Let $f:[0,1]\times [0,1]\to\mathbb{R}$ such that $f(x,y)=1 $ if $x= \frac{m}{n} ,y= \frac{q}{n}, (m,n)=(q,n)=1 $ otherwise $f(x,y) =0$ ,Now which of following options is true ?
I think because $f(x,y) =0 ,a.e$ then option 4 is true and because $f$ is not continue so we can't use fobini`s theorem is this true ?
$$1)\int_{[0,1]}\left(\int_{[0,1 ]}f(x,y)\,\text{d}y\right)\,\text{d}x=1$$
$$2)\int_{[0,1]}\left(\int_{[0,1]}f(x,y)\,\text{d}y\right)\,\text{d}x=0$$
$$3)\int_{[0,1]\times [0,1]} f(x,y)\,\text{dxdy}=1$$ $$4)\int_{[0,1]\times [0,1]} f(x,y)\,\text{dxdy}=0$$
Note that $f=0$ a.e. on $[0,1]\times [0,1]$. Conclusion ?