If $f(x,y)>0$ for almost all $(x,y)$, can we conclude that $min(f(x,y),f(y,x))>0$ for almost all...
$begingroup$
Let $(E,mathcal E,mu)$ be a measure space and $f:E^2to[0,infty)$ be $mathcal E^{otimes2}$-measurable with $$f(x,y)>0;;;text{for }mu^{otimes2}text{-almost all }(x,y)in E^2.tag1$$
Are we able to conclude that $$g(x,y):=min(f(x,y),f(y,x))>0;;;text{for }mu^{otimes2}text{-almost all }(x,y)in E^2text?tag2$$
It would be trivial, if we would have $$f(x,y)>0;;;text{for }mutext{-almost all }x,yin Etag{1'}$$ instead of $(1)$, but with $(1)$ I'm unsure how we need to argue. The problem is that by assumption there is a $mu^{otimes2}$-null set $N$ with $f(x,y)>0$ for all $(x,y)in E^2setminus N$, but $(x,y)in E^2setminus N$ doesn't imply $(y,x)in E^2setminus N$.
real-analysis measure-theory
asked Jan 15 at 10:42
0xbadf00d0xbadf00d
1,59341534
$endgroup$
add a comment |
$begingroup$
Let $(E,mathcal E,mu)$ be a measure space and $f:E^2to[0,infty)$ be $mathcal E^{otimes2}$-measurable with $$f(x,y)>0;;;text{for }mu^{otimes2}text{-almost all }(x,y)in E^2.tag1$$
Are we able to conclude that $$g(x,y):=min(f(x,y),f(y,x))>0;;;text{for }mu^{otimes2}text{-almost all }(x,y)in E^2text?tag2$$
It would be trivial, if we would have $$f(x,y)>0;;;text{for }mutext{-almost all }x,yin Etag{1'}$$ instead of $(1)$, but with $(1)$ I'm unsure how we need to argue. The problem is that by assumption there is a $mu^{otimes2}$-null set $N$ with $f(x,y)>0$ for all $(x,y)in E^2setminus N$, but $(x,y)in E^2setminus N$ doesn't imply $(y,x)in E^2setminus N$.
real-analysis measure-theory
asked Jan 15 at 10:42
0xbadf00d0xbadf00d
1,59341534
$endgroup$
add a comment |
$begingroup$
Let $(E,mathcal E,mu)$ be a measure space and $f:E^2to[0,infty)$ be $mathcal E^{otimes2}$-measurable with $$f(x,y)>0;;;text{for }mu^{otimes2}text{-almost all }(x,y)in E^2.tag1$$
Are we able to conclude that $$g(x,y):=min(f(x,y),f(y,x))>0;;;text{for }mu^{otimes2}text{-almost all }(x,y)in E^2text?tag2$$
It would be trivial, if we would have $$f(x,y)>0;;;text{for }mutext{-almost all }x,yin Etag{1'}$$ instead of $(1)$, but with $(1)$ I'm unsure how we need to argue. The problem is that by assumption there is a $mu^{otimes2}$-null set $N$ with $f(x,y)>0$ for all $(x,y)in E^2setminus N$, but $(x,y)in E^2setminus N$ doesn't imply $(y,x)in E^2setminus N$.
real-analysis measure-theory
asked Jan 15 at 10:42
0xbadf00d0xbadf00d
1,59341534
$endgroup$
Let $(E,mathcal E,mu)$ be a measure space and $f:E^2to[0,infty)$ be $mathcal E^{otimes2}$-measurable with $$f(x,y)>0;;;text{for }mu^{otimes2}text{-almost all }(x,y)in E^2.tag1$$
Are we able to conclude that $$g(x,y):=min(f(x,y),f(y,x))>0;;;text{for }mu^{otimes2}text{-almost all }(x,y)in E^2text?tag2$$
It would be trivial, if we would have $$f(x,y)>0;;;text{for }mutext{-almost all }x,yin Etag{1'}$$ instead of $(1)$, but with $(1)$ I'm unsure how we need to argue. The problem is that by assumption there is a $mu^{otimes2}$-null set $N$ with $f(x,y)>0$ for all $(x,y)in E^2setminus N$, but $(x,y)in E^2setminus N$ doesn't imply $(y,x)in E^2setminus N$.
real-analysis measure-theory
real-analysis measure-theory
asked Jan 15 at 10:42
0xbadf00d0xbadf00d
1,59341534
asked Jan 15 at 10:42
0xbadf00d0xbadf00d
1,59341534
asked Jan 15 at 10:42
0xbadf00d0xbadf00d
1,59341534
asked Jan 15 at 10:42
0xbadf00d0xbadf00d
1,59341534
asked Jan 15 at 10:42
0xbadf00d0xbadf00d
1,59341534
1,59341534
add a comment |
add a comment |
1 Answer
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If $N$ is a $mu^{otimes2}$-null set then so are $N'={(y,x)mid (x,y)in N}$ (by symmetry) and $M=Ncup N'$.
