Justify the statement that $\int_0^1\int_0^1 \frac{(x-y)\sin(xy)}{x^2+y^2} \ dx \ dy=\int_0^1\int_0^1 \frac{(x-y)\sin(xy)}{x^2+y^2} \ dy \ dx$ Attempt: 
I think I have to use the Fubini-Tonelli theorem. First, note that $[0,1]$ with the Lebesgue measure is $\sigma$ finite. Next, away from the origin, $f(x,y)=\frac{(x-y)\sin(xy)}{x^2+y^2}$ is $\mathbb{[0,1]} \times \mathbb{[0,1]}$ measurable since $f(x,y)$ is continuous. Finally,
\begin{align*}
\int_0^1\left(\int_0^1\big|f(x,y)\big| \ dx\right) \ dy &\leq\int_0^1\left(\int_0^1\bigg|\frac{x-y}{x^2+y^2}\bigg| \ dx\right) \ dy\\
&=\int_0^1\int_0^y\frac{y-x}{x^2+y^2} \ dx \ dy+\int_0^1\int_y^1 \frac{x-y}{x^2+y^2} \ dx \ dy \\
\end{align*}
I need to show that the above is finite.
Questions:
Is this the correct way to go about doing this? How can I show that $f(x,y)$ is measurable at the origin? Since I am studying for a test, I would appreciate a rigorous answer addressing any technical points I have missed.
 A: Note that the given $f$ is not defined at $(0,0)$.
The function $g(x,y) = \begin{cases} f(x,y) & (x,y) \neq 0 \\
0, & \text{otherwise} \end{cases}$ is measurable since it is
continuous everywhere (on $[0,1]^2$). Since $f=g$ ae. and the
Lebesgue measure is complete, we see that $f$ is measurable.
Note that $0 \le (x-y)^2 = x^2+y^2 - 2xy$, so
${x y \over x^2+y^2} \le 2$ for $(x,y) \neq 0$. Since
$|\sin(xy)| \le |xy|$ we see that
$|x-y| {|\sin (xy)| \over x^2+y^2 } \le 2 |x-y| \le 2$.
Since $(x,y) \to 2$ is integrable on $[0,1]^2$ we see that $f$
is integrable (with respect to the product measure).
A: Let $U$ be a small neighbourhood of the origin in $Q=[0,1]\times[0,1]$. Both $x-y$ and $\frac{\sin(xy)}{x^2+y^2}$ are differentiable functions over $Q\setminus U$ and bounded functions over $Q$, since:
$$\left|\frac{\sin(xy)}{x^2+y^2}\right|\leq\frac{|xy|}{x^2+y^2}\leq\frac{1}{2}$$
by the AM-GM inequality. So we have that $(x-y)\cdot \frac{\sin(xy)}{x^2+y^2}$ is an integrable function over $Q$ equipped with the product measure, hence Fubini's theorem holds. Then, by symmetry:
$$ \iint_{Q}\frac{(x-y)\sin(xy)}{x^2+y^2}\,dx\,dy = \iint_{Q}\frac{(y-x)\sin(xy)}{x^2+y^2}\,dx\,dy = \color{red}{0}.$$
