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I've been reading through Amol Sasane "The How and Why of One Variable Calculus" and I've got a question from one of the proofs about the sum property of integrals. (Page 199)

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I know this doesn't have all the details. We've chosen an arbitrary positive epsilon and found partitions to make this statement of the upper sums of two functions work.

So we have : $$\overline{S}(f + g) - \epsilon < \overline{S}(f) + \overline{S}(g)$$ Which I can follow when doing the proof myself. Though Sasane's next statement confuses me a little, at least with the technical details. When looking at this proof on my own a thought I had was "Can I take the limit as $\epsilon \to 0$?". I thought about it and it seemed like it might a way of doing this. Though I can't see how it would work rigorously or with change to a non-strict inequality.

Would anyone be able to shed some like on the details or find me material explaining this property?

Edit: If anyone would be so inclined, is this something really obvious and I just didn't see it? Reading through this book I feel like many students starting off would need more explanation than what was given.

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2 Answers 2

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Assume the contrary that $\bar{S}(f+g)>\bar{S}(f)+\bar{S}(g)$. Then define $\epsilon = \bar{S}(f+g)-\bar{S}(f)-\bar{S}(g)>0$. It has been shown that $$\bar{S}(f+g)-\epsilon<\bar{S}(f)+\bar{S}(g)$$ since $\epsilon>0$ is positive. However this gives us the contradiction $$\bar{S}(f+g)-\epsilon<\bar{S}(f)+\bar{S}(g)=\bar{S}(f+g)-\epsilon$$ since no number is strictly less than itself.

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  • $\begingroup$ When we assume the contrary to our hypothesis we find the contradiction so we can then assume that since ">" is false than the answer must be "< or $\leq$" and we can just choose the non-strict? I'm trying to ensure I understand everything. Thank you this helps a lot. $\endgroup$
    – Derek C.
    Commented Jan 19, 2021 at 16:59
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    $\begingroup$ Precisely, since $\bar{S}(f+g)>\bar{S}(f)+\bar{S}(g)$ can't hold we must have $$\bar{S}(f+g)\leq\bar{S}(f)+\bar{S}(g)$$ $\endgroup$
    – OgvRubin
    Commented Jan 19, 2021 at 18:38
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Starting with the statement $$\overline{S}(f + g) - \varepsilon < \overline{S}(f) + \overline{S}(g)\tag{1}$$ where $\varepsilon > 0$ is arbitrary. Suppose now that $$\overline{S}(f+g) > \overline{S}(f) + \overline{S}(g)\tag{2}.$$ In this case, set $$\overline{S}(f+g) - \overline{S}(f) - \overline{S}(g) = m > 0$$ and let $\varepsilon' = \frac{m}{2} >0$. Then we have $$\overline{S}(f+g) -\varepsilon' -\overline{S}(f) - \overline{S}(g) = m - \varepsilon' = m - \frac{m}{2} = \frac{m}{2} > 0.$$ This, however, means that $$\overline{S}(f+g) - \varepsilon' > \overline{S}(f) + \overline{S}(g),$$ which contradicts $(1)$. Thus, our assumption $(2)$ was incorrect, and we have by contradiction that $$\overline{S}(f+g) \leq \overline{S}(f) + \overline{S}(g).$$

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    $\begingroup$ This makes it very clear, thank you. $\endgroup$
    – Derek C.
    Commented Jan 19, 2021 at 17:12

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