If $f_x$ is $T$-measurable, $f^y$ is $S$-measurable, is $f$ $(Stimes T)$-measurable?
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Rudin's Real & Complex Analysis Theorem 8.5 states
Let $f$ be an $(Stimes T)$-measurable function on $X times Y$. Then:
a) $forall xin X, f_x $ is $T$-measurable function;
b) $forall yin Y, f^y $ is $S$-measurable function
where $f_x(y) = f(x,y)$, $f^y(x)=f(x,y)$.
This statement is simple and nice. I am thinking about is inverse also true? That is, if $f_x$ is $T$-measurable, $f^y$ is $S$-measurable, then $f$ is $(Stimes T)$-measurable.
Or we need some more strong condition on $f_x$ and $f^y$ to show $f$ is $(Stimes T)$-measurable.
real-analysis
asked Nov 15 at 4:02
Awoo
448
add a comment |
up vote
0
down vote
favorite
Rudin's Real & Complex Analysis Theorem 8.5 states
Let $f$ be an $(Stimes T)$-measurable function on $X times Y$. Then:
a) $forall xin X, f_x $ is $T$-measurable function;
b) $forall yin Y, f^y $ is $S$-measurable function
where $f_x(y) = f(x,y)$, $f^y(x)=f(x,y)$.
This statement is simple and nice. I am thinking about is inverse also true? That is, if $f_x$ is $T$-measurable, $f^y$ is $S$-measurable, then $f$ is $(Stimes T)$-measurable.
Or we need some more strong condition on $f_x$ and $f^y$ to show $f$ is $(Stimes T)$-measurable.
real-analysis
asked Nov 15 at 4:02
Awoo
448
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Rudin's Real & Complex Analysis Theorem 8.5 states
Let $f$ be an $(Stimes T)$-measurable function on $X times Y$. Then:
a) $forall xin X, f_x $ is $T$-measurable function;
b) $forall yin Y, f^y $ is $S$-measurable function
where $f_x(y) = f(x,y)$, $f^y(x)=f(x,y)$.
This statement is simple and nice. I am thinking about is inverse also true? That is, if $f_x$ is $T$-measurable, $f^y$ is $S$-measurable, then $f$ is $(Stimes T)$-measurable.
Or we need some more strong condition on $f_x$ and $f^y$ to show $f$ is $(Stimes T)$-measurable.
real-analysis
asked Nov 15 at 4:02
Awoo
448
Rudin's Real & Complex Analysis Theorem 8.5 states
Let $f$ be an $(Stimes T)$-measurable function on $X times Y$. Then:
a) $forall xin X, f_x $ is $T$-measurable function;
b) $forall yin Y, f^y $ is $S$-measurable function
where $f_x(y) = f(x,y)$, $f^y(x)=f(x,y)$.
This statement is simple and nice. I am thinking about is inverse also true? That is, if $f_x$ is $T$-measurable, $f^y$ is $S$-measurable, then $f$ is $(Stimes T)$-measurable.
Or we need some more strong condition on $f_x$ and $f^y$ to show $f$ is $(Stimes T)$-measurable.
real-analysis
real-analysis
asked Nov 15 at 4:02
Awoo
448
asked Nov 15 at 4:02
Awoo
448
edited Nov 15 at 4:08
asked Nov 15 at 4:02
Awoo
448
asked Nov 15 at 4:02
Awoo
448
asked Nov 15 at 4:02
Awoo
448
448
add a comment |
add a comment |
1 Answer
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The inverse statement is false. Let $E$ be a Vitali-set, definie $f(x,y) = 1_{E times E}(x,y)$. Then $f^x(y) = 1$ if $x=y$ and $y in E$, otherwise zero. Thus Borel-measurable, but $f$ is not!
We need, in fact, stronger condtions:
$S$ is a separable metric space.
$x mapsto f(x,y)$ is continuous for all $y in T$.
$y mapsto f(x,y)$ is measurable for all $x in S$.
Then $f$ is measurable. (Proof can be found in many standard books on measure theory.)
answered Nov 15 at 10:36
p4sch
3,790216
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
The inverse statement is false. Let $E$ be a Vitali-set, definie $f(x,y) = 1_{E times E}(x,y)$. Then $f^x(y) = 1$ if $x=y$ and $y in E$, otherwise zero. Thus Borel-measurable, but $f$ is not!
We need, in fact, stronger condtions:
$S$ is a separable metric space.
$x mapsto f(x,y)$ is continuous for all $y in T$.
$y mapsto f(x,y)$ is measurable for all $x in S$.
Then $f$ is measurable. (Proof can be found in many standard books on measure theory.)
answered Nov 15 at 10:36
p4sch
3,790216
add a comment |
up vote
1
down vote
The inverse statement is false. Let $E$ be a Vitali-set, definie $f(x,y) = 1_{E times E}(x,y)$. Then $f^x(y) = 1$ if $x=y$ and $y in E$, otherwise zero. Thus Borel-measurable, but $f$ is not!
We need, in fact, stronger condtions:
$S$ is a separable metric space.
$x mapsto f(x,y)$ is continuous for all $y in T$.
$y mapsto f(x,y)$ is measurable for all $x in S$.
Then $f$ is measurable. (Proof can be found in many standard books on measure theory.)
answered Nov 15 at 10:36
p4sch
3,790216
add a comment |
up vote
1
down vote
up vote
1
down vote
The inverse statement is false. Let $E$ be a Vitali-set, definie $f(x,y) = 1_{E times E}(x,y)$. Then $f^x(y) = 1$ if $x=y$ and $y in E$, otherwise zero. Thus Borel-measurable, but $f$ is not!
We need, in fact, stronger condtions:
$S$ is a separable metric space.
$x mapsto f(x,y)$ is continuous for all $y in T$.
$y mapsto f(x,y)$ is measurable for all $x in S$.
Then $f$ is measurable. (Proof can be found in many standard books on measure theory.)
answered Nov 15 at 10:36
p4sch
3,790216
The inverse statement is false. Let $E$ be a Vitali-set, definie $f(x,y) = 1_{E times E}(x,y)$. Then $f^x(y) = 1$ if $x=y$ and $y in E$, otherwise zero. Thus Borel-measurable, but $f$ is not!
We need, in fact, stronger condtions:
$S$ is a separable metric space.
$x mapsto f(x,y)$ is continuous for all $y in T$.
$y mapsto f(x,y)$ is measurable for all $x in S$.
Then $f$ is measurable. (Proof can be found in many standard books on measure theory.)
answered Nov 15 at 10:36
p4sch
3,790216
edited Nov 15 at 19:52
answered Nov 15 at 10:36
p4sch
3,790216
answered Nov 15 at 10:36
p4sch
3,790216
answered Nov 15 at 10:36
p4sch
3,790216
3,790216
add a comment |
add a comment |
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