Riemann integral interpretation step by step help. A real number $I$ is the Riemann integral of $f$ over $[a,b]$ if it satisfies the following approximation condition:
$\forall \epsilon > 0$ $\exists \delta > 0$ such that if $P,T$ is any partition pair then $mesh \, P < \delta$ implies $|R-I| < \epsilon$ where $R = R(f, P, T)$.
Thank you.
 A: Perhaps we should first clear up some notation. Fix the interval $[a,b]$. A partition is a set of values $a = x_0 < x_1 < \ldots < x_{n-1} < x_{n} = b$, so you can simply think of these numbers as dividing up the interval $[a,b]$. 
The mesh size of the partition is the maximum size of the subintervals $(x_i, x_{i+1})$ formed. So if they say that the mesh size of $P$ is less than $\delta$, then each of the $x_i$'s are spaced at most a distance $\delta $ apart. 
A set of tags $T$ is a set of values $\{t_i\}$ such that $x_i \le t_i \le x_{i+1}$. In other words, each value $t_i$ sits inside each of the divided subintervals of $[a,b]$ formed by the partition $P$.
Now, we define the Riemann-Stieltjes sum by
$$ R(f,P,T) = \sum_{i=0}^{n-1} f(t_i) (x_{i+1} - x_i)$$
Up till now, all these fancy constructions really boil down to one thing:

$R(f,P,T)$ is the total area of rectangles of width $x_{i+1} - x_i$ and height $f(t_i)$.

Now, the Riemann-Stieltjes integral $I = \int_a^b f \, dx$ can now be defined as you said above: for every $\epsilon > 0$, there exists $\delta > 0$ such that for every tagged partition $(P,T)$ with mesh size $< \delta$, $|R(f,P,T) - I| < \epsilon$. But, in plain terms, all this is saying is:

The integral $I$ exists if you can approximate it to arbitrary precision by the areas of sufficiently thin rectangles with heights given by some value of $f$ inside those rectangles, such that any set of sufficiently thin rectangles works.

This should click with your intuition about integrating continuous functions, which is the area under the curve. So we basically just use the area of these rectangles to approximate the area under the curve, and the only way for this to make sense is if the approximation is valid regardless of which rectangles you pick.
