Prob. 1, Chap. 6, in Baby Rudin: If $f(x_0)=1$ and $f(x)=0$ for all $x \neq x_0$, then $\int f\ \mathrm{d}\alpha=0$ Here is Prob. 1, Chap. 6, in the book Principles of Mathematical Analysis by Walter Rudin, 3rd edition: 

Suppose $\alpha$ increases on $[a, b]$, $a \leq x_0 \leq b$, $\alpha$ is continuous at $x_0$,$f\left( x_0 \right) = 1$, and $f(x) = 0$ if $x \neq x_0$. Prove that $f \in \mathscr{R}(\alpha)$ and that $\int f \ \mathrm{d} \alpha = 0$. 

My Attempt: 

As $\alpha$ is continuous at $x_0$, so, given $\varepsilon > 0$, we can find a $\delta > 0$ such that $$ \left\lvert \alpha(t) - \alpha \left( x_0 \right) \right\rvert < \frac{\varepsilon}{4} \tag{*} $$
  for all $t \in [a, b]$ for which $\left\lvert t-x_0 \right\rvert < \delta$.
Let $n$ be a natural number such that $n > 2$, and let $P = \left\{ t_0, t_1, \ldots, t_{n-1}, t_n \right\}$ be a partition of $[a, b]$ such that $x_0$ is one of the points of $P$, and such that $\Delta t_j < \frac{\delta}{2}$ for each $j = 1, \ldots, n$. [We have included $x_0$ in $P$ in order to account for the case when $x_0$ is either of the endpoints of $[a, b]$.] 
Then, for any point $u_j \in \left[ t_{j-1}, t_j \right]$ ($1 \leq j \leq n$), we have 
  $$
\begin{align}
& \left\lvert \sum_{j=1}^n f \left( u_j \right) \left[ \alpha \left( t_j \right) - \alpha \left( t_{j-1} \right) \right] \right\rvert \\ 
&\leq  \begin{cases}  \alpha\left( t_1 \right) - \alpha \left( t_0 \right)  \qquad \mbox{ if $x_0 = a$} \\
\left[ \alpha \left( t_{i+1} \right) - \alpha \left( t_i \right) \right] + \left[ \alpha \left( t_i \right) - \alpha \left( t_{i-1} \right) \right] = \alpha \left( t_{i+1} \right) - \alpha \left( t_{i-1} \right) \qquad \mbox{ if $x_0 \in (a, b)$ and $x_0 = t_i$ for some $i \in \{ 1, \ldots, n-1 \}$} \\
\alpha \left( t_n \right) - \alpha \left( t_{n-1} \right) \qquad \mbox{ if $x_0 = b$}
\end{cases} \\
&< \frac{\varepsilon}{4}. \qquad \mbox{ [ by (*) above ] }
\end{align}
$$
  Thus we can conclude that, for any choice of points $u_j \in \left[ t_{j-1}, t_j \right]$ ($1 \leq j \leq n$), we have 
  $$ - \frac{\varepsilon}{4} \leq \sum_{j=1}^n f \left( u_j \right) \Delta \alpha_j \leq  \frac{\varepsilon}{4}, \tag{0} $$
But 
  $$ L(P, f, \alpha) = \inf \left\{ \ \sum_{j=1}^n f\left( u_j \right) 
\left[ \alpha \left( t_j \right) - \alpha \left( t_{j-1} \right) \right] \ \colon \ u_j \in \left[ t_{j-1}, t_j \right] 
\ \mbox{ for } \ j = 1, \ldots, n \ \right\}, \tag{A}$$
  and 
  $$ U(P, f, \alpha) = \sup  \left\{ \ \sum_{j=1}^n f\left( u_j \right) 
\left[ \alpha \left( t_j \right) - \alpha \left( t_{j-1} \right) \right] \ \colon \ u_j \in \left[ t_{j-1}, t_j \right] 
\ \mbox{ for } \ j = 1, \ldots, n \ \right\}. \tag{B}$$

Here is the link to my post here on Math SE on how we can obtain (A) and (B) 
Riemann-Stieltjes Upper and Lower Sums as Suprema and Infima

So from (0) we can conclude that 
  $$ -\frac{\varepsilon}{4} \leq L(P, f, \alpha) \leq U(P, f, \alpha) \leq \frac{\varepsilon}{4}. \tag{1}$$
Now from (1) we obtain 
  $$ U(P, f, \alpha) - L(P, f, \alpha) \leq \frac{\varepsilon}{2} < \varepsilon; $$
  but as $\varepsilon$ is an arbitrary positive real number, so the last set of inequalities, by virtue of Theorem 6.6 in Baby Rudin, implies that $f \in \mathscr{R}(\alpha)$ on $[a, b]$. 
Moreover, as 
  $$ \int_a^b f \ \mathrm{d} \alpha = \sup \left\{ \ L(Q, f, \alpha) \ \colon \ \mbox{ Q is a partition of $[a, b]$ } \ \right\} =  \inf \left\{ \ U(Q, f, \alpha) \ \colon \ \mbox{ Q is a partition of $[a, b]$ } \ \right\}, $$ 
  so we must also have 
  $$ L(P, f, \alpha ) \leq  \int_a^b f \ \mathrm{d} \alpha \leq U(P, f, \alpha). \tag{2} $$ 
So from (1) and (2), we obtain 
  $$ - \frac{\varepsilon}{4} \leq  \int_a^b f \ \mathrm{d} \alpha \leq  \frac{\varepsilon}{4}, $$
  which implies that 
  $$ \left\lvert  \int_a^b f \ \mathrm{d} \alpha  \right\rvert < \varepsilon. \tag{3} $$
But $\varepsilon$ was an arbitrary positive real number. So from (3) we can conclude that 
  $$  \int_a^b f \ \mathrm{d} \alpha = 0, $$
  as required. 

