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How do I prove that a condition for (Riemann) integrability is that the difference between the upper and lower sums of the regular partition $D_n:0=a_0<a_1<...<a_n=1$ where $a_k=(k/n)$ tends to 0 as n tends to infinity? I've tried used the Riemann criterion, so for all $ε>0$ there is a dissection $D$ such that the difference between the upper and lower sums of $D$ is less than $ε$, and this is fine in the case where $D$ is composed only of rationals, but I can't figure out how to make the argument rigorous in the case that $D$ may contain irrationals. Any help would be greatly appreciated! :)

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  • $\begingroup$ Nobody can prove that a condition in some definition (which you don't give) is necessary. Formulating a definition is a free act. Some definitions are successful, others aren't. In short: You have to give more details. $\endgroup$ – Christian Blatter Apr 12 '17 at 18:13
  • $\begingroup$ Sorry I don't understand...my question is for the upper sum being the sum of the supremums of the function over the dissection, and the lower sum the infimums, with Dn being the regular dissection of the interval [0,1]. The condition for integrability is that, over all possible dissections of [0,1], the supremum of the lower sums equals the infimum of the upper sums, but the question I have is asking about regular dissections in particlar. (Sorry I've edited the wording of my question so it's clear this is about regular partitions) $\endgroup$ – user294388 Apr 12 '17 at 19:20
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First, it is a sufficient condition.

As long as $f$ is bounded on $[0,1]$, the upper and lower sums corresponding to arbitrary partitions are bounded. Let $\mathcal{P}$ denote the set of all partitions of $[0,1].$ Consequently, the sets $\{L(P,f): P \in \mathcal{P}\}$ and $\{U(P,f): P \in \mathcal{P}\}$ are bounded, and this guarantees the existence of

$$\underline{\int}_0^1 f(x) \, dx = \sup_{P \in \mathcal{P}}\, L(P,f), \\ \overline{\int}_0^1 f(x) \, dx = \inf_{P \in \mathcal{P}}\, U(P,f) , $$

which are called the lower and upper integrals.

Given any regular partition $D_n$ we have

$$L(D_n,f) \leqslant \underline{\int}_0^1 f(x) \, dx \leqslant \overline{\int}_0^1 f(x) \, dx \leqslant U(D_n,f).$$

The central inequality follows because for any partitions $P$ and $Q$ we have $L(P,f) \leqslant U(Q,f)$ (take a common refinement of the partitions to show this) and, thus $\sup_{P \in \mathcal{P}} \,L(P,f) \leqslant \inf_{Q \in \mathcal{P}} \,U(Q,f)$.

Hence,

$$0 \leqslant \overline{\int}_0^1 f(x) \, dx - \underline{\int}_0^1 f(x) \, dx \leqslant U(D_n,f) - L(D_n,f).$$

The right-hand side converges to $0$ as $n \to \infty$, by hypothesis, which implies that $f$ is integrable since we must have

$$\underline{\int}_0^1 f(x) \, dx = \overline{\int}_0^1 f(x) \, dx, $$

where the common value of lower and upper integrals is by definition the value of the integral.

To show it is a necessary condition, consider

$$\left|U(D_n,f) - L(D_n,f) \right| \leqslant \left|U(D_n,f) - \int_0^1 f(x) \, dx \right| + \left|L(D_n,f) - \int_0^1 f(x) \, dx \right|.$$

The two terms on the RHS go to zero as $n \to \infty$. This is a consequence of the equivalent condition for integrability where for arbitrary Riemann sums corresponding to tagged partitions we have

$$\tag{*}\int_0^1 f(x) \, dx = \lim_{\|P\| \to 0} S(P,f).$$

Here $\|P\| = \max_{1 \leqslant j \leqslant n} (x_j - x_{j-1})$ is the norm of the partition $P = (x_0,x_1, \ldots, x_n)$ and, clearly, $\|D_n\| \to 0$ if and only if $n \to \infty$.

It takes a bit of effort to prove the equivalence of $(*)$ to the definition of the Riemann integral in terms of partition refinement or the Darboux approach. It has been shown a number of times on this site including here.

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  • $\begingroup$ Sorry for only just responding to this but that was a really good explanation thank you!! $\endgroup$ – user294388 Apr 24 '17 at 20:29
  • $\begingroup$ You're welcome. Glad to help. $\endgroup$ – RRL Apr 25 '17 at 18:58

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