2
$\begingroup$

Prove the converse of the Cauchy criterion for integration. That is, prove that if $f$ is integrable on $[a,b]$, then for any $\epsilon>0$ there is a $\delta>0$ so that for any partitions $P_1,P_2$ with $\left \|P_1 \right \|< \delta$, $\left \|P_2 \right \|< \delta$, and $S_1$, $S_2$ sample points from $P_1,P_2$.

My work: Since $f$ is integrable on $[a,b]$ let $I=\int_{a}^{b}f$ i.e. $\forall \epsilon>0$ there is a $\delta_1>0$ so that for any partition $P_1$ with $\left \|P_1 \right \|< \delta$ and $S_1$ sample points from $P_1$ we have, $|RS(f,P_1,S_1) - I|<\frac{\epsilon }{2}$.

Similarly, $\forall \epsilon>0$ there is a $\delta_2>0$ so that for any partition $P_2$ with $\left \|P_2 \right \|< \delta$ and $S_2$ sample points from $P_2$ we have, $|RS(f,P_2,S_2) - I|<\frac{\epsilon }{2}$.

Let $\delta=\min (\delta_1,\delta_2)$. Then when $\left \|P_1 \right \|< \delta$ and $\left \|P_2 \right \|< \delta$, we have

$|RS(f,P_1,S_1) - I+I-RS(f,P_2,S_2)|\leq|RS(f,P_1,S_1) - I|+|RS(f,P_2,S_2)-I|<\frac{\epsilon }{2}+\frac{\epsilon }{2} = \epsilon$

So if $f$ is integrable then $RS(f,P,S)$ is Cauchy.

Does this look correct?

$\endgroup$
  • $\begingroup$ Yes, it looks correct. But the first paragraph of your question looks incomplete. $\endgroup$ – Daniel Fischer Nov 24 '13 at 18:27
  • $\begingroup$ What is the actual claim about $S_1,S_2$ (someting's mising)? And are you aware that there is no need to distinguish $\delta_1$ vs. $\delta_2$ (they are both "the" $\delta$ belonging to $\frac\epsilon2$)? Apart from that this looks fine $\endgroup$ – Hagen von Eitzen Nov 24 '13 at 18:29
  • $\begingroup$ $S_1$ and $S_2$ are just sample points from the partitions $P_1$ and $P_2$. $\endgroup$ – user87274 Nov 24 '13 at 23:05

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy