Suppose $\alpha$ increases on $[a,b]$, $a \leq x_{0} \leq b$, $\alpha$ is continuous at $x_{0}$, $f(x_{0})=1$, and $f(x)=0$ if $x \neq x_{0}$. Prove that $f \in \mathscr{R} (\alpha)$ and that $ \int f d\alpha =0$
My attempt:
Approach (1)
$f$ is discontinuous at $x_{0}$. By theorem 6.10,$f \in \mathscr{R} (\alpha)$. If $U(P,f,\alpha)-L(P,f,\alpha) \lt \epsilon$ holds for partition $P=\{ x_0,....,x_n\}$ and $t_i \in [x_{i-1},x_{i}]$, then $|\sum_{i=1}^n f(t_i)\Delta \alpha_i - \int_{a}^b f d\alpha| \lt \epsilon$. let $p=x_{0}$. Now there are two cases:
(1)$p \in P=\{ x_0,....,x_n\}$, i.e. $p=x_{j}$, for some $j \in J_n \cup \{0\}$. Then choose any $t_i \in (x_{i-1},x_{i}) \subset [x_{i-1},x_{i}], \forall i$. So $f(t_i)=0, \forall i \in J_{n}$. Thus $ \sum_{i=1}^n f(t_i)\Delta \alpha_i=0$. Hence, $|0 - \int_{a}^b f d\alpha| \lt \epsilon$ holds for all $\epsilon$. Thus, $ \int f d\alpha =0$.
(2)$p \notin P=\{ x_0,....,x_n\}$, i.e. $p \in (x_{j-1},x_{j})$, for some $j \in J_{n}$. Then choose any $t_i \in (x_{i-1},x_{i}), \forall i \in J_{n} \setminus \{j\}$, and $t_{j} \in \{x_{j-1},x_{j}\}$. Thus, $f(t_i)=0, \forall i \in J_n$. Hence $|0 - \int_{a}^b f d\alpha| \lt \epsilon$ holds for all $\epsilon$. Thus $ \int f d\alpha =0$.
In either case, $\int f d\alpha =0$.
Approach(2)
In this approach, I will use definition of integral(upper and lower Riemann-Stieltjes integral). Again $f \in \mathscr{R} (\alpha)$, by theorem 6.10. So $\inf U(P, f, \alpha) = \sup L(P, f, \alpha) = \int_{a}^b f d \alpha$. For this given function computing $L(P,f, \alpha)$ is easy. $L(P,f, \alpha)= \sum_{i} m_i \Delta \alpha_i$. Now take any partition $P$ of $[a,b]$. Again using the same technique as in approach (1), we can show that $m_i=0, \forall i$. Thus, $L(P,f, \alpha)= \sum_{i} m_i \Delta \alpha_i=0$, for all partition of $[a,b]$. Hence, $\sup L(P,f,\alpha)=0$. Our desired result.
Is above two proofs correct?