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I am studying real analysis and specifically the Riemann integral using the book

Foundations and Elementary Real Analysis
D. J. H. Garling, University of Cambridge

There is a Proposition (Prop 8.3.7 on page 217) that confuses me.

Prop 8.3.7: Suppose that $(D_r)_{r=1}^\infty$ is a sequence of dissections of $[a,b]$, and that $\delta(D_r)\rightarrow 0$ as $r\rightarrow \infty$. If $f$ is a bounded function on $[a,b]$, then $f$ is Riemann integrable if and only if $S_{D_r}-s_{D_r}\rightarrow 0$ as $r\rightarrow \infty$. If so, then $\int_a^b f(x)dx = \lim_{r\rightarrow \infty}S_{D_r} = \lim_{r\rightarrow \infty} s_{D_r}$.

Some notation: Here $\delta(D_r)$ is the mesh-size of the dissection $D_r$, i.e. the maximum length between points of $D_r$. Further $S_{D_r}$ and $s_{D_r}$ are the upper- and respectively lower-Riemann sums corresponding to the dissection $D_r$.

My question: Why do we need the mesh-size $\delta(D_r)$ to vanish? There is another Proposition (Prop 8.3.3 p. 215) which seems to conflict with the idea that this would be necessary.

Prop 8.3.3: Suppose that $f$ is a bounded function on $[a,b]$. Then $f$ is Riemann integrable if and only if given $\epsilon >0$ there exists a dissection $D$ such that $S_D - s_D < \epsilon$.

In the second Proposition it seems we don't need to restrict the mesh-size of the dissection. What am I overlooking here?

typedrums
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  • I don't think this question has much to do with Riemann integral, rather with the idea of limit. – lcv Sep 25 '19 at 10:14
  • @lcv Do you mean I should change the tags? – typedrums Sep 25 '19 at 10:46
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    typedrums, I'm pretty sure that you can prove that if the difference between the upper and lower Darboux sums goes to zero, then the integral is the limit of those sums. (See https://en.wikipedia.org/wiki/Riemann_integral#Riemann_sums ) However, you cannot prove the converse without assuming that the mesh size goes to zero. – irchans Sep 25 '19 at 11:58
  • Prop 8.3.3 is much easier to prove than prop 8.3.7. However there should be an intermediate proposition say prop 8.3.x which says "a bounded function $f:[a, b] \to\mathbb {R} $ is Riemann integrable on $[a, b] $ if and only if for every $\epsilon >0$ there is a $\delta>0$ such that $S_D-s_D<\epsilon $ for all dissections $D$ of $[a, b] $ with mesh less than $\delta$." The equivalence of 8.3.3 and 8.3.x is handled in this answer. – Paramanand Singh Sep 25 '19 at 16:28
  • I simply mean that when we write $\lim_{n\to \infty} a_n = L$ we say that $|a_n -L| \to 0$ but it means more precisely that for all $\epsilon>0$ there exist an $N$ such that $n >N$ implies $ |a_n -L|<\epsilon$. – lcv Sep 26 '19 at 13:55
  • ... which looks very similar to what you wrote 'mutatis mutandis' – lcv Sep 26 '19 at 13:58

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Thanks to @irchans comment I see the answer now.

The point is that we can take a "very pathological" sequence of dissections $(D_r)_r$ such that the differences of the upper and lower Riemann/Darboux sums don't tend to zero as $r\rightarrow \infty$ even if $f$ is Riemann integrable.

But if we have a sequence of dissections for which the differences of the upper and lower Riemann/Darboux sums do tend to zero as $r\rightarrow \infty$, then $f$ is Riemann integrable without having to require that the mesh-size tends to zero.

typedrums
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  • Your last paragraph is correct. At the same time if we have a sequence $D_r$ with $\delta(D_r) $ not tending to $0$ then one can't ensure that $S_{D_r} - s_{D_r} \to 0$. – Paramanand Singh Sep 27 '19 at 03:02