Help with Apostol's "Calculus, vol. 1", Section 1.18

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Homework Help Overview

The discussion revolves around understanding theorems presented in Apostol's "Calculus, vol. 1", specifically section 1.18, which addresses the area of an ordinate set expressed as an integral. Participants are examining the logic behind theorem 1.11, particularly the claim regarding the measurability of the set Q' and its area in relation to Q.

Discussion Character

  • Conceptual clarification, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants are questioning the validity of Apostol's argument that Q' is measurable and that the areas of Q and Q' are equal. There is discussion about the implications of step regions and their definitions in relation to the graph of the function f(x).

Discussion Status

Multiple interpretations of the theorems are being explored, with participants suggesting alternative definitions and approaches to the problem. Some have proposed the use of sequences to address the convergence of areas, while others emphasize the need for clarity regarding the definitions of sets involved.

Contextual Notes

Participants note that the discussion is constrained by the material covered in the book up to this point, particularly the absence of sequences in Apostol's explanations. The Least Upper Bound Property is mentioned as a relevant concept, but its application to the current problem remains unclear.

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In section 1.18 ("The area of an ordinate set expressed as an integral"), Apostol proves two theorems. the first, theorem 1.10, deals with the area of a function's ordinate set; the second, theorem 1.11, deals with the area of the graph of the function of theorem 1.10. (I have attached two excrepts from Apostol's book, one per theorem.)

I am having problems understansing Apostol's logic in theorem 1.11 where he states:

"The argument used to prove Theorem 1.10 also shows that Q’ is measurable and that a(Q’) = a(Q)."

I don see how he could argue this, being that ##Q'=\{(x,y)|a \le x \le b, 0 \le y < f(x) \}## which implies that ##S \subseteq Q'## is not true for all step regions S (since S may contain a point of the graph of ##f(x)## which Q', by definition, can't).

Thanks in advanced for any help.
 

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I think the method is the same, but the definition of S would have to change if a point in S was equal to f(x).
So for the measure of Q', you would have sets S' and T', but the result would be the same--that the area of a line (or curve) is zero.
 
You can make the points ##(x, y)## in the set ##Q'## sufficiently close to the graph of ##y = f(x)##. So the areas are indeed equal.
 
RUber, I see what you're saying, however the new step regions (S' and T') would produce functions bearing the following relationship with the function##f(x)##: ##s'(x) < f(x) \le t'(x)##. Which doesn't help since the definition of the integral requires "##\le##" for both step function inequalities.

Zondrina, The idea is making the of make the points equal since by definition of Q': ##Q \neq Q'##. So no matter how close I get too the points of ##f(x)##, I would not be able to put all points of both regions in a one-to-one correspondence in order to then argue by congruence of regions.
 
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You can define a sequence s_n(x) such that the limit as n goes to infinity of s_n(x) = f(x). In this way, you will still satisfy the requirement that the intervals will converge on the true integral.
 
I get what you're saying. The only problem is that Apostol still hasn't mentioned sequences. The only relevant thing I can think of that has been covered in the book till now is the Least Upper Bound Property of Numbers. Which would simply state that ##f(x)## is the supremum for ##Q'##
 

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