Suppose $f$ is continuous on $[a,b]$, $\alpha (x)=0$ if $x \in [a,c)$, and $\alpha (x)=1$ if $x \in [c,b]$. Show that $\int_a^bfd\alpha=f(c)$

Question

Suppose $f$ is continuous on $[a,b]$, $a \lt c \lt b $, $\\$ $\alpha (x)=0$ if $x \in [a,c)$, and $\alpha (x)=1$ if $x \in [c,b]$. Show that $\int_a^bfd\alpha=f(c)$

I started learning about integrals and I saw this problem with the solution in the photo but I don't really understand it. Does someone has another approach or can somebody help me understand it? I would really appreciate your help

Answer 1

You have just to apply the definition of Riemann-Stieltjes integral. Since $$ \Delta \alpha_k := \alpha(x_k) - \alpha(x_{k-1}) = \begin{cases} 1, & \text{if}\ k = i,\\ 0, & \text{if}\ k\neq i, \end{cases} $$ then $U(P, f, \alpha)$ and $U(P, f, \alpha)$ are the one reported in the picture.

On the other hand, since you can choose the points $x_{i-1} < c \leq x_i$ as close as you want, by continuity $$ \sup_{P}\inf_{x\in [x_{i-1}, x_i]} \{f(x):\ \ x_{i-1} < c \leq x_i\} = f(c), $$ and the same holds for the $\sup$.