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I'm confused with the terminology from my book:

LEMMA

If $P$ and $Q$ are partitions of $[a,b]$, then $L_f(P)\le U_f(Q)$.

(proof omitted)

From this lemma it follows that the set of all lower sums is bounded above and has a least upper bound $L$. s.t. $$L_f(P)\le L\le U_f(P)\quad\text{for all partitions }P,$$ and is clearly the least of such numbers. Similarly...

I don't know what is "the set of all lower sums". So I tried to take it as "the set of all partitions". Since $$L_f(P)\le L_f(A)\le U_f(A)\le U_f(P)\quad\text{for all partitions }P,$$ which $A$ is the union of all possible partitions. Then I take $L_f(A)$ as L, and $U_f(A)$ as $U$, which follows the Similarly... .

Are these correct?

Edit:

I use another theorem from my book in my proof above:

THEOREM

Suppose that f is continuous on $[a,b]$, and $P$ and $Q$ are partitions of $[a,b]$. If $P \subseteq Q$, then $$L_f(P)\le L_f(Q)\quad\text{and}\quad U_f(Q)\le U_f(P).$$

(no proof)

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    I assume that it should be $L_f(P)\le U_f(Q)$ in your first lemma. – 2017-02-02

1 Answers 1

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When the authors say "set of all lower sums", they mean the set of all lower sums. That is, they are refering to the set

$$S=\{L_f(P)| P\text{ is a partition of }[a,b]\}$$

This set is, clearly, bounded above since $U_f([a,b])$ is an (not neccesarily exact) upper bound of $S$.

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    May I ask that if I know both $$L_f(P)\le L\le U_f(P)\quad\text{for all partitions }P$$ and $$L_f(P)\le U\le U_f(P)\quad\text{for all partitions }P,$$ which *U* is the greatest lower bound, how can I get the conclusion $$L_f(P)\le L\le U\le U_f(P)\quad\text{for all partitions }P$$ in my proof? I don't know but I need the conclusion in another proof. – 2017-02-02
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    Just from those two things, you can't. But you probably define $U$ as the infimum of all upper sums, and in that case, you can prove that $U$ is indeed an upper bound for all lower sums. – 2017-02-03