Let $I = [0 , 1]$; let $Q = I \times I$. Define $f: Q \to \mathbb{R}$ by letting $f(x , y) = 1 /q$ if y is rational and $X = p/q$, where p and q are positive integers with no common factor; let $f(x, y) = 0$ otherwise.
need help with c)
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3have you done the first 2 parts? – Lost1 Jan 16 '14 at 20:22
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yes just need c) – user115722 Jan 16 '14 at 20:24
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Notice, by part 2,
$\int_{y\in [0,1]} f(x,y) \text{d}y = 0$,
you should also find that:
$\int_{x\in [0,1]} f(x,y) \text{d}x = 0$
Note that this means $\int_{x\in[0,1]}\int_{y\in [0,1]} f(x,y) \text{d}y \text{d}x = \int 0 \text{d}x = 0 = \int 0 \text{d}y = \int_{y\in [0,1]} \int_{x\in [0,1]} f(x,y) \text{d}x\text{d}y$
which verifies Fubini's theorem which says you are allowed to exchange the order of integration in this case.
Lost1
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@user115722 This was not clear in your other question, are you just doing Riemann integration or have you done Lebesgue integration? Even if you don't use Lebesgue integral, note you have shown by part (b) that the first two integrals are 0. – Lost1 Jan 16 '14 at 20:42
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@user115722 then please state in your question 1. have you done lebesgue integration? 2. what is the version of fubini's theorem you have been taught? this is a trivial question if the answer is yes to the first question. – Lost1 Jan 16 '14 at 21:02
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@user115722 then why didn't you say so in the question? notice your other integration, someone actually commented 'what is your definition of integration?' – Lost1 Jan 16 '14 at 21:12
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yes i was said riemann.hi say nothing and i deleted.thanks for your answer. – user115722 Jan 16 '14 at 21:21
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yes i solved that.for this problem i said integral exist and equal zero then b) obvious.am i right? – user115722 Jan 16 '14 at 21:34
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