How to choose $\alpha$ such that the improper integral with respect to $\alpha$ is finite?

Question

Give and example of a function $f$ and a choice of $\alpha$ that is strictly increasing such that $$\int^\infty_0 f\,dx=+\infty \quad \text{and} \quad \int^\infty_0 f\,d\alpha<\infty.$$

Answer 1

Let $f(x)=\chi_{[0,1)}(x)x+\chi_{[1,\infty)}(x)x^{-1}$ and $\alpha(x)=\displaystyle\int_{0}^{x}f(t)dt$, then $\alpha$ is strictly increasing and that $\alpha'(x)=f(x)$, so \begin{align*} \int_{0}^{\infty}f(x)d\alpha(x)&=\int_{0}^{\infty}(f(x))^{2}dx\\ &=\int_{0}^{1}x^{2}dx+\int_{1}^{\infty}\dfrac{1}{x^{2}}dx\\ &<\infty, \end{align*} but \begin{align*} \int_{0}^{\infty}f(x)dx>\int_{1}^{\infty}\dfrac{1}{x}dx=\infty. \end{align*}