# Can I remove ε from an inequality because it is arbitrary? [duplicate]

I'm looking at this proof (relating to lower/upper sums and integrals) and near the end it says this:

U(f) ≤ U(f,P) < L(f,P) + ε ≤ L(f) + ε

And then it says: "Since ε > 0 is arbitrary, we must have U(f) ≤ L(f)". I don't have any trouble understanding the rest of the proof but I don't understand how we get from the inequality above to U(f) ≤ L(f). Can someone explain to me the reason why we can remove ε and get to U(f) ≤ L(f)?

I've added a picture of the full proof to make it easier to understand my question. • Since $\varepsilon>0$ is arbitrary, you can take $\lim_{\varepsilon\to0}$ in both sides of your inequality Jan 28, 2021 at 18:26
• Remember: epsilon is an arbitrary positive number. Jan 28, 2021 at 18:26
• Does this answer your question? Intuition: If $a\leq b+\epsilon$ for all $\epsilon>0$ then $a\leq b$? Jan 28, 2021 at 18:43
• MartinR: OP tried to understand a piece of the proof; and the way OP asked the question shows that they don't know the underlying (simple) principle. This particular question was not asked and answered before. Your linked question, which asked for intuition of a known principle, and this one are closely related though.
– user9464
Jan 28, 2021 at 19:02
• Please cite your edition and page numbers? This is Theorem 29.9 on pages 273-4 in Steven Lay's Analysis with an Introduction to Proof (2005 4ed), and Theorem 1.9 on pages 296-7 in his 2015 5e of the New International Edition.
– user1124753
May 21 at 22:39

If you have two (real) quantities $$A$$ and $$B$$ such that $$A\le B+\epsilon$$ for every $$\epsilon>0$$, then you must have $$A\le B$$ since otherwise ($$A>B$$), you can find some $$\epsilon_0>0$$ such that $$A>B+\epsilon_0$$

(For instance, you can take $$\epsilon_0=\frac{A-B}{2}$$.)

In your example $$A=U(f)$$ and $$B=L(f)$$.

• This has been asked and answered many times before. With >3K reputation you have the privilege to cast close votes, e.g. as a duplicate. Compare math.stackexchange.com/help/privileges/close-questions. Jan 28, 2021 at 18:45
• MartinR: OP tried to understand a piece of the proof; and the way OP asked the question shows that they don't know the underlying (simple) principle. This particular question was not asked and answered before. Your linked question, which asked for intuition of a known principle, and this one are closely related though.
– user9464
Jan 28, 2021 at 19:01
• Maybe I'm being dumb, but if we have A ≤ C < D + ε ≤ B + ε (A, B, C, D being real numbers), wouldn't that give us A < B + ε rather than A ≤ B + ε ? Jan 28, 2021 at 19:23
• @SteveJoe $A<B+\epsilon$ implies $A\le B+\epsilon$. Remember that $X\le Y$ means "$X<Y$ or $X=Y$".
– user9464
Jan 28, 2021 at 19:25
• Ah I see. Thank you for your help Jan 28, 2021 at 19:27