Is $f: [-1,1]^{2} \rightarrow \mathbb{R}, f(x,y) = \begin{cases} \frac{x^{2}-y^{2}}{(x^{2}+y^{2})^{2}} & \text{ if } (x,y) \neq (0,0)\\ 0& \text{ if } (x,y)= (0,0) \end{cases}$ $\lambda _{2}$-integrable over $[-1,1]^{2}$?
It is Lebesgue measurable, because it is continous on $[-1,1]^{2}$ with exception on (0,0). But this one point doesn´t matter for the measurability. Is it Lebesgue-integrable, though?
No. If it were, \begin{align*} \int_{-1}^{1}\int_{-1}^{1}\dfrac{|x^{2}-y^{2}|}{(x^{2}+y^{2})^{2}}dxdy&\geq\int_{0}^{1}\int_{0}^{y}\dfrac{|x^{2}-y^{2}|}{(x^{2}+y^{2})^{2}}dxdy\\ &\geq\int_{0}^{1}\int_{0}^{y}\dfrac{y^{2}-x^{2}}{(2y^{2})^{2}}dxdy\\ &=\int_{0}^{1}\dfrac{1}{4}\dfrac{1}{y}-\dfrac{1}{4y^{4}}\dfrac{y^{3}}{3}dy\\ &=\dfrac{1}{6}\int_{0}^{1}\dfrac{1}{y}dy\\ &=\infty. \end{align*}