Mixing
What exactly do I need to show before using Fubini-Tonelli
$begingroup$
We have written in our text book:
Let $(X_{1},mathcal{A}_{1},mu_{1})$ and $(X_{2},mathcal{A}_{2},mu_{2})$ be two $sigma-$finite measures. Define $(X,mathcal{A},mu):=(X_{1}times X_{2},mathcal{A_{1}} otimes mathcal{A_{2}},mu_{1}otimesmu_{2})$
Let $f: X to bar{mathbb R}$ be $mathcal{A}-$measurable
Then for $g in {f_{-},f_{+}}$:
$X_{1}to [0,infty],x_{1}mapstoint_{X_{2}}g(x_{1},x_{2})dmu_{2}(x_{2})$ is $mathcal{A_{1}}-$measurable
and
$X_{2}to [0,infty],x_{2}mapstoint_{X_{1}}g(x_{1},x_{2})dmu_{1}(x_{1})$ is $mathcal{A_{2}}-$measurable
Then we can use Tonelli for $f geq 0$ a.e. as well as Fubini.
Problem:
I have a case, let's say $f(x,t):=e^{-xt}sin{x}$ and would like to use Fubini-Tonelli for $R> 0$ on
$int_{[0,R]}int_{[0,infty[}e^{-xt}sin{x}dlambda(x)dlambda(t)$
Part of the solution simply states:
$int_{[0,R]}int_{[0,infty[}|e^{-xt}sin{x}|dlambda(t)dlambda(x)<infty$ (drastically shortened)
Which intuitively makes sense in order to use Fubini.
However, where in this solution is shown that $f:[0,R]times[0,infty[tobar{mathbb R}$ is indeed measurable.
Does it suffice to simply state $f$ is continuous on $[0,R]$ as well as on $[0,infty[$
and therefore it (i.e. the function as a whole) is measurable?
real-analysis integration measure-theory multivariable-calculus
asked Jan 13 at 10:08
MinaThumaMinaThuma
1388
$endgroup$
add a comment |
$begingroup$
We have written in our text book:
Let $(X_{1},mathcal{A}_{1},mu_{1})$ and $(X_{2},mathcal{A}_{2},mu_{2})$ be two $sigma-$finite measures. Define $(X,mathcal{A},mu):=(X_{1}times X_{2},mathcal{A_{1}} otimes mathcal{A_{2}},mu_{1}otimesmu_{2})$
Let $f: X to bar{mathbb R}$ be $mathcal{A}-$measurable
Then for $g in {f_{-},f_{+}}$:
$X_{1}to [0,infty],x_{1}mapstoint_{X_{2}}g(x_{1},x_{2})dmu_{2}(x_{2})$ is $mathcal{A_{1}}-$measurable
and
$X_{2}to [0,infty],x_{2}mapstoint_{X_{1}}g(x_{1},x_{2})dmu_{1}(x_{1})$ is $mathcal{A_{2}}-$measurable
Then we can use Tonelli for $f geq 0$ a.e. as well as Fubini.
Problem:
I have a case, let's say $f(x,t):=e^{-xt}sin{x}$ and would like to use Fubini-Tonelli for $R> 0$ on
$int_{[0,R]}int_{[0,infty[}e^{-xt}sin{x}dlambda(x)dlambda(t)$
Part of the solution simply states:
$int_{[0,R]}int_{[0,infty[}|e^{-xt}sin{x}|dlambda(t)dlambda(x)<infty$ (drastically shortened)
Which intuitively makes sense in order to use Fubini.
However, where in this solution is shown that $f:[0,R]times[0,infty[tobar{mathbb R}$ is indeed measurable.
Does it suffice to simply state $f$ is continuous on $[0,R]$ as well as on $[0,infty[$
and therefore it (i.e. the function as a whole) is measurable?
real-analysis integration measure-theory multivariable-calculus
asked Jan 13 at 10:08
MinaThumaMinaThuma
1388
$endgroup$
$begingroup$
Yes, it is because $f$ continuous on $[0,R]$ and on $[0,infty[$ implies $f$ continuous on $[0,R] times [0,infty[$ (as a whole) and this implies that $f$ is measurable.
$endgroup$
– mathcourse
Jan 13 at 11:30
add a comment |
$begingroup$
We have written in our text book:
Let $(X_{1},mathcal{A}_{1},mu_{1})$ and $(X_{2},mathcal{A}_{2},mu_{2})$ be two $sigma-$finite measures. Define $(X,mathcal{A},mu):=(X_{1}times X_{2},mathcal{A_{1}} otimes mathcal{A_{2}},mu_{1}otimesmu_{2})$
Let $f: X to bar{mathbb R}$ be $mathcal{A}-$measurable
Then for $g in {f_{-},f_{+}}$:
$X_{1}to [0,infty],x_{1}mapstoint_{X_{2}}g(x_{1},x_{2})dmu_{2}(x_{2})$ is $mathcal{A_{1}}-$measurable
and
$X_{2}to [0,infty],x_{2}mapstoint_{X_{1}}g(x_{1},x_{2})dmu_{1}(x_{1})$ is $mathcal{A_{2}}-$measurable
Then we can use Tonelli for $f geq 0$ a.e. as well as Fubini.
