I am trying to understand a part of the following theorem:

Theorem. Assume that $f:[a,b]\to\mathbb{R}$ is bounded, and let $c\in(a,b)$. Then, $f$ is integrable on $[a,b]$ if and only if $f$ is integrable on $[a,c]$ and $[c,b]$. In this case, we have $$\int_a^bf=\int_a^cf+\int_c^bf.$$ Proof. If $f$ is integrable on $[a,b]$, then for every $\epsilon>0$ there exists a partition $P$ such that $U(f,P)-L(f,P)<\epsilon$. Because refining a partition can only potentially bring the upper and lower sums closer together, we can simply add $c$ to $P$ if it is not already there. Then, let $P_1=P\cap[a,c]$ be a partition of $[a,c]$, and $P_2=P\cap[c,b]$ be a partition of $[c,b]$. It follows that $$U(f,P_1)-L(f,P_1)<\epsilon\text{ and }U(f,P_2)-L(f,P_2)<\epsilon,$$ implying that $f$ is integrable on $[a,c]$ and $[c,b]$.


How does that last expression "follow?" Neither $P_1$ nor $P_2$ are refinements of $P$, but they are still somehow less than $\epsilon$; will that not make their difference larger? That is, $$U(f,P_i)-L(f,P_i)\geqslant U(f,P)-L(f,P),$$ for $i=1,2$? Thanks in advance!

  • 1
    $\begingroup$ May help to draw a picture. $\endgroup$ – john w. Apr 19 '12 at 14:35

(Too long for a comment.)

That the answer has been spelt out, I wanted to clarify this thing about refinements:

  1. Neither of $P_i$ is a refinement of $P$.

Yes. You're right. In fact, the containment goes this way: $P_i \subsetneq P$. So, your contention is right.

  • You claim this inequality: $$U(f,P_i)-L(f,P_i) \geqslant U(f,P)-L(f,P) \tag{$\ast$}$$

    I think the reason why you think this is true is because of the same old containment I mention there. But, that simply does not mean that $P$ is a refinement of $P_i$. First of all, $P_i$ is a partition of $[a,c]$ if $i=1$ and $[c,b]$ when $i=2$.

So, $(\ast)$ fails for a nice reason. $P$ is bigger but not a finer partition than $P_i$'s.

To see that $(\ast)$ really fails, just note that those that appear in the sum with the partition $P_i$ also appear in $P$. And, that the terms involved are non-negative gives you the inequality reversed. As one of the answers here point out, $P=P_1 \cup P_2$ also, tells you that inequality is actually reversed.

I hope this helps.

  • $\begingroup$ In ($\ast$) both, lhs and rhs are sums of no negative terms. Some of the terms of the sum of the lhs already are in the sum in the rhs. $P$ have more points that $P_i$, so the sum in the rhs, "contains" the sum on the lhs on possibly some other terms, so ($\ast$) fails. I think is worth say that $P_1\cup P_2$ is a partition of $P$. $\endgroup$ – leo Apr 19 '12 at 18:25
  • $\begingroup$ @leo I agree. I thought this was there in the other answer too. But, I'll edit to add this... Thank you for your comments. $\endgroup$ – user21436 Apr 19 '12 at 18:27

Think about how $U(f,P_1)-L(f,P_1)$ relates to $U(f,P)-L(f,P)$. Suppose $P_1=\{ t_0,t_1,\ldots,t_n \}$ and $P = \{ t_0, t_1, \ldots, t_n,\ldots,t_{n+k}\}$. Then we have

$$ U(f,P_1)-L(f,P_1) = \sum_{i=1}^n (\sup\limits_{[t_{i-1},t_i]}(f)-\inf\limits_{[t_{i-1},t_i]}(f))|t_i-t_{i-1}| \leq \sum_{i=1}^{n+k} (\sup\limits_{[t_{i-1},t_i]}(f)-\inf\limits_{[t_{i-1},t_i]}(f))|t_i-t_{i-1}| = U(f,P)-L(f,P) < \epsilon $$ where we used the fact that $\sup(f) - \inf(f)$ is nonnegative on each partition interval.


i look your question, and i'm thinking the following. By Definition, with the partition $P$ you have: $ U(f,P) - L(f,P) <\epsilon$ so the difference will be near zero, because we say in terms inf and sup for sums Riemman, and $U(f,P)\geq L(f,P)$.

So, If $P_1=P\cap [a,c]$ and $P_2 = P \cap [c,b]$ should be $P = P1 \cup P2$ and $P$ was you partition for $[a,b]$, then $$ \begin{align*} L(f,P_1) + L(f,P_2)) &\leq U(f,P_1)+U(f,P_2)\\ L(f,P_i) &\leq U(f,P_i)\\ &\text{well you get by subtracting }\\ L(f,P) - L(f,P_i) &\leq U(f,P) -U(f,P_i)\\ U(f,P_i)−L(f,P_i) &\leq U(f,P)−L(f,P) \end{align*} $$

Can not be, $U(f,P_i)−L(f,P_i)\geq U(f,P)−L(f,P)$.


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