Proving Existence of Riemann Integral via Darboux I am trying to prove the existence of the Riemann integral for continuous functions defined over the closed interval $[a,b].$ I am doing this with Darboux upper and lower integrals. To show that a function $f:[a,b]\to \mathbb{R}$ has a Riemann integral $I=\int_{a}^{b} f(x) dx,$ we must show for every $\varepsilon > 0$ there exists a $\delta >0$ such that any partition $P$ that has partition norm $||\Delta x|| < \delta$ will satisfy
$|\sum_{i=1}^{n} f(c_i) \Delta x_i - I| < \varepsilon ,$ where $c_i$ is any element in $[x_{i-1},x_{i}]$ and $n$ is the number of intervals resulting from the partition. 
We have a lemma establishing that for any partitions $P$ and $Q$ of $[a,b]$ with $Q \subset P,$ $L(f,Q) \leq L(f,P) \leq I_L \leq I_U \leq U(f,P) \leq U(f,Q)$ where $U$ and $L$ denote the upper and lower sums and $I_L, I_U$ denote the upper and lower integrals, with $I_U = \inf\limits_{P} U(f,P)$ where $P$ runs over all partitions of $[a,b].$ 
Assuming that $f$ is continuous, I have shown that $I_U$ exists and this is my candidate for what the Riemann integral $I$ of $f$ should be. I know that $f$ is uniformly continuous over its compact domain so given $\varepsilon >0$ I can choose $\delta > 0$ so that $|x-y| < \delta$ implies $|f(x) - f(y)| <\frac{\varepsilon}{3(b-a)}$. Let $P$ be any partition of $[a,b]$ with norm $||\Delta x|| < \delta$. Then by the triangle inequality
\begin{align*}
| \sum_{i=1}^n f(c_i) \Delta x_i - I_U|
& \leq
| \sum_{i=1}^n f(c_i) \Delta x_i - U(f,P) | + | U(f,P) - I_U |
&\\=
|\sum_{i=1}^n (f(c_i) - M_i) \Delta x_i| + | U(f,P) - I_U |,
\end{align*}
where $M_i$ is the maximum value that $f$ attains over $[x_{i-1},x_i].$ By our choice of $P$ and $\delta$ we are guaranteed that $|f(c_i) - M_i| < \frac{\varepsilon}{3(b-a)}$ so the sum involving this term is less than $\varepsilon /3.$ For the remaining term we have by the infimal nature of $I_U$ that there exists a partition $P_2$ such that $|I_U - U(f,P_2)| < \varepsilon/3,$ and this must be true of $Q= P\cup P_2$ as well by the inequalities in our lemma. Hence we can apply the triangle inequality again to conclude that
$$
 | U(f,P) - I_U |
\leq 
|U(f,Q) - I_U| + |U(f,P) - U(f,Q)|,
$$
where we have said already that $|U(f,Q) - I_U| <\varepsilon/3.$ Now here is where I am stuck. I want to be able to conclude that the remaining term, $|U(f,P) - U(f,Q)|,$ is or can be made arbitrarily small. My issue is that these upper sums are over different partitions so they have differing numbers of terms and differing lengths of each subinterval. 
My question is does what I have so far look correct and is it true that this final term I have can be made arbitrarily small?
 A: Any proof of Riemann integrability is essentially a demonstration that upper and lower Darboux sums can be made arbitrarily close.  This argument is constructive since the upper and lower Darboux integrals must be equal and provide the value of the integral. 
You have chosen to look only at the upper integral as a candidate for the integral.
One minor point -- you state, "Assuming that $f$ is continuous, I have shown that $I_U$ exists ..." .  Note that you need only boundedness to conclude that the upper integral exists.  
Your argument using the uniform continuity of $f$ on the compact interval $[a,b]$ is correct, leading to
$$ |\sum_{i=1}^n f(c_i) \Delta x_i - I_U|
 \leq \frac{2 \epsilon}{3}+ | U(f,P) - U(f,Q) |.$$
The partition $P_2$ was chosen such that $I_U \leqslant U(f,P_2) < I_U + \epsilon/3$.  Since $Q = P \cup P_2$ is a refinement of $P_2$, we also have 
$$I_U \leqslant U(f,Q) \leqslant U(f,P_2) < I_U + \epsilon/3 \implies |U(f,Q) - I_U| < \epsilon/3.$$
The partition $P$ was chosen with sufficiently small mesh such that $U(f,P)$ is within $\epsilon/3$ of any tagged Riemman sum, that is $S(f,P) \leqslant U(f,P) \leqslant S(f,P) + \epsilon/3$. At this point, you also have that $I_U \leqslant U(f,Q) < I_U + \epsilon/3$.  However, you can't immediately conclude that $|U(f,P) - U(f,Q)| < \epsilon/3$ , since you have shown nothing about the distance of $S(f,P)$ from $I_U$.
