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Let $I=[a,b]\subset\mathbb{R}$ be a closed and bounded intervall and let $f:I\to\mathbb{R}$ be a bounded function. For a subdivision $\sigma$ of $I$ (i.e. a finite strictly increasing sequence $(x_0,...,x_N)$ with $x_0=a$ and $x_N=b$) we define $$ \overline{S}_{\sigma}(f)=\sum_{i=1}^{N}(x_i-x_{i-1})\sup_{x\in[x_{i-1},x_i]}f(x)\\ \underline{S}_{\sigma}(f)=\sum_{i=1}^{N}(x_i-x_{i-1})\inf_{x\in[x_{i-1},x_i]}f(x) $$ and thereby $$ \overline{S}(f)=\inf\{\overline{S}_\sigma(f)\ :\ \sigma\text{ subdivision of } I\}\\ \underline{S}(f)=\sup\{\underline{S}_\sigma(f)\ :\ \sigma\text{ subdivision of } I\} $$ Using this, we define the space of Riemann integrable functions over $I$, denoted by $\mathcal{R}(I)$, as $$ \mathcal{R}(I)=\{f:I\to\mathbb{R}\ :\ f \text{ bounded},\ \overline{S}(f)=\underline{S}(f)\} $$ and for $f\in\mathcal{R}(I)$ $$ \int_I f=\overline{S}(f). $$ I have three questions:

(1) If $\|\cdot\|$ denotes the uniform function norm over $I$ (i.e. $\|f\|=\sup_{x\in I} f(x)$), then is it true that $(\mathcal{R}(I),\|\cdot\|)$ is complete?

(2) Is it equivalent to the above definition when we consider only the uniform subdivisions $\tau_{N}=(x_0,...,x_N)$ with $x_i=a+i\frac{b-a}{N}$? More precisely, is it always true that $$ \overline{S}(f)=\inf\{\overline{S}_{\tau_N}(f)\ :\ N\in\mathbb{N}\}\\ \underline{S}(f)=\sup\{\underline{S}_{\tau_N}(f)\ :\ N\in\mathbb{N}\}? $$

(3) For a subdivision $\sigma=(x_0,...,x_N)$ of $I$ we define $\Delta(\sigma)=\max_{i=1}^{N}x_i-x_{i-1}$. Is it true that for a Riemann integrable function $f$ and a sequence of subdivisions $\{\sigma_n\}_{n\in\mathbb{N}}$ with $\lim_{n\to\infty}\Delta(\sigma_n)=0$ we have $$ \int_I f=\lim_{n\to\infty}\overline{S}_{\sigma_n}(f)=\lim_{n\to\infty}\underline{S}_{\sigma_n}(f)? $$ This is true if $f$ is continuous (note that $C^0(I)\subset\mathcal{R}(I)$), however I fail to see if it is true for non-continuous functions.

How to prove the answers to the three questions? Questions (2) and (3) are somewhat related, although not completely equivalent. If you think it would be better to create one post for one question each, then leave a comment and I will change it.

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  • $\begingroup$ (1) is true, but I need to use the fact that a function is Riemann integrable iff it is bounded and the set of discontinuities has measure zero. In that case, the $f_n$ are continuous on a set of full measure, hence they converge to a $f$ that is continuous on the same set, hence it is Riemann integrable. $\endgroup$ – copper.hat Dec 15 '16 at 18:46
  • $\begingroup$ Ideally these questions should have been discussed in the textbook while discussing the theory of Riemann integration, but unfortunately some books avoid these. Apostol's Mathematical Analysis handles these in good detail. $\endgroup$ – Paramanand Singh Dec 16 '16 at 7:41
  • $\begingroup$ @Paramanad Singh: I couldn't agree more; I learnt the basics of Riemann Integration one year ago in my first semester at university, but it is only now that I start to think more thoroughly about it such that I realized that we didn't discuss these natural question (or only parts of them). That's why I asked them here; I will check out the book you've mentioned, thanks. $\endgroup$ – Redundant Aunt Dec 16 '16 at 8:01
  • $\begingroup$ You can also have a look at my blog post paramanands.blogspot.com/2012/07/… which is based on material presented in Apostol's book. $\endgroup$ – Paramanand Singh Dec 16 '16 at 8:22
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Part (1)

