$(\Omega,\mathscr{F},\mu):=((-1,1),\mathscr{B}((-1,1)),\lambda)$ is $f(x,y):=\dfrac{xy}{(x^2+y^2)^2}$measurable?

Question

Let $(\Omega,\mathscr{F},\mu):=((-1,1),\mathscr{B}((-1,1)),\lambda)$ be a measure space, where $\lambda$ denotes the Lebesgue measure on $\Omega$, and define the function by

$$ f(x,y):= \begin{cases} \dfrac{xy}{(x^2+y^2)^2}\hspace{0,7cm} \text{if} (x,y)\neq(0,0)\\ 0\hspace{2,6cm}\text{else} \end{cases} $$

Show that $f$ is $\mathscr{F}\otimes\mathscr{F}/\mathbb{R}$-measurable.

We know that continuous maps between $\mathbb{R}^d$ and $\mathbb{R}^n$ are measurable. Since $\mathscr{B}((-1,1))$ is also a topological space we could in case of $f$ being continuous also apply the aforementioned result. However $f$ isn't, so I'm not quite sure how to procced.

Edit: Wouldn't my idea actually work if we were to use $\infty$ instead of $0$ for the "else" case ?

Answer 1

Let $f_n(x,y) =\frac {xy} {(x^{2}+y^2)^{2}}$ for $\|(x,y\| >\frac 1 n $ and $f_n(x,y) =n^{4}xy$ for $\|(x,y\| \leq \frac 1 n $. Then $f_n$ is continuous, hence Borel measurable. Further $f_n(x,y) \to f(x,y)$ for all $(x,y)$. Since pointwise limits of measurable functions is measurable it follows that $f$ is measurable.

Answer 2

The function is almost everywhere continuous and the measure of $\{(0,0)\}$ is zero.

So the function is continuous a.e,so it is measurable.