By definition, prove that, if $ \alpha$ in$ [a,b]$ is defined as
$\alpha(x) = \begin{cases} 1 & x = c \\ 0 & x \neq c \end{cases} $
where$ c \in (a,b)$ and f is continuous on c, then$\ \int_a^b fd\alpha =0$
Attempt
My definition for a Riemman-Stieltjes sum is, for P a partition of [a,b] then $$\ S(P,f,\alpha) = \sum_{k=q}^ n f(t_k)\Delta\alpha_k $$
Since $c\in(a,b)$ we take a partition of $ [a,b]$, $$P=(a,x_1,...,x_{j-1},c,x_{j+1},...,x_{n-1},b)$$
From this, we see that
$$\ S(P,f,\alpha) = \sum_{k=q}^ n f(t_k)\Delta\alpha_k = 0$$
Since f is continuous in c, by definition we have that $$ |x-c|<\delta \Rightarrow |f(x)-f(c)|<\epsilon$$
So we choose $P_{\epsilon}$ such that $ ||P_{\epsilon}||<\delta$ and we'll have that $ ||P||≤||P_{\epsilon}||<\delta$
If we choose $ \int_a^b fd\alpha = 0$ then
$$ |S(P,f,\alpha) - \int_a^b fd\alpha| = 0 < \epsilon$$
therefore, $ \int_a^b fd\alpha = 0$.
Is this a correct aprroach?
Your basic approach and statement about $\delta$ and $\epsilon$ is correct, but it is not clear how this implies that the integral is $0$. Furthermore, $S(P,f,\alpha) \neq 0$, in general, for your chosen partition.
If $c$ is not an endpoint of a subinterval $[x_{k-1},x_k]$ of the partition $P$, then for all $k = 1,\ldots,n$ we have $\alpha(x_k) - \alpha(x_{k-1}) = 0$. In this case, $S(P,f,\alpha) = 0$ and $|S(P,f,\alpha)-0| < \epsilon$ for any $\epsilon > 0.$
Otherwise, suppose $c = x_j$. We then have
$$S(P,f,\alpha) = f(t_j)(\alpha(c) - \alpha(x_{j-1})) + f(t_{j+1})(\alpha(x_{j+1}) - \alpha(c)) = f(t_j) - f(t_{j+1})$$
Now fill in the steps to show that the continuity of $f$ implies
$$|S(P,f,\alpha)-0| = |f(t_j) - f(t_{j+1})| < \epsilon$$
if $\|P\|$ is sufficiently small.