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.
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.