# Lower sum of partition and upper sum of another partition

Given $P_1$, $P_2$ partitions of [a, b], $$s(P_1, f) \leq S(P_2, f)$$ where f is the function, and $$s(P, f) = \sum\limits_{k=1}^n(P_k - P_{k-1})m_k\text{ (lower sum)}$$ $$S(P, f) = \sum\limits_{k=1}^n(P_k - P_{k-1})M_k\text{ (upper sum)}$$ with $m_k = \inf\limits_{[P_{k-1}, P_k]}f$

and $M_k = \sup\limits_{[P_{k-1}, P_k]}f$

I can't understand this theorem, how can it be possible for every partition? I think it should be $$P_1 \subseteq P_2$$ Or something like that...

"sandwich" another Partition $P_3:=P_1\cup P_2$ in between:
(By that notation I mean the coarsest partition being finer than $P_1$ and $P_2$)
Then by definition $s(P_3,f)\leq S(P_3,f)$ and you easily see that $s(P_1,f)\leq s(P_3,f)$ and $S(P_3,f)\leq S(P_2,f)$.
• But if you have something like this: math.feld.cvut.cz/mt/txtd/1/gifa1/pc3da1ai.gif and $P_1$ = "the third partition", $P_2$ = "the first partition": it doesn't work. The lower sum of the third partition is greater than the upper sum of the first partition. – moonknight Jan 8 '17 at 16:50
• watching your picture I see twice the same partition, so maybe you accidently uploaded the wrong picture? anyway, if that helps you: instead of what I chose you might as well choose any partition as $P_3$ that is a refinement for both $P_1$ and $P_2$ simultaneously. – Max Jan 8 '17 at 16:55
• I meant $P_2$ = $xx_1$ and $P_1$ = $x_2x_3$. Maybe I might have the wrong concept of partitions? – moonknight Jan 8 '17 at 17:07
• both partitions must have the same end points on the very left and very right (the boundaries of your integration interval) so with your notation in that case you can only have $P_2=axx_1b$ and $P_1=ax_2x_3b$. the coarsest partition being finer than both then would be $P_3=axx_1x_2x_3b$. – Max Jan 8 '17 at 17:12