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Let $S = \{(x,y) \in \mathbb{R}^2 \colon 0 <|x|, |y|<1\}$ be a "punctured square." Consider the measure space $(S, \mathcal{L}_2, m)$. Here, $\mathcal{L}_2$ denotes the Lebesgue measurable subsets of $S$, and $m$ denotes Lebesgue measure on $S$. Define $f : S \to \mathbb{R}$ by $f(x,y) = \frac{xy}{(x^2 +y^2)^2}$.

Now, $f$ is clearly continuous. Hence $f$ is Lebesgue measurable on $S$, at least so long as we agree to define $f : S \to \mathbb{R}$ is measurable $\iff$ for all $\alpha$, $f^{-1}((\alpha, \infty)) \in \mathcal{L}_2$ . So we are free to define the integral $\displaystyle \int_S f dm \in \mathbb{R} \cup \{\infty\}$. I would like to show that $\displaystyle \int_S f dm = \infty$.

I have not gotten very far with this problem, but I do have some thoughts. Perhaps I could try to construct a sequence of characteristic functions that lie below $f$ yet whose integrals diverge? That would do the trick. I have a sneaking suspicion that the proof will hinge on the fact that $\int_0^\epsilon \frac{1}{r^2}dr = \infty$ for all $\epsilon > 0$, but I'm not sure how to apply this yet.

Hints or solutions are greatly appreciated.

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Does your title match your function? Which is it? –  Euler....IS_ALIVE Mar 1 '13 at 19:00
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have you tried polar coordinates? Afterwards you should be able to split the integral into one small ball (thats probably where $\varepsilon$ comes from) and one larger ball that covers the rest of the domain –  Quickbeam2k1 Mar 1 '13 at 19:06
    
Hint: Your function is greater than $C {1 \over x^2 + y^2}$ for some constant $C$ on "most" of your domain. –  Zarrax Mar 1 '13 at 19:07
    
Got something from the answer below? –  Did Mar 29 '13 at 8:03
    
@Did--Thanks yes this answer makes perfect sense. I did edit the inequality you had for $f(x,y)$, since $x^2 + y^2 < 5x^2$ implies that the denominator $(x^2 + y^2)^2$ is less than $25x^4$. –  jtms88 Mar 30 '13 at 17:51

1 Answer 1

up vote 1 down vote accepted

Consider the triangle $T\subset S$ with vertices $(0,0)$, $(1/2,1/2)$, $(1/2,1)$. Thus, $T$ is defined by the inequalities $0\lt x\lt y\lt 2x\lt 1$. For every $(x,y)$ in $T$, $xy\gt x^2$ and $x^2+y^2\lt5x^2$, hence $$ f(x,y)=\frac{xy}{(x^2+y^2)^2}\gt\frac{x^2}{(5x^2)^2}=\frac1{25x^2}. $$ Thus, $$ \iint_Tf(x,y)\mathrm dm(x,y)\geqslant\int_0^{1/2}\int_x^{2x}\frac1{25x^2}\mathrm dy\mathrm dx=\int_0^{1/2}\frac1{25x}\mathrm dx, $$ which is infinite. Hence the function $f$ is not integrable on $T$, and a fortiori not on $S$ either.

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Note that I do not understand why you say that $f$ is clearly nonnegative. –  Did Mar 22 '13 at 8:29
    
Yeah I edited out the statement. I was subconsciously thinking that we could solve the problem by restricting our attention to the first quadrant, where $f$ is nonnegative. Sometimes my mind gets ahead of me. –  jtms88 Mar 30 '13 at 17:47

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