Let $A\subset \mathbb R^n$ be a non-empty open set, and let $D_n\ (n\in\mathbb N)$ be a sequence of subsets of $A,$ then if $D_n$ satisfys
$1. D_n\subset \mathrm{Int}(D_{n+1}),\ \forall\ n\in\mathbb N;$
$2. \displaystyle\bigcup_{n=1}^\infty D_n=A;$
$3.$ Each $D_n$ is a compact Jordan measurable set;
then we call sequence $D_n$ a exhaustion sequence of open set $A.$ It could be proved that every non-emptyset open set has an exhaustion sequece. (The proof could be found in the book Analysis on manifold by Munkres .)
Question: Assume that $f:A\to\mathbb R$ is a locally Riemann integrable function, and there exists a real number $a,$ such that for any exhaustion sequence $D_n\ (n\in\mathbb N)$ of $A,$ we have $\displaystyle\lim_{n\to\infty}\int_{D_n} f(t)\ \mathrm{dt}=a.$ Then can we say that limit $\displaystyle\lim_{n\to\infty}\int_{D_n} |f(t)|\ \mathrm{dt}$ also exist?
Background: In the very book Munkres has showed that we can define the improper Riemann integral of a non-negative loccaly Riemann integrable function $f:A\to \mathbb R$ to be the limit $$\displaystyle\mathrm{(Improper)}\int f(t)\ \mathrm{dt}:=\lim_{n\to\infty}\int_{D_n} f(t)\ \mathrm{dt},$$ if the limit exists. Munkres proved that this limit doesn't rely on the choose of exhaustion of $A.$
And similarly for any loccaly Riemann integrable function $f:A\to \mathbb R,$ we can define its improper integral as $$\displaystyle\mathrm{(Improper)}\int f(t)\ \mathrm{dt}:=\mathrm{(Improper)}\int f^+(t)\ \mathrm{dt}-\mathrm{(Improper)}\int f^-(t)\ \mathrm{dt}.$$ (Note that this definition cannot handle the integral $\displaystyle\int_0^\infty {\sin x\over x}\mathrm{dx}$ properly.)
A plausible idea is that we simply define that improper integral of any locally Riemann integrable function $f:A\to \mathbb R$ to be the limit $\displaystyle\lim_{n\to\infty}\int_{D_n} f(t)\ \mathrm{dt},$ as long as $|f(t)|$ is improper integrable. However, we can see that this limit generally depend on the specific choose of exhaustion sequence.
For example, consider the function $f:\mathbb R\to\mathbb R,\ x\mapsto \sin(\pi x),$ then the integral along exhaustion sequences $C_n:=[-n,n],\ D_n=[-2n,2n+1]$ will end up to be $\displaystyle \displaystyle \lim_{n\to\infty}\displaystyle\int_{C_n} f\ \mathrm{dv}=0,$ and $ \displaystyle \displaystyle \lim_{n\to\infty}\displaystyle\int_{D_n} f\ \mathrm{dv}={2\over \pi}. $
So what if we assume that the limit exists and independent of the choose of exhaustion sequence? Will $|f(t)|$ be improperly integrable under this condition?