Let $g$ be continuous on [0,1] and assume that $f$ is continuous on (0,1]. Assume further that for any $\epsilon\in (0,1)$, we have \begin{align*} \int_{\epsilon}^{1} f(s)\, ds=\int_{h(\epsilon)}^{1} g(u)\,du \end{align*} where $h(\epsilon)\in (0,1)$ and $h(\epsilon)\to 0$ as $\epsilon \to 0$.
I want to show that the improper integral \begin{align*} \int_{0}^{1}f(s)\,ds \end{align*} is well defined.
My question is that how to show an improper integral is well defined? Do I need to use Cauthy's criterion to that this integral is bounded by some other well defined integral?
The improper integral is well defined if $\displaystyle \lim_{\epsilon \to 0} \int_e^1 f(s) \,ds$ exists.
Since $g$ is continuous and integrable on $[0,1]$ you can take your pick of ways to prove this.
Using the Cauchy criterion we have
$$\left|\int_\alpha^\beta f(s) \, ds \right| = \left|\int_{h(\alpha)}^{h(\beta)} g(s) \, ds \right| \leqslant \sup_{[0,1]}|g|\,|h(\beta) - h(\alpha)| \longrightarrow _{\alpha,\beta \to 0} 0$$