Convergence theorems for improper integrals

Question

For a sequence of $\mathbb{R}$-valued (Borel-)measurable functions $\{h_n\}_{n=1,2,\cdots}$ let us consider $$ \int_0^1\frac{h_n(x)}{1-x}dx:=\lim_{\epsilon\to0}\int_0^{1-\epsilon}\frac{h_n(x)}{1-x}dx. $$ Here we regard the integral in the right-hand side as Lebesgue integral. Assume that the limit exists for each $n=1,2,\cdots$. Then is the assumptions of Fatou's lemma enough to establish $$ \liminf_{n\to\infty}\int_0^1\frac{h_n(x)}{1-x}dx\ge\int_0^1\liminf_{n\to\infty}\frac{h_n(x)}{1-x}dx? $$

I'm confused to answer this question because I have to change the order $\lim_{\epsilon\to0}$ and $\liminf_{n\to\infty}$. However, in this case, it's not always that $\lim_{\epsilon\to0}$ or $\liminf_{n\to\infty}$ have any uniformity (such as Moore-Osgood Theorem). I don't know whether well-known convergence theorems are valid for improper integrals or not.

Please tell me ideas, answers or references if you have. Thank you in advance.

Answer 1

Consider $\{h_n\}$ to be positive measurable functions (otherwise you can't use Fatou's lemma), then \begin{align} f_n(x) = \frac{h_n(x)}{1-x} \end{align} is also a sequence of positive measurable functions. Hence it follows \begin{align} f(x):= \limsup_{n\rightarrow \infty} \frac{h_n(x)}{1-x} \end{align} is also measurable. Then by Fatou's lemma, we have that \begin{align} \int^1_0 f(x)\ dx \leq \liminf_{n\rightarrow \infty} \int^1_0 f_n(x)\ dx. \end{align} This inequality allows for both left and right side to be infinity.