You just need to extend the consequence of the sufficiently small mesh  to show that $U(f,P)-L(f,P) < \epsilon/3.$
Then we have
$$L(f,P) \leqslant I_L \leqslant I_U \leqslant U(f,P)  \implies I_U \leqslant U(f,P) < I_U + \epsilon/3$$
Since $Q$ is also a refinement of $P$, we also have 
$$I_U \leqslant U(f,Q) \leqslant U(f,P) < I_U + \epsilon/3,$$
and you can conclude that  $|U(f,P) - U(f,Q)| < \epsilon/3$ and finish.
Alernate Proof
If $f$ is merely bounded, then it is true that the upper and lower Darboux integrals exist, and, in terms of upper and lower Darboux sums, they satisfy:
$$\overline{\int}_a^bf(x) \, dx= \inf_{P}U(P,f),\\ \underline{\int}_a^bf(x) \, dx= \sup_{P}L(P,f),\\\underline{\int}_a^bf(x) \, dx \leqslant \overline{\int}_a^bf(x) \, dx .$$
Assuming that $f$ is continuous on $[a,b]$, then it follows that $f$ is uniformly continuous. For any $\epsilon > 0$ there exists $\delta(\epsilon) > 0$ such that $|x-y| < \delta(\epsilon)$ implies $|f(x)-f(y)| < \epsilon$.
Suppose $P_{\epsilon}$ has $||P_{\epsilon}|| < \delta(\epsilon/(b-a))$ where $\delta$ arises in the definition of uniform continuity.
We have 
$$U(P_{\epsilon},f) - L(P_{\epsilon},f) = \sum_{j=1}^n(M_j-m_j)(x_j-x_{j-1}),$$
where 
$$M_j = \sup\{f(x): x \in [x_{j-1},x_j]\}, \\ m_j = \inf\{f(x): x \in [x_{j-1},x_j]\}.\\$$
Since $||P_{\epsilon}|| < \delta(\epsilon/(b-a))$, then for $x,y \in [x_{j-1},x_j]$ we have $M_j-m_j = \sup \{ |f(x)-f(y)|:x,y \in [x_{j-1},x_j] \} < \epsilon/(b-a)$, and
$$U(P_{\epsilon},f) - L(P_{\epsilon},f) < \epsilon.$$
As
$$L(P_{\epsilon},f) \leqslant \underline{\int}_a^bf(x) \, dx \leqslant \overline{\int}_a^bf(x) \, dx \leqslant U(P_{\epsilon},f),$$
it follows that for any $\epsilon > 0$,
$$\overline{\int}_a^bf(x) \, dx - \underline{\int}_a^bf(x) \, dx \leqslant \epsilon,$$
and
$$\overline{\int}_a^bf(x) \, dx = \underline{\int}_a^bf(x) \, dx  := I.$$
We have established that the lower and upper integrals are equal.  
Using this common value $I$, for any $\epsilon > 0$, if $P$ is any partition with $||P|| < \delta = \delta(\epsilon/(b-a))$ and $S(P,f)$ is any corresponding tagged Riemann sum, then as before,
$$U(P,f)-L(P,f) < \epsilon.$$
We also have,
$$L(P,f) \leqslant I \leqslant U(P,f),$$
and
$$L(P,f) \leqslant S(P,f) \leqslant U(P,f).$$
Since, $I \leqslant U(P,f)$ and $-S(P,f) \leqslant - L(P,f)$, it follows that
$$I - S(P,f) \leqslant U(P,f) - L(P,f) < \epsilon.$$
Since, $-I \leqslant -L(P,f)$ and $S(P,f) \leqslant U(P,f)$, it follows that
$$S(P,f)-I \leqslant U(P,f) - L(P,f) < \epsilon \implies -\epsilon < I - S(P,f).$$
Whence, if $||P|| < \delta$, then
$$|S(P,f) - I| < \epsilon.$$
Therefore, continuity implies that $f$ is Riemann integrable. 
