\mathbb{R}^d) = A\)
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2: Proof
Whole Strategy: Step 1: see that \(B (\mathbb{R}^d)\) is generated by \(S := \{(r_1, r'_1] \times ... \times (r_d, r'_d]\}\); Step 2: see that \(A\) is generated by a set that contains \(S\), and conclude that \(B (\mathbb{R}^d) \subseteq A\); Step 3: for each \(a_j \in B (\mathbb{R})\), see that \(a_1 \times ... \times a_d \in B (\mathbb{R}^d)\); Step 4: see that \(B (\mathbb{R}^d)\) is generated by a set that contains \(\{a_1 \times ... \times a_d\}\), and conclude that \(A \subseteq B (\mathbb{R}^d)\); Step 5: conclude the proposition.
Step 1:
\(B (\mathbb{R}^d)\) is generated by \(S := \{(r_1, r'_1] \times ... \times (r_d, r'_d] \subseteq \mathbb{R}^d\}\), which we accept as a well-known fact.
Step 2:
\(A\) is generated by \(S' := \{a_1 \times ... \times a_d \vert a_j \in B (\mathbb{R})\}\), by the definition of product \(\sigma\)-algebra.
\(S \subseteq S'\), because \((r_j, r'_j] \in B (\mathbb{R})\).
So, \(B (\mathbb{R}^d) \subseteq A\).
Step 3:
Let \(a_j \in B (\mathbb{R})\) be any.
Let us see that \(a_1 \times ... \times a_d \in B (\mathbb{R}^d)\).
Let \(\pi_j: \mathbb{R}^d \to \mathbb{R}\) be the projection into the \(j\)-th component.
\(\pi_j\) is continuous, by the proposition that the \(d\)-dimensional Euclidean topological space is homeomorphic to the product of any combination of some lower-dimensional Euclidean spaces whose (the product's) dimension equals \(d\) and the proposition that for any product topological space, any projection is continuous, so, is measurable, by the proposition that any continuous map between any topological spaces with the Borel \(\sigma\)-algebras is measurable.
\(a_1 \times ... \times a_d = \cap_{l \in \{1, ..., d\}} S_{l, 1} \times ... \times S_{l, d}\) where \(S_{l, j} = a_l\) when \(j = l\) and \(S_{l, j} = \mathbb{R}\) otherwise, by the proposition that any product subset of any product set is the intersection of the product subsets each of which takes a component subset and the other whole sets.
But \(S_{l, 1} \times ... \times S_{l, d} = \mathbb{R} \times ... \times \mathbb{R} \times a_l \times \mathbb{R} \times ... \times \mathbb{R} = {\pi_l}^{-1} (a_l)\).
So, \(a_1 \times ... \times a_d = \cap_{l \in \{1, ..., d\}} {\pi_l}^{-1} (a_l)\).
As \(\pi_l\) is measurable, \({\pi_l}^{-1} (a_l) \in B (\mathbb{R}^d)\), so, \(a_1 \times ... \times a_d \in B (\mathbb{R}^d)\).
Step 4:
So, \(B (\mathbb{R}^d)\) is generated by a set that contains \(S' = \{a_1 \times ... \times a_d \vert a_j \in B (\mathbb{R})\}\).
But \(A\) is generated by \(S'\).
So, \(A \subseteq B (\mathbb{R}^d)\).
Step 5:
So, \(B (\mathbb{R}^d) = A\).
3: Note
This cannot be easily generalized to the \(B (T^d)\) and \(A\) for any topological space, \(T\), case, because \(B (T^d)\) is generated by \(S := \{\cup_{j \in J} U_{j, 1} \times ... \times U_{j, d}\}\) where \(J\) is any possibly uncountable index set, \(A\) is generated by \(S' := \{a_1 \times ... \times a_d \vert a_j \in B (T)\}\), and \(\cup_{j \in J} U_{j, 1} \times ... \times U_{j, d}\) is not any element of \(S'\).
