Rudin's Real and Complex Analysis -upto Section 1.9 review:
At this point (if you are following this from previous post, Rudin introduces a Theorem 1.7 that is fundamental and very useful.
Stated informally,
1. Continuous function of continuous function is continuous.
2. Continuous function of measurable function is measurable.
(2) requires Y,Z are topological spaces for continuous functions and X is measurable. Of course (1) requires all spaces be topological.
The proof relies on definitions of continuity and measurability. Measurability requires f−1(V) of any open set V be measurable.
Theorem 1.8 is very puzzling at first. There is no explanation why we need a plane and what it is doing here. Fortunately, reading next section (1.9) makes it clear. Also it uses a fact from real analysis - every open set in plane is a countable union of rectangles with sides parallel to axis.
The proof is somewhat complicated - but not too bad. Given u(x),v(x) are measurable, it starts off by setting f=(u(x),v(x)) as mapping of X (measurable space) into a plane and setting h=Φ∘f and thus requiring measurability of f to complete the proof. Since Φ is defined to be continuous function, the measurability of f completes the proof. To show f is measurable all that's needed to show is f−1(V) for any open set V should be measurable.
At this point (if you are following this from previous post, Rudin introduces a Theorem 1.7 that is fundamental and very useful.
Stated informally,
1. Continuous function of continuous function is continuous.
2. Continuous function of measurable function is measurable.
(2) requires Y,Z are topological spaces for continuous functions and X is measurable. Of course (1) requires all spaces be topological.
The proof relies on definitions of continuity and measurability. Measurability requires f−1(V) of any open set V be measurable.
Theorem 1.8 is very puzzling at first. There is no explanation why we need a plane and what it is doing here. Fortunately, reading next section (1.9) makes it clear. Also it uses a fact from real analysis - every open set in plane is a countable union of rectangles with sides parallel to axis.
The proof is somewhat complicated - but not too bad. Given u(x),v(x) are measurable, it starts off by setting f=(u(x),v(x)) as mapping of X (measurable space) into a plane and setting h=Φ∘f and thus requiring measurability of f to complete the proof. Since Φ is defined to be continuous function, the measurability of f completes the proof. To show f is measurable all that's needed to show is f−1(V) for any open set V should be measurable.
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