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;

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


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

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.

  • $\begingroup$ 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? $\endgroup$ – Katlus Dec 20 '12 at 23:08

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.