Problem:
I have a case, let's say $f(x,t):=e^{-xt}sin{x}$ and would like to use Fubini-Tonelli for $R> 0$ on
$int_{[0,R]}int_{[0,infty[}e^{-xt}sin{x}dlambda(x)dlambda(t)$
Part of the solution simply states:
$int_{[0,R]}int_{[0,infty[}|e^{-xt}sin{x}|dlambda(t)dlambda(x)<infty$ (drastically shortened)
Which intuitively makes sense in order to use Fubini.
However, where in this solution is shown that $f:[0,R]times[0,infty[tobar{mathbb R}$ is indeed measurable.
Does it suffice to simply state $f$ is continuous on $[0,R]$ as well as on $[0,infty[$
and therefore it (i.e. the function as a whole) is measurable?
real-analysis integration measure-theory multivariable-calculus
asked Jan 13 at 10:08
MinaThumaMinaThuma
1388
$endgroup$
We have written in our text book:
Let $(X_{1},mathcal{A}_{1},mu_{1})$ and $(X_{2},mathcal{A}_{2},mu_{2})$ be two $sigma-$finite measures. Define $(X,mathcal{A},mu):=(X_{1}times X_{2},mathcal{A_{1}} otimes mathcal{A_{2}},mu_{1}otimesmu_{2})$
Let $f: X to bar{mathbb R}$ be $mathcal{A}-$measurable
Then for $g in {f_{-},f_{+}}$:
$X_{1}to [0,infty],x_{1}mapstoint_{X_{2}}g(x_{1},x_{2})dmu_{2}(x_{2})$ is $mathcal{A_{1}}-$measurable
and
$X_{2}to [0,infty],x_{2}mapstoint_{X_{1}}g(x_{1},x_{2})dmu_{1}(x_{1})$ is $mathcal{A_{2}}-$measurable
Then we can use Tonelli for $f geq 0$ a.e. as well as Fubini.
Problem:
I have a case, let's say $f(x,t):=e^{-xt}sin{x}$ and would like to use Fubini-Tonelli for $R> 0$ on
$int_{[0,R]}int_{[0,infty[}e^{-xt}sin{x}dlambda(x)dlambda(t)$
Part of the solution simply states:
$int_{[0,R]}int_{[0,infty[}|e^{-xt}sin{x}|dlambda(t)dlambda(x)<infty$ (drastically shortened)
Which intuitively makes sense in order to use Fubini.
However, where in this solution is shown that $f:[0,R]times[0,infty[tobar{mathbb R}$ is indeed measurable.
Does it suffice to simply state $f$ is continuous on $[0,R]$ as well as on $[0,infty[$
and therefore it (i.e. the function as a whole) is measurable?
real-analysis integration measure-theory multivariable-calculus
real-analysis integration measure-theory multivariable-calculus
asked Jan 13 at 10:08
MinaThumaMinaThuma
1388
asked Jan 13 at 10:08
MinaThumaMinaThuma
1388
asked Jan 13 at 10:08
MinaThumaMinaThuma
1388
asked Jan 13 at 10:08
MinaThumaMinaThuma
1388
asked Jan 13 at 10:08
MinaThumaMinaThuma
1388
1388
$begingroup$
Yes, it is because $f$ continuous on $[0,R]$ and on $[0,infty[$ implies $f$ continuous on $[0,R] times [0,infty[$ (as a whole) and this implies that $f$ is measurable.
$endgroup$
– mathcourse
Jan 13 at 11:30
add a comment |
$begingroup$
Yes, it is because $f$ continuous on $[0,R]$ and on $[0,infty[$ implies $f$ continuous on $[0,R] times [0,infty[$ (as a whole) and this implies that $f$ is measurable.
$endgroup$
– mathcourse
Jan 13 at 11:30
$begingroup$
Yes, it is because $f$ continuous on $[0,R]$ and on $[0,infty[$ implies $f$ continuous on $[0,R] times [0,infty[$ (as a whole) and this implies that $f$ is measurable.
$endgroup$
– mathcourse
Jan 13 at 11:30
$begingroup$
Yes, it is because $f$ continuous on $[0,R]$ and on $[0,infty[$ implies $f$ continuous on $[0,R] times [0,infty[$ (as a whole) and this implies that $f$ is measurable.
