How to prove a function of two variables is Lebesgue measurable? 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?
 A: 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.
