Question about Riemann-Stieltjes integral This might be a silly question  but I need to ask anyway.
I don't understand why the integral with respect to a constant $c=\alpha(x)$ is always $0$.For example, Let f be riemann-stieltjes integrable on [a,b] and $c=\alpha(x)$ .Then
$$\int_a^b f(x)\, d \alpha(x) =0$$
 A: As Dunham noted, you can write it out from definition and it is clearly 0. Intuitively, your $\alpha(x)$ is some sort of measure of change. If $\alpha$ is constant, you are not sensitive to changes of any sort anywhere, which is why you get zero.
UPDATE
There seems to be a confusion about the definition of Riemann-Stieltjes integration. You define using the partition $P$
$$
\int_a^b f d\alpha := \lim_{P \to 0} \sum_{i=0}^{n-1} f(c_i)[g(x_{i+1}) - g(x_i)]
$$
and since $g$ is constant, what is the value of $g(x_{i+1}) - g(x_i)$ in every term of the summation? What can you conclude about the value of the entire summation?
A: Consider the following definition:
If there is a number $A$ such that for every $\epsilon>0$ there is a partition $P_{\epsilon}$ such that for every partition $P=\{a=x_0<x_1<\cdots<x_n=b\}$ that refines $P_\epsilon$
\begin{equation}
\left| \sum_{k=1}^{n}f(t_k)(\alpha(x_{k})-\alpha(x_{k-1}))
-
A
\right|
<
\epsilon
, \quad 
with\
t_k \in [x_{k-1},x_k]
\end{equation}
then we define $\int_a^b f(x)d\alpha(x) =A$.
Now, if we let $\alpha$ be a constant function, the sum is always zero, so the partitions are irrelevant and the definition simplifies in the following way:
If there is a number $A$ such that for every $\epsilon>0$ 
\begin{equation}
|A|
<
\epsilon
\end{equation}
then we define $\int_a^b f(x)d\alpha(x) =A$.
The only nonnegative number that is smaller than every positive number is $0$. Hence the integral is always $0$.
