# About the double integral and fubini's theorem

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 ?

$$1)\int_{[0,1]}\left(\int_{[0,1 ]}f(x,y)\,\text{d}y\right)\,\text{d}x=1$$

$$2)\int_{[0,1]}\left(\int_{[0,1]}f(x,y)\,\text{d}y\right)\,\text{d}x=0$$

$$3)\int_{[0,1]\times [0,1]} f(x,y)\,\text{dxdy}=1$$ $$4)\int_{[0,1]\times [0,1]} f(x,y)\,\text{dxdy}=0$$

• $(m,n)=(q,n)=1$? – d.k.o. Feb 15 '19 at 8:02
• You don't need continuity to apply Fubini's Theorem. 2) and 4) are both true and 1) and 3) are false. – Kavi Rama Murthy Feb 15 '19 at 8:04
• @d.k.o (a,b) is greatest common divisor (gcd) of a,b. – 1200785626 Feb 15 '19 at 8:06
• @KaviRamaMurthy . Why 2 is true ? If x be fix and x be irrational we have another answer . – 1200785626 Feb 15 '19 at 8:09
• If $x$ is irrational then $f(x,y)=0$ for all $y$. – Kavi Rama Murthy Feb 15 '19 at 8:11

Note that $$f=0$$ a.e. on $$[0,1]\times [0,1]$$. Conclusion ?