description/proof of that for map from measurable space into \(1\)-dimensional complex Euclidean topological space with Borel \(\sigma\)-algebra, iff map is measurable, real and imaginary parts maps are 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.
- The reader knows a definition of complex Euclidean topological space.
- The reader knows a definition of Borel \(\sigma\)-algebra of topological space.
- 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 for any measurable maps between any arbitrary subspaces of any measurable spaces, the composition is measurable.
- The reader admits 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.
- The reader admits the proposition that some para-product maps of measurable maps are measurable.
Target Context
- The reader will have a description and a proof of the proposition that for any map from any measure space into the \(1\)-dimensional complex Euclidean topological space with the Borel \(\sigma\)-algebra, if and only if the map is measurable, the real and imaginary parts maps are 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:
\((M, A)\): \(\in \{\text{ the measurable spaces }\}\)
\(\mathbb{C}\): \(= \text{ the complex Euclidean topological space }\) with the Borel \(\sigma\)-algebra
\(f\): \(: M \to \mathbb{C}\)
\(f_1\): \(: M \to \mathbb{R}, m \mapsto Re (f (m))\)
\(f_2\): \(: M \to \mathbb{R}, m \mapsto Im (f (m))\)
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Statements:
\(f \in \{\text{ the measurable maps }\}\)
\(\iff\)
\(f_1, f_2 \in \{\text{ the measurable maps }\}\)
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2: Proof
Whole Strategy: Step 1: take the canonical homeomorphism, \(g: \mathbb{R}^2 \to \mathbb{C}\), the projections, \(g_1: \mathbb{R}^2 \to \mathbb{R}\) and \(g_2: \mathbb{R}^2 \to \mathbb{R}\), and \(h: M \to \mathbb{R}^2, m \mapsto (f_1 (m), f_2 (m))\); Step 2: see that \(f_1 = g_1 \circ g^{-1} \circ f\), \(f_2 = g_2 \circ g^{-1} \circ f\), and \(f = g \circ h\); Step 3: suppose that \(f\) is measurable and see that \(f_1\) and \(f_2\) are measurable; Step 4: suppose that \(f_1\) and \(f_2\) are measurable and see that \(f\) is measurable.
Step 1:
Let \(\mathbb{R}^2\) be the Euclidean topological space with the Borel \(\sigma\)-algebra.
Let \(\mathbb{R}\) be the Euclidean topological space with the Borel \(\sigma\)-algebra.
Let \(g: \mathbb{R}^2 \to \mathbb{C}\) be the canonical homeomorphism mentioned in the definition of complex Euclidean topological space.
Let \(g_1: \mathbb{R}^2 \to \mathbb{R}, (r_1, r_2) \mapsto r_1\) be the projection.
Let \(g_2: \mathbb{R}^2 \to \mathbb{R}, (r_1, r_2) \mapsto r_2\) be the projection.
Let \(h: M \to \mathbb{R}^2, m \mapsto (f_1 (m), f_2 (m))\).
Step 2:
\(f_1 = g_1 \circ g^{-1} \circ f\).
\(f_2 = g_2 \circ g^{-1} \circ f\).
\(f = g \circ h\).
Step 3:
Let us suppose that \(f\) is measurable.
Let us think of \(g_1 \circ g^{-1}: \mathbb{C} \to \mathbb{R}\).
\(g_1\) 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, \(g_1 \circ g^{-1}\) is continuous.
So, \(g_1 \circ g^{-1}\) is measurable, by the proposition that any continuous map between any topological spaces with the Borel \(\sigma\)-algebras is measurable.
So, \(f_1 = g_1 \circ g^{-1} \circ f\) is measurable, by the proposition that for any measurable maps between any arbitrary subspaces of any measurable spaces, the composition is measurable.
\(f_2 = g_2 \circ g^{-1} \circ f\) is measurable, likewise.
Step 4:
Let us suppose that \(f_1\) and \(f_2\) are measurable.
\(h\) is measurable, by 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 and the proposition that some para-product maps of measurable maps are measurable.
\(g\) is measurable, by the proposition that any continuous map between any topological spaces with the Borel \(\sigma\)-algebras is measurable.
So, \(f = g \circ h\) is measurable, by the proposition that for any measurable maps between any arbitrary subspaces of any measurable spaces, the composition is measurable.
