Calculate $U(p,f)$ and $L(p,f)$
$(\alpha(x) = x), x \in [-1,1] \\ p =\{-1, \frac{-1}{2},\frac{1}{2},1 \} \\ f(x) = 1+x^2$
Using Riemann Stieljes Integral
$\text{We define upper sum } \quad U(f,p,\alpha) = \sum_{j=1}^n M_j \triangle\alpha_ j \\ \text{We define lowersum } \quad L(f,p,\alpha) = \sum_{j=1}^n m_j \triangle\alpha_ j \\ \triangle \alpha_j = \alpha(x_j)-\alpha(x_{j-1}) \\ \text{p is a partition } \{x_0,x_1,.....x_n \} \\ M_j =\text{sup}f(x)\\ m_j =\text{inf} f(x)$
$$\triangle\alpha_1 = \alpha(\frac{-1}{2})-\alpha(-1) = \frac{-3}{2} \\ \triangle\alpha_2= 1 \\ \triangle\alpha_3 =\frac{1}{2} \\ M_1 =\ f(-1) =2 \\ M_2=f(-1)=2\\M_3=f(-\frac{1}{2})= \frac{5}{4} \\ U(p,f,\alpha) = M_1\triangle\alpha_1+M_2\triangle\alpha_2+M_3\triangle\alpha_3 \\ (2)(-3/2)+2(1)+5/4(1/2) = \frac{-3}{8} ≈ -.4 \\ m_1 =f(-1) =2 \\ m_2 = f(\frac{-1}{2}) = \frac{5}{4}\\ m_3 = f(\frac{1}{2}) = \frac{5}{4} \\ L(f,p,\alpha) = m_1 \triangle\alpha_1+m_2\triangle\alpha_2+m_3\triangle\alpha_3 \\ (2)(-3/2)+(5/4)(1)+5/4(1/2) = \frac{-9}{8} ≈ -1.125 \\ $$
My question for this problem is how does one know which points to chose for M and m? I know that M1 is the maximum of the function and m1 is the minimum of the function. Is there a pattern used to choose the points of M and m? Does $U(f,p,\alpha) $ need to be greater than $L(f,p,\alpha)$?
$M_j$ is the maximum value of $f$ on $[x_{j-1},x_j]$ and $m_j$ is the minimum value on the interval. To find these values we can find the points, if any, where the derivative is $0$. $M_j$ is then the maximum of the values of $f$ at these points together with the values at $x_{j-1}$ and $x_j$. In this example $f'(x)=0$ only when $x=0$. $M_1$, for example, is the maximum of $f(-1), f(-\frac 1 2)$ (since there is no point in this interval where $f'(x)=0$).
The upper sum is always greater then or equal to the lower sum.