Any sequence of functions $f_n \in (\mathcal{R}(I),\|\cdot\|) $ that is a Cauchy sequence with respect to the uniform norm is uniformly convergent to some function $f:I \to \mathbb{R}.$

It follows that $f$ is bounded. To see this note that for all $x \in I$ $$|f(x)| \leqslant |f_n(x) - f(x)| + |f_n(x)|,$$

and

$$\|f(x)\| = \sup_{x \in I}|f(x)| \leqslant \sup_{x \in I}|f_n(x) - f(x)| + \sup_{x \in I}|f_n(x)| = \|f_n - f\| + \|f_n\|,$$

By uniform convergence, there exists $N$ such that $\|f_N-f\| < 1$ and because each $f_n$ is bounded there exists $M$ such that $\|f_N\| < M$. Hence,

$$\|f(x)\| \leqslant \|f_N - f\| + \|f_N\| < 1 +M.$$

Furthermore, for any $\epsilon > 0$ there exists $N$ such that $=\|f_N - f\| \leqslant \hat{\epsilon} =\epsilon/(2(b-a))$ and for all $x \in I$

$$f_N(x) - \hat{\epsilon} \leqslant f(x) \leqslant f_N(x) - \hat{\epsilon}. $$

Since $f_N$ is integrable, we have

$$\int_a^bf_N(x) \,dx - \hat{\epsilon}(b-a) \leqslant \underline\int_a^bf(x) \, dx \leqslant \overline\int_a^bf(x) \, dx \leqslant \int_a^bf_N(x) \,dx + \hat{\epsilon}(b-a).$$

Thus

$$ 0 \leqslant \overline\int_a^b f(x) \,dx - \underline\int_a^b f(x) \,dx \leqslant 2(b-a) \hat{\epsilon} = \epsilon.$$

Since $\epsilon$ can be arbitrarily close to $0$ it follows that the upper and lower integrals are equal and $f \in \mathcal{R}(I)$. Therefore, the space is complete.

Part (2)

The collection of uniform partitions is a subset of the collection of all partitions.

Hence,

$$\inf_{\sigma} \overline S_{\sigma}(f) \leqslant \inf\{\overline{S}_{\tau_N}(f)\ :\ N\in\mathbb{N}\}.$$

A similar argument applies to the lower sums.

Proving equality requires more work. See comments by Paramanand.

Part (3)

It can be shown that the definition of Riemann integrability in terms of equality of upper and lower Darboux integrals is equivalent to

$$\lim_{\Delta(\sigma) \to 0}S_{\sigma}(f) = \int_I f,$$

where $S_\sigma(f)$ is any Riemann sum with respect to the partition and any choice of intermediate points. In particular, this includes lower and upper sums.

For a proof see here.

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    $\begingroup$ Just one point, how do we guarantee that given any non uniform partition there is a refinement which is uniform? Uniform partition has rational points and a non uniform partition may have irrational points, assuming the interval under consideration is $[0,1]$. $\endgroup$ – Paramanand Singh Dec 16 '16 at 7:56
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    $\begingroup$ I think part 2 of the problem is actually a subset of part 3 and can be handled directly in the manner given in math.stackexchange.com/a/2047959/72031 $\endgroup$ – Paramanand Singh Dec 16 '16 at 8:03
  • $\begingroup$ @ParamanandSingh: I was just trying to see if that might work. Thanks for identifying the flaw. I'm going to edit accordingly. If you have a complete answer to part (2) you should post it - upvote from me. $\endgroup$ – RRL Dec 16 '16 at 8:07
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    $\begingroup$ Discussion of such nuance is generally missing in textbooks as far as I know. In fact, proof of the equivalence between the Darboux integral and the limit of Riemann sums with diminishing partition norm seems to be not well known. $\endgroup$ – RRL Dec 16 '16 at 8:16
  • $\begingroup$ In the linked answer I have shown that infimum of upper sums is also the limit of upper sums where limit is of two types, one via refinement of partitions, another via decreasing norm to zero. The case of uniform partitions belongs to the limit when norm of partition tends to zero. And yes I agree that such subtle topics are not dealt with properly in many textbooks. Maybe a first course in real analysis does not need so much in depth analysis. $\endgroup$ – Paramanand Singh Dec 16 '16 at 8:44

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