# Order of taking limits on Riemann-Stieltjes Integral (Improper Integral)

Let $\alpha$ be a monotonically increasing function on $(0,\infty)$. Let $g:(0,\infty)\rightarrow \mathbb{R}$ be a function such that $g\in\mathscr{R}(\alpha)$ on any bounded closed connected set. (That is, $[a,b]$)

Let $a>0$ be a real.

What i have learned is;

$$\int_0^a g \, d\alpha=\lim_{t\to 0} \int_t^a g \, d\alpha$$

And

$$\int_a^\infty g \, d\alpha=\lim_{t\to\infty} \int_a^t g \, d\alpha$$

And for $b>0$, $\int_a^b g \, d\alpha=\lim_{t\to b}\int_a^t g \, d\alpha$ if $\alpha$ is continuous at $b$.

(If these limits exist)

Here, what is the definition of $\int_0^\infty g \, d\alpha$ ?

Which limit should i take first? And what constraint gurantees that order of taking limits is irrelevant?

You can write $$\int_{t_1}^{t_2} g d\alpha = \int_{t_1}^a g d\alpha + \int_a^{t_2} g d\alpha$$ and then you see that the limits act on different terms and thus the order of the limits does not matter.
• I don't remember what it is called, but i remember in pre-calculus class i learned some kind of integral which makes $\int_{-a}^{a} \frac{1}{x}dx=0$, which cannot be true in your definition. Is your definition is generally used for Stieltjes Integral? And what is the name of integral i just mentioned? – Katlus Dec 20 '12 at 23:08