If $f \in R(\alpha)$ and if $\int_a ^b f~ d \alpha = 0$ then, prove that $\alpha$ must be constant If $f \in R(\alpha)$  on $[a,b]$ and if for every monotonic function $f : $ 
$$\int_a ^b f~ d \alpha = 0 $$
then, prove that $\alpha$ must be constant on $[a,b]$
Proof:
By integration by parts : 
$\int_a ^b f~ d \alpha + \int_a ^b \alpha~ df = f(b) \alpha (b) -f(a) \alpha (a) $ . Substituting $\int_a ^b f~ d \alpha = 0 $ , we get :
$\int_a ^b \alpha~ df = f(b) \alpha (b) -f(a) \alpha (a) $ 
Given any point $c \in [a, b)$, we may choose a monotonic function $f$ defined as follows : 
$f(x) =   \begin{cases} 
      0 & x \leq c \\
      1 &  x > c 
   \end{cases}
$ 
*So, we have $\int_a ^b \alpha ~df= \alpha(c) = \alpha(b)$, that is, $ \alpha$ is constant in $[a,b]$
I don't understand why choosing that function yields the result I marked with *
 A: Since $f$ is integrable with respect to $\alpha$, it follows that $\alpha$ is integrable with respect to $f$ -- this is part of the theorem that justifies the integration by parts used in this proof.  Thus, for any $\epsilon > 0$ there exists $\delta > 0$ such that for every partition $P = (x_0,x_1, \ldots,x_n)$ with $\|P\| < \delta$ and any Riemann-Stieltjes sum
$$S(P,\alpha,f) = \sum_{k=1}^n\alpha(t_k)[f(x_k) - f(x_{k-1})],$$
we have (no matter how the intermediate points $t_k \in [x_{k-1},x_k]$ are chosen)
$$\tag{*}\left|S(P,\alpha,f) - \int_a^b \alpha \, df\right| < \epsilon$$
For any such $P$, we can assume that $P$ has $c$ as one of the partition points, say $c = x_j$. Otherwise, add $c$ to the partition and $\|P\| < \delta$ still holds.
Given that $f(x_k) = 0$ for $x_k \leqslant x_j = c$ and $f(x_k) = 1$ for $x_k > x_j$, we have
$$S(P,\alpha,f) \\ = \sum_{k \leqslant j}\alpha(t_k)[f(x_k) - f(x_{k-1})] + \alpha(t_{j+1})[f(x_{j+1}) - f(x_j)] + \sum_{k > j+1}\alpha(t_k)[f(x_k) - f(x_{k-1})] \\ = \alpha(t_{j+1})$$
It follows that $\left|\alpha(t_{j+1}) - \int_a^b \alpha \, df \right| < \epsilon$ and since (*) holds for any $t_{j+1} \in [x_j, x_{j+1}] = [c, x_{j+1}]$,we can choose $t_{j+1} = c$ to obtain $\left|\alpha(c) - \int_a^b \alpha \, df \right| < \epsilon$.  Since $\epsilon$ can be arbitrarily close to $0$ it follows that
$$\alpha(c) = \int_a^b \alpha \, df$$
