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.

  • $\begingroup$ Does your title match your function? Which is it? $\endgroup$ – Euler....IS_ALIVE Mar 1 '13 at 19:00
  • 4
    $\begingroup$ 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 $\endgroup$ – Quickbeam2k1 Mar 1 '13 at 19:06
  • $\begingroup$ Hint: Your function is greater than $C {1 \over x^2 + y^2}$ for some constant $C$ on "most" of your domain. $\endgroup$ – Zarrax Mar 1 '13 at 19:07
  • $\begingroup$ Got something from the answer below? $\endgroup$ – Did Mar 29 '13 at 8:03
  • $\begingroup$ @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 $25x^4$. $\endgroup$ – JZS Mar 30 '13 at 17:51

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.

  • $\begingroup$ Note that I do not understand why you say that $f$ is clearly nonnegative. $\endgroup$ – Did Mar 22 '13 at 8:29
  • $\begingroup$ 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. $\endgroup$ – JZS Mar 30 '13 at 17:47
  • $\begingroup$ @Did, I'm a newbie to Lebesgue integral. Fubini's Theorem discusses about squared shaped areas. Could you say how Fubini's Theorem can be applied for Triangular areas ? $\endgroup$ – Fardad Pouran Apr 10 '15 at 18:38
  • 1
    $\begingroup$ @FardadPouran No, Fubini's Theorem does not "discuss about squared shaped areas" only. I do not know what led you to this erroneous conclusion -- but this could make a good, separate, math.SE question. $\endgroup$ – Did Apr 10 '15 at 18:49
  • $\begingroup$ Thanks. Someone has asked my question here and characteristic functions solve my problem ! : math.stackexchange.com/questions/36558/… $\endgroup$ – Fardad Pouran Apr 10 '15 at 19:18

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.