Riemann Stieltjes and Riemann sum This is not a question rather I am asking for a clarification. Ok, what are the difference between these two formulas to find $U(P,f)$ and $L(P,f)$: 
$U(P,f) = \sum_{i = 1}^{n} M_j\cdot \Delta x$ and $\sum_{i = 1}^{n} f(x_i)\cdot \Delta x, \Delta x = \frac{x_{i}-x_{i-1}}{n}$ 
$L(P,f) = \sum_{i = 1}^{n} m_j\cdot \Delta x$ and $\sum_{i = 1}^{n} f(x_i - \Delta x)\cdot \Delta x$ 
I mean I have calculated the same problem with two different formulas and got two different results. Now, I am confused when to use what. 
The problem I am talking about is this: Find $U(P,f)$ and $L(P,f)$ when $f(x) = x, x\in [-1,1], \Delta \alpha x =\Delta x,  P = \{-1,-1/2,0,1/2,1\}$. Just a side note: When $ \alpha x = x,\Delta \alpha x = \Delta x$. Can anyone please help me understand these concepts better? Thanks. 
 A: The difference between the formulae you have given lies in the definition of $m_j$ and $M_j$, that is:
$$M_j:=\sup_{x\in [x_{i-1}, x_i]} \{f(x)\} \\ 
m_j:=\inf_{x\in [x_{i-1}, x_i]} \{f(x)\}.$$
Here the definition of sumpremum and infimum of a set. The upper and lower Riemann sums are defined in terms of $M_j$ and $m_j$ (see, for example, Rudin's Principles of Mathematical Analysis). 
That is to say:
$$U(f, P, \alpha):= \sum M_j \Delta \alpha(x_j)\\
L(f, P, \alpha):= \sum m_j \Delta \alpha(x_j),$$
where $\Delta\alpha(x_j)=\alpha(x_j)-\alpha(x_{j-1})$.
To understand these definitions you need to know what a partition $P$ is and the role of the function $\alpha$. In fact, in your case, you simply set:
$$\alpha(x)=x$$
for all $x$ in the $[a, b]$ (the interval over which you want to integrate the function). Thus clearly:
$$\Delta\alpha(x_j)=\alpha(x_j)-\alpha(x_{j-1})=x_j-x_{j-1}=:\Delta x_j.$$
Finally, the Riemann-Stieltjes integral is defined as:
$$\int_a^b f(x) d\alpha (x)=\inf U(f, P, \alpha)=\sup L(f, P, \alpha),$$
where the supremum and infimum are taken over the set of all partitions of $[a,b]$. 
Notice that the equality
$$\inf U(f, P, \alpha)=\sup L(f, P, \alpha)$$
does not hold for all functions. Indeed a function is Riemann integrable if and only if this equality holds. 
Finally, to go to you case, your interval is $[-1,1]$, $\alpha(x)=x$, and your partition is given by:
$$P=\{x_0, x_1, x_2, x_3, x_4\}$$
where $x_0=-1, x_1=-\frac{1}{2}, x_2=0, x_3=\frac{1}{2}, x_4=1$.
Also, since your function is continuous on a bounded interval, you have that, for example:
$$m_1=-1,  M_1=-\frac{1}{2}.$$
Two possibly useful links: a b
A: The idea is that neither gives a perfect answer for the area "under" $f$ (technically the "signed area" under $f$, meaning the area between $f$ and the $x$-axis, where the area counts as negative when $f$ is negative). But if $f$ is a Riemann integrable function, then if you take $P$ to be a finer and finer partition, the upper and lower sums $U(P,f)$ and $L(P,f)$ become closer and closer to one another and to the actual area under $f$.
