I'm practicing my $\delta$-$\epsilon$ proofs by verifying that the Riemann and Darboux integrals are equivalent for functions on a closed interval. Here are the definitions I'm working with (taken from Tao's Introduction to Measure Theory) for reference:
Riemann Integral: Let $[a,b]$ be an interval of positive length, and $f : [a,b] \to \mathbb{R}$ be a function. A tagged partition $\mathcal{P} = ((x_0,\ldots,x_n), (x_1^*,\ldots,x_n^*))$ of $[a,b]$ is a finite sequence of real numbers $a = x_0 < x_1 < \cdots < x_n = b$, together with additional numbers $x_{i-1} \leq x_i^* \leq x_i$ for each $i=1,\ldots,n$. We abbreviate $x_i - x_{i-1}$ as $\delta x_i$. The quantity $\Delta(\mathcal{P}) := \sup_{1 \leq i \leq n}\delta x_i$ will be called the norm of the tagged partition. The Riemann sum $\mathcal{R}(f,\mathcal{P})$ of $f$ with respect to the tagged partition $\mathcal{P}$ is defined as $$ \mathcal{R}(f,\mathcal{P}) := \sum_{i=1}^n f(x_i^*)\delta x_i. $$ We say that $f$ is Riemann-integrable on $[a,b]$ if there exists a real number, denoted $\int_a^b f(x)\,dx$ and referred to as the Riemann integral of $f$ on $[a,b]$, for which we have $$ \int_a^b f(x)\,dx = \lim_{\Delta(\mathcal{P}) \to 0} \mathcal{R}(f,\mathcal{P}) $$ by which we mean that for every $\epsilon > 0$ there exists $\delta > 0$ such that $|\mathcal{R}(f,\mathcal{P}) - \int_a^bf(x)\,dx| \leq \epsilon$ for every tagged partition $\mathcal{P}$ with $\Delta(\mathcal{P}) \leq \delta$.
Darboux Integral: For a bounded function $f : [a,b] \to \mathbb{R}$, we define the lower and upper Darboux integrals by $$ \underline{\int_a^b}f(x)\,dx := \sup_{g\leq f, \text{ piecewise constant}}\int_a^b g(x)\,dx $$ and $$ \overline{\int_a^b}f(x)\,dx := \inf_{h\geq f, \text{ piecewise constant}}\int_a^b h(x)\,dx, $$ where we assume integrals of piecewise constant functions have already been defined. If the two integrals are equal, we say that $f$ is Darboux integrable and the common value is called the Darboux integral of $f$ on $[a,b]$.
I think I've proven one direction of the equivalence between the two integrals:
Proposition. If $f : [a,b] \to \mathbb{R}$ is a bounded Riemann integrable function, then $f$ is Darboux integrable and the integrals are equal.
Is the following proof correct?
Proof. Let $R$, $D_*$ and $D^*$ denote the Riemann, lower Darboux, and upper Darboux integrals of $f$ respectively. Fix $\epsilon > 0$ and pick $\delta > 0$ such that $|\mathcal{R}(f,\mathcal{P}) - R| \leq \epsilon/2$ whenever $\Delta(\mathcal{P}) \leq \delta$. Let $a = x_0 < \cdots < x_n = b$ be any (untagged) partition of norm less than or equal to $\delta$. For $i = 1,\ldots,n-1$, let $I_i = [x_{i-1},x_i)$ and let $I_n = [x_{n-1},x_n]$. For each $i = 1,\ldots,n$, let $$ c_i := \sup_{t \in I_i}f(t), $$ and let $g$ be the piecewise constant function taking value $c_i$ on $I_i$ for each $i$.
Introduce tags for the partition by picking $x_i^*$ for each $i = 1,\ldots,n$ such that $$ f(x_i^*) \geq c_i - \frac{\epsilon}{2(b-a)}. $$ Call the tagged partition $\mathcal{P}$. We then have \begin{align*} \int_a^b g(x)\,dx &= \sum_{i=1}^n c_i |I_i| \\ &\leq \sum_{i=1}^n \left(f(x_i^*) + \frac{\epsilon}{2(b-a)}\right)|I_i| \\ &= \left(\sum_{i=1}^nf(x_i^*)\,\delta x_i\right) + \frac{\epsilon}{2} \\ &= \mathcal{R}(f,\mathcal{P}) + \frac{\epsilon}{2}. \end{align*}
Since $|\mathcal{R}(f,\mathcal{P}) - R| \leq \epsilon/2$, we then have $$ D^* \leq \int_a^b g(x)\,dx \leq \mathcal{R}(f, \mathcal{P}) + \frac{\epsilon}{2} \leq R + \epsilon. $$
Since $\epsilon$ was arbitrary, it follows that $D^* \leq R$. A completely analogous argument gives us $R \leq D_*$, so we have $R \leq D_* \leq D^* \leq R$, which forces equalities all the way through.
