Mixing
![Show the set is connected and open (Complex Analysis) [closed]](https://lh5.googleusercontent.com/-XMmhbX0MoUs/AAAAAAAAAAI/AAAAAAAAACA/XJ0JlQppQgQ/s72-c/photo.jpg?sz=32)
show that $Delta in mathcal{B} otimes mathcal{P}$
$begingroup$
let $lambda$ be the Lebesgue measure on the sigma algebra $ mathcal{B}
= mathcal{B} ([0,1])$ and let $mu$ be the counting measure on the sigma algebra $ mathcal{P} = mathcal{P}([0,1])$
let $Delta = {(x,x), mid , x in [0,1] }$
show that $Delta in mathcal{B} otimes mathcal{P}$
one way to prove it would be to write it as $Delta_1 times Delta_2$ such that $Delta_1 in mathcal{B} $ and $Delta_2 in mathcal{P} $
which I couldn't do
my attempt :
$Delta = bigcup_{x in [0,1]} {(x,x)} $
and ${(x,x)} = {x} times {x } in mathcal{B} times mathcal{P} $
now if the union was countable I would be done but it's not.
hints or any kind help are very welcome.
measure-theory lebesgue-measure
edited Jan 13 at 23:09
Bernard
121k740116
asked Jan 13 at 23:05
rapidracimrapidracim
1,7241419
$endgroup$
add a comment |
$begingroup$
let $lambda$ be the Lebesgue measure on the sigma algebra $ mathcal{B}
= mathcal{B} ([0,1])$ and let $mu$ be the counting measure on the sigma algebra $ mathcal{P} = mathcal{P}([0,1])$
let $Delta = {(x,x), mid , x in [0,1] }$
show that $Delta in mathcal{B} otimes mathcal{P}$
one way to prove it would be to write it as $Delta_1 times Delta_2$ such that $Delta_1 in mathcal{B} $ and $Delta_2 in mathcal{P} $
which I couldn't do
my attempt :
$Delta = bigcup_{x in [0,1]} {(x,x)} $
and ${(x,x)} = {x} times {x } in mathcal{B} times mathcal{P} $
now if the union was countable I would be done but it's not.
hints or any kind help are very welcome.
measure-theory lebesgue-measure
edited Jan 13 at 23:09
Bernard
121k740116
asked Jan 13 at 23:05
rapidracimrapidracim
1,7241419
$endgroup$
1
$begingroup$
$Delta$ is closed in $[0,1]^2$ so is contained in $mathcal{B}([0,1]^2)=mathcal{B} otimes mathcal{B} subset mathcal{B} otimes mathcal{P}$.
$endgroup$
– Mindlack
Jan 13 at 23:41
add a comment |
$begingroup$
let $lambda$ be the Lebesgue measure on the sigma algebra $ mathcal{B}
= mathcal{B} ([0,1])$ and let $mu$ be the counting measure on the sigma algebra $ mathcal{P} = mathcal{P}([0,1])$
let $Delta = {(x,x), mid , x in [0,1] }$
show that $Delta in mathcal{B} otimes mathcal{P}$
one way to prove it would be to write it as $Delta_1 times Delta_2$ such that $Delta_1 in mathcal{B} $ and $Delta_2 in mathcal{P} $
which I couldn't do
my attempt :
$Delta = bigcup_{x in [0,1]} {(x,x)} $
and ${(x,x)} = {x} times {x } in mathcal{B} times mathcal{P} $
now if the union was countable I would be done but it's not.
hints or any kind help are very welcome.
measure-theory lebesgue-measure
edited Jan 13 at 23:09
Bernard
121k740116
asked Jan 13 at 23:05
rapidracimrapidracim
1,7241419
$endgroup$
let $lambda$ be the Lebesgue measure on the sigma algebra $ mathcal{B}
= mathcal{B} ([0,1])$ and let $mu$ be the counting measure on the sigma algebra $ mathcal{P} = mathcal{P}([0,1])$
let $Delta = {(x,x), mid , x in [0,1] }$
show that $Delta in mathcal{B} otimes mathcal{P}$
one way to prove it would be to write it as $Delta_1 times Delta_2$ such that $Delta_1 in mathcal{B} $ and $Delta_2 in mathcal{P} $
which I couldn't do
my attempt :
$Delta = bigcup_{x in [0,1]} {(x,x)} $
and ${(x,x)} = {x} times {x } in mathcal{B} times mathcal{P} $
now if the union was countable I would be done but it's not.
hints or any kind help are very welcome.
measure-theory lebesgue-measure
measure-theory lebesgue-measure
edited Jan 13 at 23:09
Bernard
121k740116
asked Jan 13 at 23:05
rapidracimrapidracim
1,7241419
edited Jan 13 at 23:09
Bernard
121k740116
asked Jan 13 at 23:05
rapidracimrapidracim
1,7241419
edited Jan 13 at 23:09
Bernard
121k740116
edited Jan 13 at 23:09
Bernard
121k740116
edited Jan 13 at 23:09
Bernard
121k740116
121k740116
asked Jan 13 at 23:05
rapidracimrapidracim
1,7241419
asked Jan 13 at 23:05
rapidracimrapidracim
1,7241419
asked Jan 13 at 23:05
rapidracimrapidracim
1,7241419
1,7241419
1
$begingroup$
$Delta$ is closed in $[0,1]^2$ so is contained in $mathcal{B}([0,1]^2)=mathcal{B} otimes mathcal{B} subset mathcal{B} otimes mathcal{P}$.
