Mixing
Is the function Riemann integrable
$begingroup$
$f(x,y)=1$ if $x$ is rational and $1leqslant x^2+y^2leqslant10$ and $f(x,y)=0$ in all other cases.
This really resembles the Dirichlet function. I think it is not integrable because if we take $x$ to be rational there is a an open ball with center $(x,y)$ where the function has value $0$ and the same if x is irrational. So the function is nowhere continuous then is no integrable.
Am I right? And how can I write it more clearly.
calculus
asked Jan 27 at 10:04
spyerspyer
1188
$endgroup$
add a comment |
$begingroup$
$f(x,y)=1$ if $x$ is rational and $1leqslant x^2+y^2leqslant10$ and $f(x,y)=0$ in all other cases.
This really resembles the Dirichlet function. I think it is not integrable because if we take $x$ to be rational there is a an open ball with center $(x,y)$ where the function has value $0$ and the same if x is irrational. So the function is nowhere continuous then is no integrable.
Am I right? And how can I write it more clearly.
calculus
asked Jan 27 at 10:04
spyerspyer
1188
$endgroup$
$begingroup$
If you already have the theorem in hand that Riemann integrable functions are continuous almost everywhere, then yes, this works. But if that theorem is not in your current toolbox, then your argument is not available. Instead, you could show that that for any partition, the upper and lower Riemann sums have values that converge to $24pi$ and $0$ respectively. Since they do not converge to the same value, the functions is not integrable.
$endgroup$
– Paul Sinclair
Jan 27 at 17:07
add a comment |
$begingroup$
$f(x,y)=1$ if $x$ is rational and $1leqslant x^2+y^2leqslant10$ and $f(x,y)=0$ in all other cases.
This really resembles the Dirichlet function. I think it is not integrable because if we take $x$ to be rational there is a an open ball with center $(x,y)$ where the function has value $0$ and the same if x is irrational. So the function is nowhere continuous then is no integrable.
Am I right? And how can I write it more clearly.
calculus
asked Jan 27 at 10:04
spyerspyer
1188
$endgroup$
$f(x,y)=1$ if $x$ is rational and $1leqslant x^2+y^2leqslant10$ and $f(x,y)=0$ in all other cases.
This really resembles the Dirichlet function. I think it is not integrable because if we take $x$ to be rational there is a an open ball with center $(x,y)$ where the function has value $0$ and the same if x is irrational. So the function is nowhere continuous then is no integrable.
Am I right? And how can I write it more clearly.
calculus
calculus
asked Jan 27 at 10:04
spyerspyer
1188
asked Jan 27 at 10:04
spyerspyer
1188
asked Jan 27 at 10:04
spyerspyer
1188
asked Jan 27 at 10:04
spyerspyer
1188
asked Jan 27 at 10:04
spyerspyer
1188
1188
$begingroup$
If you already have the theorem in hand that Riemann integrable functions are continuous almost everywhere, then yes, this works. But if that theorem is not in your current toolbox, then your argument is not available. Instead, you could show that that for any partition, the upper and lower Riemann sums have values that converge to $24pi$ and $0$ respectively. Since they do not converge to the same value, the functions is not integrable.
$endgroup$
– Paul Sinclair
Jan 27 at 17:07
add a comment |
$begingroup$
If you already have the theorem in hand that Riemann integrable functions are continuous almost everywhere, then yes, this works. But if that theorem is not in your current toolbox, then your argument is not available. Instead, you could show that that for any partition, the upper and lower Riemann sums have values that converge to $24pi$ and $0$ respectively. Since they do not converge to the same value, the functions is not integrable.
$endgroup$
– Paul Sinclair
Jan 27 at 17:07
$begingroup$
If you already have the theorem in hand that Riemann integrable functions are continuous almost everywhere, then yes, this works. But if that theorem is not in your current toolbox, then your argument is not available. Instead, you could show that that for any partition, the upper and lower Riemann sums have values that converge to $24pi$ and $0$ respectively. Since they do not converge to the same value, the functions is not integrable.
$endgroup$
– Paul Sinclair
Jan 27 at 17:07
$begingroup$
If you already have the theorem in hand that Riemann integrable functions are continuous almost everywhere, then yes, this works. But if that theorem is not in your current toolbox, then your argument is not available. Instead, you could show that that for any partition, the upper and lower Riemann sums have values that converge to $24pi$ and $0$ respectively. Since they do not converge to the same value, the functions is not integrable.
$endgroup$
– Paul Sinclair
Jan 27 at 17:07
add a comment |
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$begingroup$
If you already have the theorem in hand that Riemann integrable functions are continuous almost everywhere, then yes, this works. But if that theorem is not in your current toolbox, then your argument is not available. Instead, you could show that that for any partition, the upper and lower Riemann sums have values that converge to $24pi$ and $0$ respectively. Since they do not converge to the same value, the functions is not integrable.
$endgroup$
– Paul Sinclair
Jan 27 at 17:07