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$f:[0,1]\times [0,1]\to\mathbb R,$ defined by $$f(x,y)= \begin{cases}1,\quad \ \ y\in\mathbb R\text{\\}\mathbb Q\\2x,\quad\text{otherwise}\end{cases}$$.

$1.1$: $\int_0^1f(x,y)dx$ exists for every $y\in[0,1]$ and is equal to $1$.

$1.2$: The iterated integral $\int_0^1(\int_0^1f(x,y)dx)dy$ exists and is $1$.

$1.3$: The double integral $\int_If(x,y)d(x,y)$ does not exist.

I am struggling with solving iterated integrals in general and with this one I don't even know where to start since the values kind of jump from 1 to 2x constantly.

Edit: Got an idea for 1.1.: I made two cases, one for an irrational y and one for the rest. Giving me $\int_0^11dx$ which is 1 and $\int_0^12xdx$ which also is 1.

Could someone give me a short explanation about them and some hints on how to approach these exercises?

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  • $\begingroup$ For 1.1, $y$ is fixed for the integration, so either $f(x,y) = 1$ for all $x$ of $f(x,y) = 2x$ for all $x$. There is no jumping around. At least, I'm not jumping around. $\endgroup$ – copper.hat May 19 '15 at 17:25
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    $\begingroup$ Do you have to take Riemann or Lebesgue integrals? $\endgroup$ – Stephan Kulla May 19 '15 at 17:27
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You will probably want to use Lebesgue's criterion for Riemann integrability: it works quickly for 1.1 and 1.2. For 1.3 you might want to note that there is a version of this criterion for functions of several values too (which essentially says the same: a bounded function is Riemann-integrable $\iff$ it is continuous almost everywhere). Therefore, in fact, your problem is less one on integration and more one on continuity.

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First notes:

(1) have you solved correctly.

For (2) you can plug in your solution $\int_0^1f(x,y)dx=1$ into the integral $\int_0^1\int_0^1f(x,y)dxdy$ to get $\int_0^11dy$ which you can solve easily...

For (3) have a look at Is Dirichlet function Riemann integrable? and https://math.stackexchange.com/a/1167830/32951 (I guess you have to consider the Riemann not the Lebesgue integral)

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  • $\begingroup$ On (3): So basically, because the irrational and rational numbers are too "dense" on each interval $[a,b]$ and the supremum therefore is 1 and the infimum is 0 and they are therefore not equal, it is not Rimann integrable? $\endgroup$ – Rab May 19 '15 at 17:41
  • $\begingroup$ @Rab: Yes. Note, that in your case you will end up with $\int_I 2x dxdy$ as a lower bound, not zero... $\endgroup$ – Stephan Kulla May 19 '15 at 17:44
  • $\begingroup$ How come it's $\int_I2xdxdy$ and not 2x then? Isn't the integral 1 on $I$? $\endgroup$ – Rab May 19 '15 at 17:49
  • $\begingroup$ Sorry, a mistake... I meant $\int_I min(2x,1)dxdy$... $\endgroup$ – Stephan Kulla May 19 '15 at 17:52
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    $\begingroup$ Then the upper bound also isn't 1 but $\int_Imax(2x,1)dxdy$? $\endgroup$ – Rab May 19 '15 at 17:54

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