$endgroup$
– Mindlack
Jan 13 at 23:41
add a comment |
1
$begingroup$
$Delta$ is closed in $[0,1]^2$ so is contained in $mathcal{B}([0,1]^2)=mathcal{B} otimes mathcal{B} subset mathcal{B} otimes mathcal{P}$.
$endgroup$
– Mindlack
Jan 13 at 23:41
1
1
$begingroup$
$Delta$ is closed in $[0,1]^2$ so is contained in $mathcal{B}([0,1]^2)=mathcal{B} otimes mathcal{B} subset mathcal{B} otimes mathcal{P}$.
$endgroup$
– Mindlack
Jan 13 at 23:41
$begingroup$
$Delta$ is closed in $[0,1]^2$ so is contained in $mathcal{B}([0,1]^2)=mathcal{B} otimes mathcal{B} subset mathcal{B} otimes mathcal{P}$.
$endgroup$
– Mindlack
Jan 13 at 23:41
add a comment |
1 Answer
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$begingroup$
Can you verify that $Delta^{c}=(cup_q [(y>qcap (x<q)]) cup (cup_q [(x>qcap (y<q)])$ where $q$ ranges over rational numbers and conclude that $Delta^{c}$, hence $Delta$ belongs to $mathcal B otimes mathcal P$?. [$Delta^{c}$ is the complement of $Delta $]
answered Jan 13 at 23:48
Kavi Rama MurthyKavi Rama Murthy
60.2k42161
$endgroup$
add a comment |
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1 Answer
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1 Answer
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$begingroup$
Can you verify that $Delta^{c}=(cup_q [(y>qcap (x<q)]) cup (cup_q [(x>qcap (y<q)])$ where $q$ ranges over rational numbers and conclude that $Delta^{c}$, hence $Delta$ belongs to $mathcal B otimes mathcal P$?. [$Delta^{c}$ is the complement of $Delta $]
answered Jan 13 at 23:48
Kavi Rama MurthyKavi Rama Murthy
60.2k42161
$endgroup$
add a comment |
$begingroup$
Can you verify that $Delta^{c}=(cup_q [(y>qcap (x<q)]) cup (cup_q [(x>qcap (y<q)])$ where $q$ ranges over rational numbers and conclude that $Delta^{c}$, hence $Delta$ belongs to $mathcal B otimes mathcal P$?. [$Delta^{c}$ is the complement of $Delta $]
answered Jan 13 at 23:48
Kavi Rama MurthyKavi Rama Murthy
60.2k42161
$endgroup$
add a comment |
$begingroup$
Can you verify that $Delta^{c}=(cup_q [(y>qcap (x<q)]) cup (cup_q [(x>qcap (y<q)])$ where $q$ ranges over rational numbers and conclude that $Delta^{c}$, hence $Delta$ belongs to $mathcal B otimes mathcal P$?. [$Delta^{c}$ is the complement of $Delta $]
answered Jan 13 at 23:48
Kavi Rama MurthyKavi Rama Murthy
60.2k42161
$endgroup$
Can you verify that $Delta^{c}=(cup_q [(y>qcap (x<q)]) cup (cup_q [(x>qcap (y<q)])$ where $q$ ranges over rational numbers and conclude that $Delta^{c}$, hence $Delta$ belongs to $mathcal B otimes mathcal P$?. [$Delta^{c}$ is the complement of $Delta $]
answered Jan 13 at 23:48
Kavi Rama MurthyKavi Rama Murthy
60.2k42161
answered Jan 13 at 23:48
Kavi Rama MurthyKavi Rama Murthy
60.2k42161
answered Jan 13 at 23:48
Kavi Rama MurthyKavi Rama Murthy
60.2k42161
answered Jan 13 at 23:48
Kavi Rama MurthyKavi Rama Murthy
60.2k42161
60.2k42161
add a comment |
add a comment |
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$begingroup$
$Delta$ is closed in $[0,1]^2$ so is contained in $mathcal{B}([0,1]^2)=mathcal{B} otimes mathcal{B} subset mathcal{B} otimes mathcal{P}$.
$endgroup$
– Mindlack
Jan 13 at 23:41