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

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In section 1.18 of Apostol's "Calculus, vol. 1," the discussion centers on the proofs of theorems 1.10 and 1.11 regarding the area of a function's ordinate set and its graph. The confusion arises in understanding how theorem 1.11 asserts that the area of Q' is equal to the area of Q, despite Q' being defined in a way that suggests not all step regions S can be contained within it. Participants argue that while the method may be similar, the definition of step regions must change to accommodate points on the graph of f(x). The conversation highlights the challenge of establishing a one-to-one correspondence between regions Q and Q' and the necessity of sequences to approach the integral's limit. Ultimately, the discussion emphasizes the importance of understanding the properties of measurable sets and the implications of the Least Upper Bound Property in this context.
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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'##
 
First, I tried to show that ##f_n## converges uniformly on ##[0,2\pi]##, which is true since ##f_n \rightarrow 0## for ##n \rightarrow \infty## and ##\sigma_n=\mathrm{sup}\left| \frac{\sin\left(\frac{n^2}{n+\frac 15}x\right)}{n^{x^2-3x+3}} \right| \leq \frac{1}{|n^{x^2-3x+3}|} \leq \frac{1}{n^{\frac 34}}\rightarrow 0##. I can't use neither Leibnitz's test nor Abel's test. For Dirichlet's test I would need to show, that ##\sin\left(\frac{n^2}{n+\frac 15}x \right)## has partialy bounded sums...