Understanding a Property of the Riemann Integral 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 $U(f,P)-L(f,P)&lt\epsilon$. 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
  $$U(f,P_1)-L(f,P_1)&lt\epsilon\text{ and }U(f,P_2)-L(f,P_2)&lt\epsilon,$$
  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!
 A: (Too long for a comment.)
That the answer has been spelt out, I wanted to clarify this thing about refinements:


*

*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.
A: 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
$$
U(f,P_1)-L(f,P_1) = \sum_{i=1}^n (\sup\limits_{[t_{i-1},t_i]}(f)-\inf\limits_{[t_{i-1},t_i]}(f))|t_i-t_{i-1}| 
\leq \sum_{i=1}^{n+k} (\sup\limits_{[t_{i-1},t_i]}(f)-\inf\limits_{[t_{i-1},t_i]}(f))|t_i-t_{i-1}|  = U(f,P)-L(f,P) &lt \epsilon
$$
where we used the fact that $\sup(f) - \inf(f)$ is nonnegative on each partition interval.
A: i look your question, and i'm thinking the following.
By Definition, with the partition $P$ you have:
$ U(f,P) - L(f,P) &lt\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, $U(f,P_i)−L(f,P_i)\geq U(f,P)−L(f,P)$.
