Let $E = [0,1]\times[0,1]$ and $f: E \rightarrow \mathbb{R}$ given by $f(x,y) = 2^n, \ \text{if}\ x \in [\frac{1}{2^n}, \frac{1}{2^{n-1}}] \ \text{and} \ y \in [\frac{1}{2^n}, \frac{1}{2^{n-1}}); \ \ f(x,y) = -2^{2n +1}, \ \text{if} \ x \in [\frac{1}{2^{n+1}}, \frac{1}{2^{n}}) \ \text{and} \ y \in [\frac{1}{2^n}, \frac{1}{2^{n-1}}); f(x,y) = 0, \ \text{otherwise}.$
How to compute $\int_{0}^{1} \Big(\int_{0}^{1} f(x,y)dx \Big)dy$ and $\int_{0}^{1} \Big(\int_{0}^{1} f(x,y)dy \Big)dx$ and show that it gives us $0$ and $1/4$, respectively? I tried to explicit the sets $E_y = \{x \in [0,1]: (x,y) \in E\}$ and $E_x$ to compute these integrals, but i got completely confused with that and didn't get the result.
Can anyone help me with this?