I was working out a proof in Spivak's Calculus (2008) - pg 279. The following is a screenshot of the portion of the proof I'm having trouble with.

enter image description here

My question is in working out combining steps 1,2, and 3 correctly. I want to arrive at

$$\bigg|\sum_{i = 1}^{n}f(x_{i})(t_{i}-t_{i-1}) - \int_{a}^{b}f(x)dx \bigg| < \epsilon \\ \Rightarrow\ -\epsilon < \sum_{i = 1}^{n}f(x_{i})(t_{i}-t_{i-1}) - \int_{a}^{b}f(x)dx < \epsilon$$

Fiddling around with equation 2, I would get something of the form

$$ 0 \leq \sum_{i = 1}^{n}f(x_{i})(t_{i}-t_{i-1}) - L(f,P) \leq U(f,P) - L(f,P) < \epsilon$$

The same would occur for $\int_{a}^{b}f(x) dx$. Now using this idea I get something of the form:

$$\epsilon > U(f,P) - L(f,P) \geq \sum_{i = 1}^{n}f(x_{i})(t_{i}-t_{i-1}) - L(f,P) \geq \sum_{i = 1}^{n}f(x_{i})(t_{i}-t_{i-1}) - \int_{a}^{b}f(x) \geq ?? $$

Here is my issue, I can't say for sure that $\sum_{i = 1}^{n}f(x_{i})(t_{i}-t_{i-1}) - \int_{a}^{b}f(x) \geq 0$. Nothing that I have can imply such and as a result of this I can't conclude that $\sum_{i = 1}^{n}f(x_{i})(t_{i}-t_{i-1}) - \int_{a}^{b}f(x) > -\epsilon$. Which would allow me to finish this part of the proof. From experience I know it is a minor algebraic thing I'm missing, but I suppose I'm mentally fatigued and not seeing it. Some assistance would be nice.

  • 2
    $\begingroup$ Inequalities such as these can be easily managed if you use geometric language. Let $a=L(f, P), b=U(f, P) $ then $[a, b] $ is an interval of length less than $\epsilon $. Now distance between any two points lying in this interval can't exceed length of this interval and hence will be less than $\epsilon$. Now notice that any Riemann sum as well as the Riemann integral lie in this interval. $\endgroup$
    – Paramanand Singh
    Commented Aug 20, 2020 at 10:24
  • $\begingroup$ @ParamanandSingh By that reasoning, the result follows for any $P$. Why there is a stipulation about the lengths of $t_i - t_{i-1}$? $\endgroup$
    – yellowcat
    Commented Nov 16, 2020 at 15:10
  • $\begingroup$ @yellowcat: perhaps you have misinterpreted my comment. My comment assumes that the inequality $0\leq U(P, f) - L(P, f) <\epsilon$ holds. But this inequality does not hold for all partitions $P$ but rather for those partitions for which the points are very close ie $t_i-t_{i-1}<\delta$. $\endgroup$
    – Paramanand Singh
    Commented Nov 17, 2020 at 14:31
  • $\begingroup$ @ParamanandSingh I was wrong with "any $P$". But nevertheless, since $f$ is integrable, $\forall \epsilon >0\ \exists P$ s.t. $U(f,P)-L(f,P)<\epsilon$ (which is equivalent for integrability). We also have $L(f,P) \leq \int_{a}^{b}f \leq U(f,P)$ by definition of integral, and $L(f,P) \leq \sum_{i = 1}^{n}f(x_{i})(t_{i}-t_{i-1}) \leq U(f,P)$, and the rest of the proof is the same. Why do we need the restriction $t_i-t_{i-1}<\delta$? PS I'm fine with your first comment, just asking different, but related to this particular problem, question. $\endgroup$
    – yellowcat
    Commented Nov 18, 2020 at 16:55
  • $\begingroup$ @yellowcat: here is the deep theorem: if for every $\epsilon>0$ there is a partition $P_{\epsilon} $ such that $U(f, P_{\epsilon}) - L(f, P_{\epsilon}) <\epsilon $, then for every $\varepsilon>0$ there is a corresponding $\delta>0$ such that $U(f, P) - L(f, P) <\varepsilon$ for every partition $P$ with subintervals having lengths less than $\delta $. For proof see math.stackexchange.com/a/1834341/72031 $\endgroup$
    – Paramanand Singh
    Commented Nov 19, 2020 at 2:17

2 Answers 2


Hint: Multiply equation $(3)$ by $-1$ and add to equation $(2)$ to get:
$-(U(f,p) -L(f,P))\leq -\int_{a}^{b}f(x)dx+\sum_{i = 1}^{n}f(x_{i})(t_{i}-t_{i-1}) \leq U(f,P) - L(f,P) $
In other words, we have $-\epsilon\lt y\lt \epsilon$, whence $|y|\lt \epsilon$


$(2)$ and $(3)$ imply that both the sum and the integral are between $L(f,P)$ and $U(f,P)$ so the absolute difference between them cannot be more than $U(f,P)-L(f,P)$ and by $(1)$ this latter expression is less than $\epsilon.$


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