What does Infimum of Upper Sum and Supremum of Lower Sums mean? I'm trying to figure out the Darboux Integral definition as it states: 
$f$ is integrable if $inf${U($f$,$P$)} = $sup${L($f$,$P$)}, where U($f$,$P$) is the upper sum, L($f$,$P$) is the lower sum of $f$ and $P$ is a parition of $f$. 
I'm not understanding what the $sup$/$inf$ of the sums mean. When calculating, they are a finite value (ie. L($f$,$P$) = $\sum_{i=1}^n m_i (x_i - x_{i-1})$) 

Consider: 

$$f(x) = 2x, x \in [0,1];~ P = \{0,\frac{1}{4},\frac{1}{2},1\}.$$

Since $P$ has 4 elements, n=3, thus 3 subintervals of $[0,1]$, can you definte these 3 subintervals however you want as long as they range [0,1]? 
ie. $$[0,\tfrac{1}{4}] \,\cup\, [\tfrac{1}{4},\tfrac{1}{2}] \,\cup\,[\tfrac{1}{2},1] \text{ or } [0,\tfrac{1}{3}] \,\cup\, [\tfrac{1}{3},\tfrac{2}{3}] \,\cup\,[\tfrac{2}{3},1] $$ 
So to calculate $U(f,P)$ for the first set of subintervals: 
$$U(f,P) = \sum_{i=1}^n M_i (x_i - x_{i-1})$$
$$ = f(\frac{1}{4})(\frac{1}{4} - 0)\;+ f(\frac{1}{2})(\frac{1}{2} - \frac{1}{4})\;+  f(1)(1 - \frac{1}{2})$$
$$ = \frac{1}{8} + \frac{1}{4} + 1= \frac{11}{8}$$
Calculate $L(f,P)$:
$$L(f,P) = \sum_{i=1}^n m_i (x_i - x_{i-1})$$
$$ = f(0)(\frac{1}{4} - 0)\;+ f(\frac{1}{4})(\frac{1}{2} - \frac{1}{4})\;+  f(\frac{1}{2})(1 - \frac{1}{2})$$
$$ = 0 + \frac{1}{8} + \frac{1}{2}= \frac{5}{8}$$

First off, can someone confirm that my calculations for upper and lower sums are correct? 
Secondly, back to the main question, what is the $inf${U($f$,$P$)} and $sup${L($f$,$P$)} in this? as my $U$($f$,$P$) = $\frac{11}{8}$ and $L$($f$,$P$) = $\frac{5}{8}$. 
As I already know that $f$ is integrable, $U$($f$,$P$) = $L$($f$,$P$) only if $f$ is constant, but what is the set in which I'm supposed to take the $inf$ and $sup$ of? As per the Darboux Integral definition, $sup${$L(f,P)$} = $inf${$U(f,P)$} for this function. 
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If someone could clear this up for me it would be greatly appreciated
 A: Your calculations are fine and that's a good start. The idea of inf/sup of these upper and lower sums is easy to grasp in theory (but not in practice). Thus the value $\inf U(f, P) $ requires you to evaluate all the upper sums for each partition $P$ and then take infimum of all these sums. You can see that it is practically not possible to evaluate these sums for all partitions $P$ (because there are infinitely many partitions of a given interval). The practical method of finding the infimum here is based on the following deep and difficult theorem:

Theorem: Let $f:[a, b] \to\mathbb{R} $ be a bounded function and let $$A= \{U(f, P) \mid P\text{ is a partition of }[a, b] \}, I=\inf\, A$$ then $$I=\lim_{|P|\to 0}U(f,P)$$ where $|P|$ denotes the length of largest subinterval created by $P$.

Now one can take a uniform partition with $n$ subintervals of equal length and evaluate the upper sum over this partition and then take limit as $n\to\infty$ to get the desired infimum.
In your case let $P=\{x_0,x_1,\dots,x_n\}$ where $x_i=i/n$ and then we have $$U(f, P) =\sum_{i=1}^{n}M_i(x_i-x_{i-1})=\sum_{i=1}^{n}\frac{2i}{n}\cdot\frac{1}{n}=\frac{n(n+1)}{n^2}$$ and thus when we take limit as $n\to\infty$ we get the desired infimum as $1$.
Similarly $$L(f, P) =\sum_{i=1}^{n}\frac{2(i-1)}{n}\cdot\frac{1}{n}=\frac{n(n-1)}{n^2}$$ and hence on taking limit the supremum of lower sums is also $1$ and therefore the function $f(x) =2x$ is Riemann integrable on $[0,1]$ with integral $1$.
A: Take infimum (greatest common lower bound) /supremum (lowest common upper bound)  of all the arising values while $P$ ranges over all partitions. 
Your calculations are correct and they show
$$\frac58\ \le\ \int_0^12x\, dx=1\ \le\ \frac{11}8$$
