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

$$\int_a^b f(x)\, d \alpha(x) =0$$

• Write out the definition of the integral and it follows immediately. – Dunham May 8 '17 at 19:39
• Every Riemann Stieltjes sum is $0$ here and hence the integral is also $0$. It is as simple as that and one should not try to think it as more difficult than it actually is. – Paramanand Singh May 8 '17 at 20:10

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 $0$.

• I understand now, thanks. – user441848 May 9 '17 at 1:58

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?

• My definition is |$\sum_{k=1}^n f(t_k)(\alpha(x_k)-\alpha(x_{k-1}))$- $\int_a^b f(x)\, d \alpha(x)$|<$\epsilon$ – user441848 May 8 '17 at 20:03
• and therefore the whole summatory goes to zero. right? – user441848 May 8 '17 at 20:04
• but what happen with the integral |$\int_a^b f(x)\, d \alpha(x)$|<$\epsilon$? this is not zero yet – user441848 May 8 '17 at 20:06
• @AnneliseToft see the update - you don't have the correct definition – gt6989b May 8 '17 at 20:12
• Even in the generalized case (of the first comment here), the sum is zero and the inequality must hold for all $\epsilon>0$. Hence the integral must be 0. – Dunham May 8 '17 at 21:58