Riemann Integral Upper vs Lower Estimate. Inf vs sup? We're currently going through the Riemann integral in my real analysis course. We've defined $$\overline{\int}f \,dx = \inf\ U(P,f)$$
$$\underline{\int}f \,dx = \sup\ L(P,f)$$
Where $U(P,f)$ is the upper approximation and $L(P,f)$ is the lower approximation.
I'm not entirely sure why we're taking the $\inf$ of the upper approximation and the $\sup$ of the lower approximation. It makes sense intuitively but I've been trying to think through it geometrically and I can't quite figure it out. 
 A: Consider the following image:

The expression $U(f,P)$ approximates the area under the graph of $f$ using rectangles whose heights on each interval $[x_i,x_{i+1}]$ are given by $\sup_{x \in [x_i,x_{i+1}]} f(x)$. Hence, by definition the heights of the rectangles always exceed (or maybe equal to) the value of the function on each interval so you get an overestimate of the area (whatever it may be). As the partition gets finer and finer, you get better approximations which still generally are overestimating the area under the graph of $f$ and so you it makes sense to consider the infimum of all such approximations. In the image, you can see two such overestimates on the left which correspond to two different partitions of $[a,b]$.
Similarly, the expression $L(f,P)$ approximates the area under the graph of $f$ by using rectangles whose heights on each interval $[x_i,x_{i+1}]$ are given by $\inf_{x \in [x_i,x_{i+1}]} f(x)$. Hence, by definition the heights of the rectangles are always less than (or maybe equal to) the value of the function on each interval so you get an underestimate of the area. As the partition gets finer and finer, you get better approximations which still generally are underestimates of the area and so in this case it makes sense to consider the supremum of all such approximations. In the image, two such underestimates are calculated on the right using two different partitions.
