Mixing
Computing double integrals with counting measures
$begingroup$
Exercise: Consider the space $(OmegatimesOmega,mathscr{A}otimesmathscr{A})$ , where $Omega=mathbb{N}$,$mathscr{A}$ is the collections of all subsets of $mathbb{N}$ and $nu=mu$ denote counting measure, that means, $mu(A)=nu(A)=$#A for all $Ainmathscr{A}$. Consider the function $f:OmegatimesOmegatomathbb{R}$ given by:
$f(x,y)=delta_x(E)=begin{cases}
x,::::mbox{ if }y=x:,::xin mathbb{N}\
-x, :mbox{ if }y=x+1:,::xin mathbb{N}\0::::::::text{otherwise}
end{cases},
$
Compute $$intint f dmu:dnu$$ and $$intint f dnu:dmu$$
I computed the integrals but I guess is not right once the question ahead ask me to comment the results. This is how I computed them:
$int f_x(y)dnu(y)=xtimesnu(x)+(-x)nu(x+1)+0nu(emptyset)=x-x+0=0$
$int f_y(x)dmu(x)=xtimesmu(y)+(-x)nu(y-1)+0mu(emptyset)=x-x+0=0$
Then $intint f dmu:dnu=intint f dnu:dmu=0$
Since the counting measure the Fubini's theorem should not hold. However according to my calculations it holds.
Questions:
Are my integral computations right? If not why not? How should I compute the integrals?
Thanks in advance!
integration measure-theory proof-verification
asked Jan 14 at 19:24
Pedro GomesPedro Gomes
1,8522721
$endgroup$
add a comment |
$begingroup$
Exercise: Consider the space $(OmegatimesOmega,mathscr{A}otimesmathscr{A})$ , where $Omega=mathbb{N}$,$mathscr{A}$ is the collections of all subsets of $mathbb{N}$ and $nu=mu$ denote counting measure, that means, $mu(A)=nu(A)=$#A for all $Ainmathscr{A}$. Consider the function $f:OmegatimesOmegatomathbb{R}$ given by:
$f(x,y)=delta_x(E)=begin{cases}
x,::::mbox{ if }y=x:,::xin mathbb{N}\
-x, :mbox{ if }y=x+1:,::xin mathbb{N}\0::::::::text{otherwise}
end{cases},
$
Compute $$intint f dmu:dnu$$ and $$intint f dnu:dmu$$
I computed the integrals but I guess is not right once the question ahead ask me to comment the results. This is how I computed them:
$int f_x(y)dnu(y)=xtimesnu(x)+(-x)nu(x+1)+0nu(emptyset)=x-x+0=0$
$int f_y(x)dmu(x)=xtimesmu(y)+(-x)nu(y-1)+0mu(emptyset)=x-x+0=0$
Then $intint f dmu:dnu=intint f dnu:dmu=0$
Since the counting measure the Fubini's theorem should not hold. However according to my calculations it holds.
Questions:
Are my integral computations right? If not why not? How should I compute the integrals?
Thanks in advance!
integration measure-theory proof-verification
asked Jan 14 at 19:24
Pedro GomesPedro Gomes
1,8522721
$endgroup$
$begingroup$
Saying that Fubini's theorem doesn't hold simply means that you can't use Fubini's theorem to prove the result, not that the result isn't true. It is clearly the case (by your arguments and several others) that if the integral exists, it must be zero. So, the question is: does it exist?
$endgroup$
– Nick Peterson
Jan 14 at 20:08
add a comment |
$begingroup$
Exercise: Consider the space $(OmegatimesOmega,mathscr{A}otimesmathscr{A})$ , where $Omega=mathbb{N}$,$mathscr{A}$ is the collections of all subsets of $mathbb{N}$ and $nu=mu$ denote counting measure, that means, $mu(A)=nu(A)=$#A for all $Ainmathscr{A}$. Consider the function $f:OmegatimesOmegatomathbb{R}$ given by:
$f(x,y)=delta_x(E)=begin{cases}
x,::::mbox{ if }y=x:,::xin mathbb{N}\
-x, :mbox{ if }y=x+1:,::xin mathbb{N}\0::::::::text{otherwise}
end{cases},
$
Compute $$intint f dmu:dnu$$ and $$intint f dnu:dmu$$
I computed the integrals but I guess is not right once the question ahead ask me to comment the results. This is how I computed them:
$int f_x(y)dnu(y)=xtimesnu(x)+(-x)nu(x+1)+0nu(emptyset)=x-x+0=0$
$int f_y(x)dmu(x)=xtimesmu(y)+(-x)nu(y-1)+0mu(emptyset)=x-x+0=0$
Then $intint f dmu:dnu=intint f dnu:dmu=0$
Since the counting measure the Fubini's theorem should not hold. However according to my calculations it holds.
