Hartman and Mikusinski's book "The Theory of Lebesgue Measure and Integration" make an interesting remark on improper integrals in multiple dimensions:
In the case of one variable, we introduced, besides the concept of the Lebesgue integral on an infinite interval, the further concept of an improper integral. A similar generalization does not emerge for integrals of functions of two variables. For, if we assume only the existence of $(6)$, and do not assume the existence of $(5)$, then limits $(6)$ can have different values for different sequences $U_n$, e.g. they may have one value for the sequence of circles $x^2+y^2<n^2$ and another for the sequence of squares $|x| < n, |y| < n$. Because of this situation, the value of the integral would be undefined. If, however, we make the additional assumption that limits $(6)$ have the same value for all increasing nested sequences ${U_n}$ with union $P$, then the existence of limit $(5)$ would already follow from this and we would not obtain any generalization.
Here $(5)$ refers to $\lim_{n\rightarrow\infty}\int\int_{U_n}|f(x,y)|dxdy$ and $(6)$ refers to $\lim_{n\rightarrow\infty}\int\int_{U_n}f(x,y)dxdy$ where $(U_n)$ is an increasing sequence of bounded measurable sets whose union is an unbounded measurable set $P$.
My question is, what is the proof of the two assertions that Hartmann and Mikusinski make in the quote above? First of all, what is an example of a function $f(x,y)$ such that the integral of $f$ over the disc $x^2+y^2<n^2$ and the integral of $f(x,y)$ over the square $|x| < n, |y| < n$ converge to different values as $n\rightarrow\infty$?
Second of all, how do you prove that if $\lim_{n\rightarrow\infty}\int\int_{U_n}f(x,y)dxdy$ has the same value for every increasing sequence of bounded measurable sets $(U_n)$ whose union is $\mathbb{R}^2$, then $\lim_{n\rightarrow\infty}\int\int_{U_n}|f(x,y)|dxdy$ exists (i.e. $f$ is Lebesgue integrable on $\mathbb{R}^2$)?
Best Answer
I will answer your second question. So we assume that the integral satisfies the property on the limits that you stated.
Consider the functions $f^+$ and $f^-$, we know that $|f|=f^++f^-$. If $|f|$ is not Lebesgue integrable, it means that either $\int f^+d\mu=+\infty $ or $\int f^-d\mu =+\infty$.
Moreover, if one is infinite and the other is finite, then it is easy to see that $$\lim_{k\to\infty}\int_{A_k}fd\mu=\int fd\mu\in\{+\infty,-\infty\},$$ for any increasing sequence of bounded measurable sets $(A_k)_{k\in\mathbb N}$.
So we must have that $$ \int f^+d\mu=\int f^-d\mu =+\infty. $$ Consider $E=f^{-1}([0,+\infty))$ and put: $F=E^c$, $E_n=B(0,n)\cap E$ and $F_n=B(0,n)\cap F$. Fix $\alpha\in\mathbb R$. We define $A_1=E_1$, and then, for each $k> 1$, we define $A_k$ to be $$\mbox{the union of $A_{k-1}$ with the least indexed remaining set of the family $\{E_n\}$, if $\int_{A_{k-1}}fd\mu<\alpha$;}\\ \mbox{the union of $A_{k-1}$ with the least indexed remaining set of the family $\{F_n\}$, if $\int_{A_{k-1}}fd\mu\geq\alpha$.} $$ Then certainly $(A_k)$ is a sequence of bounded measurable sets such that the union is $\mathbb R^2$ and the sequence $\int_{A_k} fd\mu$ "osciles around $\alpha$", since it increases if $\int_{A_k} fd\mu<\alpha$ and decreases if $\int_{A_k} fd\mu\geq \alpha$. More precisely, we have that $$ \lim_{n\to\infty} \int_{E_n} fd\mu = \int f^+d\mu = +\infty\ \ \mbox{ and }\ \ \lim_{n\to\infty} \int_{F_n} fd\mu = -\int f^-d\mu = -\infty, $$ (this is an immediate consequence of the monotone convergence theorem).
Therefore, the sequence $\int_{A_k} f d\mu$ will have infinite elements greater than $\alpha$ and infinite elements less than $\alpha$. This also ensures that every set of the families $\{E_n\}$ and $\{F_n\}$ is picked by the sequence $A_k$.
By hypothesis, $\displaystyle\lim_{k\to\infty}\int_{A_k}fd\mu$ must exist and, since $f$ "osciles around $\alpha$", we must have $\displaystyle\lim_{k\to\infty}\int_{A_k}fd\mu=\alpha$.
But since $\alpha$ was arbitrarily chosen, it contradicts our assumption on the uniqueness of the limits $\displaystyle\lim_{k\to\infty}\int_{A_k}fd\mu=\alpha$.