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Consider the function $f: [0,1]\times [0,1] \rightarrow [-\infty,\infty]$ given by

$$f(x,y) = \frac{x^2-y^2}{(x^2+y^2)^2}, (x,y)\neq (0,0)$$

and $f(x,y) = 0$ when $(x,y) = (0,0)$.

I am trying to show that $f$ is $M\times M$ measurable, where $M$ is the $\sigma$-algebra of Lebesgue measurable sets.

However I am not sure how to show this. I thought maybe if $f$ where continous we would be done, but $f$ is not even continous since as $(x,y)\rightarrow (0,0)$ the function $f$ does not approach $0$.

Must I show that for any $t\in\mathbb{R}$, $f^{-1}(t,\infty)\in M\times M$? This is quite difficult.

How can I show this?

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  • $\begingroup$ Hint: Try to use $f_n(x,y)$ : $$f_n(x,y)=\frac{x^2-y^2}{(x²+y²)^2+1/n}$$ $\endgroup$ – Kelenner Jun 24 '17 at 12:52
  • $\begingroup$ Is the idea that these are continous functions and so they are $M\times M$ measurable and they are tending to $f$? so $f$ must also be $M\times M$ measurable? $\endgroup$ – fosho Jun 24 '17 at 12:54
  • $\begingroup$ Yes this is the idea $\endgroup$ – Kelenner Jun 24 '17 at 12:55
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You're right that $f$ is only continuous on $\mathbb R^2 - \{ (0,0) \}$, but the omission of the point $\{(0,0) \}$ should not cause any major difficulty.

Let me spell out the logic. Suppose we pick a half-open interval $(a, \infty]$, for some $a \in \mathbb R$. Our task is to show that $f^{-1}(a,\infty]$ is a measurable with respect to the product measure on $\mathbb R^2$. There are two cases to consider:

  • $a > 0$: In this case, $f^{-1}(a,\infty] = (f|_{\mathbb R^2 - \{ (0,0)\} })^{-1} (a, \infty]$, which is open because $f|_{\mathbb R^2 - \{ (0,0) \}}$ is a continuous function, hence is a Borel set, hence is measurable with respect to the product of the Lebesgue measures on the two $\mathbb R$ factors.

  • $a \leq 0$: In this case, $f^{-1}(a, \infty] = (f|_{\mathbb R^2 - \{ (0,0) \} })^{-1}(a, \infty]\cup \{ (0,0) \}$, which is the union of an open set and the singleton set $\{(0,0)\}$. Both these sets are Borel sets, hence also measurable with respect to the product measure, and therefore, their union is measurable with respect to the product measure too.

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  • $\begingroup$ Wait. How do we know that $f^{-1} (a,\infty) = f(a,\infty)$? $\endgroup$ – fosho Jun 24 '17 at 13:00
  • $\begingroup$ Sorry, that was a typo! $\endgroup$ – Kenny Wong Jun 24 '17 at 13:02
  • $\begingroup$ Ah! Thanks a lot, makes much more sense now. Thanks for the help :) $\endgroup$ – fosho Jun 24 '17 at 13:03

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