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To go from the definition of the Riemann integral ($f$ is Riemann integrable on $[a,b]$ if there exists a real $A$ such that $\forall \epsilon >0, \exists \delta>0$ such that $\forall D=\{([a_i,a_{i+1}],x_i)\}$ tagged partition of $[a,b]$ $h_i=a_{i+1}-a_i < \delta \implies |S_D(f) - A| < \epsilon$, where $S_D(f)$ is the Riemann sum on $D$) to the definition of the KH integral, all we do is replace the constant $\delta$ by a ("variable") gauge function $\delta(x_i)$.

However, the definition of the Riemann integral allows arbitrary tagged partitions, as long as $h_i<\delta$. Isn't that as flexible as the definition of the KH integral ($h_i < \delta(x_i)$)? For a given $\epsilon$, we can first find a tagged partition that will make the Riemann sums converge, then take $\delta = max({h_i})$ and satisfy the Riemann definition? In other words, any tagged partition that is allowed in the definition of the KH integral also seems to be allowed in the definition of the Riemann integral, and making the $\delta$ variable doesn't seem to be "essential". Of course, I am missing something. What am I missing?

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  • $\begingroup$ That KH allows to vary the jauge, that is, to replace a fixed level $\delta$ by a function $\delta:[a,b]\to(0,+\infty)$. $\endgroup$ – Did Aug 20 '17 at 23:02
  • $\begingroup$ @Did - yes I see that. How does it matter, since in any case, arbitrary tagged partitions are already allowed in the Riemann integral definition (in particular, tagged partitions with variable steps are allowed already)? $\endgroup$ – Frank Aug 20 '17 at 23:04
  • $\begingroup$ It does matter: Riemann asks more since it allows more partitions. $\endgroup$ – Did Aug 20 '17 at 23:06
  • $\begingroup$ I certainly don't doubt it matters - I'm trying to understand how. I am not understanding how "more" partitions would be allowed on any one side, since I am only required to exhibit some $\delta$, I can choose $\delta$, so I could choose any partition I want first, then "retrieve" a suitable $\delta$ or a $\delta(x_i)$, in both cases. In fact, the more I think about it, the less I see the actual difference (again, I'm sure there is of course one). I'm about to really learn something here :-) $\endgroup$ – Frank Aug 20 '17 at 23:09
  • $\begingroup$ My advice: Show in details that $f(x)=1/\sqrt{x}$ is KH-integrable on $(0,1)$. $\endgroup$ – Did Aug 20 '17 at 23:12

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