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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 ?

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

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

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

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

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