Function $f(x,y)$ is defined as follow: $$f(x,y)=\left\{ \begin{array}{cl} \frac{y^2-x^2}{(x^2+y^2)^2} &, \,0<x, y\le1\\ 0 &,\,x=y=0 \end{array}\right.$$ You know, $$\frac{\pi}{4}=\int_0^1\left( \int_0^1 f(x,y)dx\right)dy\ne\int_0^1\left( \int_0^1 f(x,y)dy\right)dx=-\frac{\pi}{4}$$ So, that means $I(y)=\int_0^1f(x,y)dx$ is integrable on $[0,1]$, but can't change the order of integration. That's what I wonder.
It's related to $f(x)$ isn't undefined at $x=y=0$, isn't it? Because, by definition,
If the funtion $f: P \rightarrow \mathbb{R}$ is continuous in the rectangle $P=\{(x,y)\in\mathbb{R}^2\mid a\le x\le b \text{ and } c\le x \le d\}$, then the integral is integrable over closed interval $[c,d]$ and the following equality holds:$$\int_c^ddy\int_a^bf(x,y)dx=\int_a^bdx\int_c^df(x,y)dy$$
Anyway, I still don't understand the problem. Please explain to me. Thank you.
The function $f$ is defined at $(0,0)$; in fact, $f(0,0)=0$. But it is not continuous at $(0,0)$ and, since the limite $\lim_{(x,y)\to(0,0)}f(x,y)$ does not exist, there is no way of extending it to a continuous function from $[0,1]^2$ into $\Bbb R$. So, your theorem does not apply.