Arzelà's Dominated Convergence Theorem for improper integrals?

Question

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

Answer 1

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.