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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?

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  • $\begingroup$ Are we allowing $\infty$ as a value of the integral? If not, don't you get problems with $\int_0^{\infty} \frac{\sin{x}}{x} \, dx$? $\endgroup$ – Chappers May 9 '18 at 12:27
  • $\begingroup$ @Chappers: Munkres requires the limit to be a real number, and of course, his definition can not handle conditionally integrable functions. $\endgroup$ – painday May 9 '18 at 12:48

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