And if: $$f(x,y)leq 0implies (x,y)in N$$ then: $$f(y,x)leq 0implies (x,y)in N'$$ and: $$g(x,y)=min(f(x,y),f(y,x))leq0implies (x,y)in M$$
answered Jan 15 at 10:51
drhabdrhab
104k545136
$endgroup$
$begingroup$
Why is $N'$ a null set?
$endgroup$
– 0xbadf00d
Jan 15 at 11:56
1
$begingroup$
Because $(muotimesmu)(N)=(muotimesmu)(N')$ by symmetry. We have $(muotimesmu)(Atimes B)=mu(A)mu(B)=mu(B)mu(A)=(muotimesmu)(Btimes A)$. This can be extended to $(muotimesmu)(N)=(muotimesmu)(N')$ where $N'={(y,x)mid (x,y)in N}$.
$endgroup$
– drhab
Jan 15 at 12:00
$begingroup$
Sure, trivial. Thank you.
$endgroup$
– 0xbadf00d
Jan 15 at 12:05
$begingroup$
You are welcome.
$endgroup$
– drhab
Jan 15 at 12:06
add a comment |
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1 Answer
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$begingroup$
If $N$ is a $mu^{otimes2}$-null set then so are $N'={(y,x)mid (x,y)in N}$ (by symmetry) and $M=Ncup N'$.
And if: $$f(x,y)leq 0implies (x,y)in N$$ then: $$f(y,x)leq 0implies (x,y)in N'$$ and: $$g(x,y)=min(f(x,y),f(y,x))leq0implies (x,y)in M$$
answered Jan 15 at 10:51
drhabdrhab
104k545136
$endgroup$
$begingroup$
Why is $N'$ a null set?
$endgroup$
– 0xbadf00d
Jan 15 at 11:56
1
$begingroup$
Because $(muotimesmu)(N)=(muotimesmu)(N')$ by symmetry. We have $(muotimesmu)(Atimes B)=mu(A)mu(B)=mu(B)mu(A)=(muotimesmu)(Btimes A)$. This can be extended to $(muotimesmu)(N)=(muotimesmu)(N')$ where $N'={(y,x)mid (x,y)in N}$.
$endgroup$
– drhab
Jan 15 at 12:00
$begingroup$
Sure, trivial. Thank you.
$endgroup$
– 0xbadf00d
Jan 15 at 12:05
$begingroup$
You are welcome.
$endgroup$
– drhab
Jan 15 at 12:06
add a comment |
$begingroup$
If $N$ is a $mu^{otimes2}$-null set then so are $N'={(y,x)mid (x,y)in N}$ (by symmetry) and $M=Ncup N'$.
And if: $$f(x,y)leq 0implies (x,y)in N$$ then: $$f(y,x)leq 0implies (x,y)in N'$$ and: $$g(x,y)=min(f(x,y),f(y,x))leq0implies (x,y)in M$$
answered Jan 15 at 10:51
drhabdrhab
104k545136
$endgroup$
$begingroup$
Why is $N'$ a null set?
$endgroup$
– 0xbadf00d
Jan 15 at 11:56
1
$begingroup$
Because $(muotimesmu)(N)=(muotimesmu)(N')$ by symmetry. We have $(muotimesmu)(Atimes B)=mu(A)mu(B)=mu(B)mu(A)=(muotimesmu)(Btimes A)$. This can be extended to $(muotimesmu)(N)=(muotimesmu)(N')$ where $N'={(y,x)mid (x,y)in N}$.
$endgroup$
– drhab
Jan 15 at 12:00
$begingroup$
Sure, trivial. Thank you.
$endgroup$
– 0xbadf00d
Jan 15 at 12:05
$begingroup$
You are welcome.