$endgroup$
– mathcourse
Jan 13 at 11:30
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Yes it sufficies to state that $f$ is continous, since a continous function is measurable (with respect to the Borel sigma algebra). Therefore to use Fubini it is enough to show
$$int_{[0,R]}int_{[0,infty[}|e^{-xt}sin{x}|dlambda(t)dlambda(x)<infty.$$
answered Jan 13 at 11:26
eddieeddie
515110
$endgroup$
add a comment |
$begingroup$
You are asking if it is enough to show that the function is continuous on $[0,R]$ as well as $[0,infty)$. It is not clear what kind of continuity you are talking about. It is important to use the fact that the function is JOINTLY continuous. Any function which is continuous from $[0,R] times[0,infty)$ to $mathbb R$ is Borel measurable.
answered Jan 13 at 12:15
Kavi Rama MurthyKavi Rama Murthy
59.6k42161
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
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oldest
votes
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oldest
votes
$begingroup$
Yes it sufficies to state that $f$ is continous, since a continous function is measurable (with respect to the Borel sigma algebra). Therefore to use Fubini it is enough to show
$$int_{[0,R]}int_{[0,infty[}|e^{-xt}sin{x}|dlambda(t)dlambda(x)<infty.$$
answered Jan 13 at 11:26
eddieeddie
515110
$endgroup$
add a comment |
$begingroup$
Yes it sufficies to state that $f$ is continous, since a continous function is measurable (with respect to the Borel sigma algebra). Therefore to use Fubini it is enough to show
$$int_{[0,R]}int_{[0,infty[}|e^{-xt}sin{x}|dlambda(t)dlambda(x)<infty.$$
answered Jan 13 at 11:26
eddieeddie
515110
$endgroup$
add a comment |
$begingroup$
Yes it sufficies to state that $f$ is continous, since a continous function is measurable (with respect to the Borel sigma algebra). Therefore to use Fubini it is enough to show
$$int_{[0,R]}int_{[0,infty[}|e^{-xt}sin{x}|dlambda(t)dlambda(x)<infty.$$
answered Jan 13 at 11:26
eddieeddie
515110
$endgroup$
Yes it sufficies to state that $f$ is continous, since a continous function is measurable (with respect to the Borel sigma algebra). Therefore to use Fubini it is enough to show
$$int_{[0,R]}int_{[0,infty[}|e^{-xt}sin{x}|dlambda(t)dlambda(x)<infty.$$
answered Jan 13 at 11:26
eddieeddie
515110
answered Jan 13 at 11:26
eddieeddie
515110
answered Jan 13 at 11:26
eddieeddie
515110
answered Jan 13 at 11:26
eddieeddie
515110
515110
add a comment |
add a comment |
$begingroup$
You are asking if it is enough to show that the function is continuous on $[0,R]$ as well as $[0,infty)$. It is not clear what kind of continuity you are talking about. It is important to use the fact that the function is JOINTLY continuous. Any function which is continuous from $[0,R] times[0,infty)$ to $mathbb R$ is Borel measurable.
answered Jan 13 at 12:15
Kavi Rama MurthyKavi Rama Murthy
59.6k42161
$endgroup$
add a comment |
$begingroup$
You are asking if it is enough to show that the function is continuous on $[0,R]$ as well as $[0,infty)$. It is not clear what kind of continuity you are talking about. It is important to use the fact that the function is JOINTLY continuous. Any function which is continuous from $[0,R] times[0,infty)$ to $mathbb R$ is Borel measurable.
answered Jan 13 at 12:15
Kavi Rama MurthyKavi Rama Murthy
59.6k42161
$endgroup$
add a comment |
$begingroup$
You are asking if it is enough to show that the function is continuous on $[0,R]$ as well as $[0,infty)$. It is not clear what kind of continuity you are talking about. It is important to use the fact that the function is JOINTLY continuous. Any function which is continuous from $[0,R] times[0,infty)$ to $mathbb R$ is Borel measurable.
answered Jan 13 at 12:15
Kavi Rama MurthyKavi Rama Murthy
59.6k42161
$endgroup$
You are asking if it is enough to show that the function is continuous on $[0,R]$ as well as $[0,infty)$. It is not clear what kind of continuity you are talking about. It is important to use the fact that the function is JOINTLY continuous. Any function which is continuous from $[0,R] times[0,infty)$ to $mathbb R$ is Borel measurable.
answered Jan 13 at 12:15
Kavi Rama MurthyKavi Rama Murthy
59.6k42161
answered Jan 13 at 12:15
Kavi Rama MurthyKavi Rama Murthy
59.6k42161
answered Jan 13 at 12:15
Kavi Rama MurthyKavi Rama Murthy
59.6k42161
answered Jan 13 at 12:15
Kavi Rama MurthyKavi Rama Murthy
59.6k42161
59.6k42161
add a comment |
add a comment |
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$begingroup$
Yes, it is because $f$ continuous on $[0,R]$ and on $[0,infty[$ implies $f$ continuous on $[0,R] times [0,infty[$ (as a whole) and this implies that $f$ is measurable.
$endgroup$
– mathcourse
Jan 13 at 11:30