Questions:
Are my integral computations right? If not why not? How should I compute the integrals?
Thanks in advance!
integration measure-theory proof-verification
asked Jan 14 at 19:24
Pedro GomesPedro Gomes
1,8522721
$endgroup$
Exercise: Consider the space $(OmegatimesOmega,mathscr{A}otimesmathscr{A})$ , where $Omega=mathbb{N}$,$mathscr{A}$ is the collections of all subsets of $mathbb{N}$ and $nu=mu$ denote counting measure, that means, $mu(A)=nu(A)=$#A for all $Ainmathscr{A}$. Consider the function $f:OmegatimesOmegatomathbb{R}$ given by:
$f(x,y)=delta_x(E)=begin{cases}
x,::::mbox{ if }y=x:,::xin mathbb{N}\
-x, :mbox{ if }y=x+1:,::xin mathbb{N}\0::::::::text{otherwise}
end{cases},
$
Compute $$intint f dmu:dnu$$ and $$intint f dnu:dmu$$
I computed the integrals but I guess is not right once the question ahead ask me to comment the results. This is how I computed them:
$int f_x(y)dnu(y)=xtimesnu(x)+(-x)nu(x+1)+0nu(emptyset)=x-x+0=0$
$int f_y(x)dmu(x)=xtimesmu(y)+(-x)nu(y-1)+0mu(emptyset)=x-x+0=0$
Then $intint f dmu:dnu=intint f dnu:dmu=0$
Since the counting measure the Fubini's theorem should not hold. However according to my calculations it holds.
Questions:
Are my integral computations right? If not why not? How should I compute the integrals?
Thanks in advance!
integration measure-theory proof-verification
integration measure-theory proof-verification
asked Jan 14 at 19:24
Pedro GomesPedro Gomes
1,8522721
asked Jan 14 at 19:24
Pedro GomesPedro Gomes
1,8522721
asked Jan 14 at 19:24
Pedro GomesPedro Gomes
1,8522721
asked Jan 14 at 19:24
Pedro GomesPedro Gomes
1,8522721
asked Jan 14 at 19:24
Pedro GomesPedro Gomes
1,8522721
1,8522721
$begingroup$
Saying that Fubini's theorem doesn't hold simply means that you can't use Fubini's theorem to prove the result, not that the result isn't true. It is clearly the case (by your arguments and several others) that if the integral exists, it must be zero. So, the question is: does it exist?
$endgroup$
– Nick Peterson
Jan 14 at 20:08
add a comment |
$begingroup$
Saying that Fubini's theorem doesn't hold simply means that you can't use Fubini's theorem to prove the result, not that the result isn't true. It is clearly the case (by your arguments and several others) that if the integral exists, it must be zero. So, the question is: does it exist?
$endgroup$
– Nick Peterson
Jan 14 at 20:08
$begingroup$
Saying that Fubini's theorem doesn't hold simply means that you can't use Fubini's theorem to prove the result, not that the result isn't true. It is clearly the case (by your arguments and several others) that if the integral exists, it must be zero. So, the question is: does it exist?
$endgroup$
– Nick Peterson
Jan 14 at 20:08
$begingroup$
Saying that Fubini's theorem doesn't hold simply means that you can't use Fubini's theorem to prove the result, not that the result isn't true. It is clearly the case (by your arguments and several others) that if the integral exists, it must be zero. So, the question is: does it exist?
$endgroup$
– Nick Peterson
Jan 14 at 20:08
add a comment |
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$begingroup$
Saying that Fubini's theorem doesn't hold simply means that you can't use Fubini's theorem to prove the result, not that the result isn't true. It is clearly the case (by your arguments and several others) that if the integral exists, it must be zero. So, the question is: does it exist?
$endgroup$
– Nick Peterson
Jan 14 at 20:08