$endgroup$
– drhab
Jan 15 at 12:06
add a comment |
$begingroup$
If $N$ is a $mu^{otimes2}$-null set then so are $N'={(y,x)mid (x,y)in N}$ (by symmetry) and $M=Ncup N'$.
And if: $$f(x,y)leq 0implies (x,y)in N$$ then: $$f(y,x)leq 0implies (x,y)in N'$$ and: $$g(x,y)=min(f(x,y),f(y,x))leq0implies (x,y)in M$$
answered Jan 15 at 10:51
drhabdrhab
104k545136
$endgroup$
If $N$ is a $mu^{otimes2}$-null set then so are $N'={(y,x)mid (x,y)in N}$ (by symmetry) and $M=Ncup N'$.
And if: $$f(x,y)leq 0implies (x,y)in N$$ then: $$f(y,x)leq 0implies (x,y)in N'$$ and: $$g(x,y)=min(f(x,y),f(y,x))leq0implies (x,y)in M$$
answered Jan 15 at 10:51
drhabdrhab
104k545136
answered Jan 15 at 10:51
drhabdrhab
104k545136
answered Jan 15 at 10:51
drhabdrhab
104k545136
answered Jan 15 at 10:51
drhabdrhab
104k545136
104k545136
$begingroup$
Why is $N'$ a null set?
$endgroup$
– 0xbadf00d
Jan 15 at 11:56
1
$begingroup$
Because $(muotimesmu)(N)=(muotimesmu)(N')$ by symmetry. We have $(muotimesmu)(Atimes B)=mu(A)mu(B)=mu(B)mu(A)=(muotimesmu)(Btimes A)$. This can be extended to $(muotimesmu)(N)=(muotimesmu)(N')$ where $N'={(y,x)mid (x,y)in N}$.
$endgroup$
– drhab
Jan 15 at 12:00
$begingroup$
Sure, trivial. Thank you.
$endgroup$
– 0xbadf00d
Jan 15 at 12:05
$begingroup$
You are welcome.
$endgroup$
– drhab
Jan 15 at 12:06
add a comment |
$begingroup$
Why is $N'$ a null set?
$endgroup$
– 0xbadf00d
Jan 15 at 11:56
1
$begingroup$
Because $(muotimesmu)(N)=(muotimesmu)(N')$ by symmetry. We have $(muotimesmu)(Atimes B)=mu(A)mu(B)=mu(B)mu(A)=(muotimesmu)(Btimes A)$. This can be extended to $(muotimesmu)(N)=(muotimesmu)(N')$ where $N'={(y,x)mid (x,y)in N}$.
$endgroup$
– drhab
Jan 15 at 12:00
$begingroup$
Sure, trivial. Thank you.
$endgroup$
– 0xbadf00d
Jan 15 at 12:05
$begingroup$
You are welcome.
$endgroup$
– drhab
Jan 15 at 12:06
$begingroup$
Why is $N'$ a null set?
$endgroup$
– 0xbadf00d
Jan 15 at 11:56
$begingroup$
Why is $N'$ a null set?
$endgroup$
– 0xbadf00d
Jan 15 at 11:56
1
1
$begingroup$
Because $(muotimesmu)(N)=(muotimesmu)(N')$ by symmetry. We have $(muotimesmu)(Atimes B)=mu(A)mu(B)=mu(B)mu(A)=(muotimesmu)(Btimes A)$. This can be extended to $(muotimesmu)(N)=(muotimesmu)(N')$ where $N'={(y,x)mid (x,y)in N}$.
$endgroup$
– drhab
Jan 15 at 12:00
$begingroup$
Because $(muotimesmu)(N)=(muotimesmu)(N')$ by symmetry. We have $(muotimesmu)(Atimes B)=mu(A)mu(B)=mu(B)mu(A)=(muotimesmu)(Btimes A)$. This can be extended to $(muotimesmu)(N)=(muotimesmu)(N')$ where $N'={(y,x)mid (x,y)in N}$.
$endgroup$
– drhab
Jan 15 at 12:00
$begingroup$
Sure, trivial. Thank you.
$endgroup$
– 0xbadf00d
Jan 15 at 12:05
$begingroup$
Sure, trivial. Thank you.
$endgroup$
– 0xbadf00d
Jan 15 at 12:05
$begingroup$
You are welcome.
$endgroup$
– drhab
Jan 15 at 12:06
$begingroup$
You are welcome.
$endgroup$
– drhab
Jan 15 at 12:06
add a comment |
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