Is my proof good enough? Or, are there any issues with its logic, rigor, or presentation? 
 A: Here's a modification to your proof:

Fix $\varepsilon > 0$. As $\alpha$ is continuous at $x_0$, We can find a $\delta > 0$ such that 
  $$ 
\left\lvert \alpha(t) - \alpha \left( x_0 \right) \right\rvert < \frac{\varepsilon}{4} \tag{*} 
$$
  for all $t \in [a, b]$ for which $\left\lvert t-x_0 \right\rvert < \delta$.
Let $n$ be a natural number such that $n > 2$, and let $P = \left\{ t_0, t_1, \ldots, t_{n-1}, t_n \right\}$ be a partition of $[a, b]$ such that $x_0 = t_k$ is one of the points in $P$, and such that $\Delta t_j < \frac{\delta}{2}$ for each $j = 1, \ldots, n$. [We have included $x_0$ in $P$ in order to account for the case when $x_0$ is either of the endpoints of $[a, b]$.] 
Then, for any point $u_j \in \left[ t_{j-1}, t_j \right]$ ($1 \leq j \leq n$), we note that $f(u_j) = 0$ unless $j = k$ or $j-1 = k$. Moreover, $\alpha$ is increasing, so $\alpha(t_j) - \alpha(t_{j-1}) \geq 0$, and $f(x) \geq 0$. Thus,
  we have 
  $$
\begin{align}
0<& \sum_{j=1}^n f \left( u_j \right) \left[ \alpha \left( t_j \right) - \alpha \left( t_{j-1} \right) \right]\\ 
&\leq  
\begin{cases}  \alpha\left( t_1 \right) - \alpha \left( t_0 \right)  \qquad \mbox{ if $k=0$} \\
\alpha \left( t_n \right) - \alpha \left( t_{n-1} \right) \qquad \mbox{ if $k=n$}\\
\left[ \alpha \left( t_{i+1} \right) - \alpha \left( t_i \right) \right] + \left[ \alpha \left( t_i \right) - \alpha \left( t_{i-1} \right) \right] = \alpha \left( t_{i+1} \right) - \alpha \left( t_{i-1} \right) 
\\ \qquad \text{otherwise}
\end{cases} \\
&< \frac{\varepsilon}{4}. \qquad \mbox{ [ by (*) above ] }
\end{align}
$$
  Thus we can conclude that, for any choice of points $u_j \in \left[ t_{j-1}, t_j \right]$ ($1 \leq j \leq n$), we have 
  $$ 0 \leq \sum_{j=1}^n f \left( u_j \right) \Delta \alpha_j \leq  \frac{\varepsilon}{4}, \tag{0} $$

[Note: so far I haven't changed much.  I might have handled the $x_0 \in \{a,b\}$ case differently but so far so good].

But 
  $$ L(P, f, \alpha) = \inf \left\{ \ \sum_{j=1}^n f\left( u_j \right) 
\left[ \alpha \left( t_j \right) - \alpha \left( t_{j-1} \right) \right] \ \colon \ u_j \in \left[ t_{j-1}, t_j \right] 
\ \mbox{ for } \ j = 1, \ldots, n \ \right\}, \tag{A}$$
  and 
  $$ U(P, f, \alpha) = \sup  \left\{ \ \sum_{j=1}^n f\left( u_j \right) 
\left[ \alpha \left( t_j \right) - \alpha \left( t_{j-1} \right) \right] \ \colon \ u_j \in \left[ t_{j-1}, t_j \right] 
\ \mbox{ for } \ j = 1, \ldots, n \ \right\}. \tag{B}$$
Here is the link to my post here on Math SE on how we can obtain (A) and (B) 
Riemann-Stieltjes Upper and Lower Sums as Suprema and Infima

[Note: what you have above is not strictly necessary for this proof.  I think it's safe to assume the reader knows what $L(P,f,\alpha)$ and $U(P,f,\alpha)$ are.  This is a matter of taste, though].
Now, I condense the rest of your proof to two lines:

So from (0) we can conclude that 
  $$0 \leq L(P, f, \alpha)\leq \int_a^b f\,d\alpha \leq U(P, f, \alpha) \leq \frac{\varepsilon}{4}. \tag{1}$$
But $\varepsilon$ was an arbitrary positive real number. So from (1) we can conclude that 
  $$  \int_a^b f \ \mathrm{d} \alpha = 0, $$
  as required. 

