# Arzelà's Dominated Convergence Theorem for improper integrals?

Consider the Arzela's Dominated Convergence Theorem for Riemann integrals:

Take a function $f_n$ defined on a bounded and closed interval $[a,b]$ and Riemann integrable on $[a,b]$ $\forall n \in \mathbb{N}$. Let $|f_n(x)|\leq M$ with $M>0$ $\forall x \in [a,b]$. Let $\lim_{n\rightarrow \infty}f_n(x)=f(x)$ $\forall x \in [a,b]$ with $f$ Riemann integrable on $[a,b]$. Then, $$\lim_{n\rightarrow \infty}\int_a^b |f_n(x)-f(x)|dx=0$$

Question: Is there a version of this theorem for improper integrals? Specifically, I am looking for something like

Take a function $h_n$ with domain $(-\infty, \infty)$ and Riemann integrable on $(-\infty, \infty)$$^{(*)} \forall n \in \mathbb{N}. Let |h_n(x)|\leq M with M>0 \forall x \in (-\infty, \infty). Let \lim_{n\rightarrow \infty}h_n(x)=h(x) \forall x \in (-\infty, \infty) with h Riemann integrable on (-\infty, \infty). Then,$$ \lim_{n\rightarrow \infty}\int_{-\infty}^{\infty} |h_n(x)-h(x)|dx=0 $$(*)\lim_{x\rightarrow -\infty} \int_{x}^ah_n(t)dt+\lim_{x\rightarrow \infty} \int_{a}^xh_n(t)dt exists and is finite ## 1 Answer No. Let h_n be the characteristic function of the interval [n,n+1]. For each n\in\Bbb N, h_n is uniformly bounded, Riemman integrable and \lim_{n\to\infty}h_n(x)=0 for all x\in\Bbb R, but$$ \lim_{n\to\infty}\int_{\Bbb R}h_n(x)\,dx=1\ne0.$$If you do not like the fact that the$h_n$are discontinuous, it is easy to modify the example to make the$h_n$continuous, even smooth. • Thank you professor. Do you know about existence of any dominated convergence theorem for improper Riemann integrals (maybe with stronger conditions of the ones I suggested in my question)? – TEX Apr 25, 2018 at 16:09 • For proper Riemann integrals, the condition of uniform boundless implies that there is an integrable majorant, so that it is a particular case of the Dominated Convergence Theorem. For improper integrals, uniform convergence of$h_n$to$h$should be enough. Of course, the existence of an integrable function$g$such that$|h_n(x)|\le g(x)$would imply the result, but I am not sure it can be proved only with the techniques of Riemann integrals, without recourse to measure theory. Apr 25, 2018 at 16:21 • Ok, thanks professor. For the second part of your comment ("Of course, the existence of an integrable function$g\$ such that ..."), this document here noncommutativeanalysis.wordpress.com/2014/05/11/… (Theorem 2 at the end) may provide a proof. Even though, it seems just for positive functions?
– TEX
Apr 25, 2018 at 16:23