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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

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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.

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  • $\begingroup$ 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)? $\endgroup$ – TEX Apr 25 '18 at 16:09
  • $\begingroup$ 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. $\endgroup$ – Julián Aguirre Apr 25 '18 at 16:21
  • $\begingroup$ 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? $\endgroup$ – TEX Apr 25 '18 at 16:23

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