Mixing
the truth of $bigcap_n L^1(mu_n)=L^1(mu)$ where the measures $mu_n$ converge to $mu$
$begingroup$
Let $mu_n$ be a sequence of positive measures on a measurable space $(X,M)$ converging to a measure $mu$,i.e. $lim_{nrightarrow infty} mu_n(A)=mu(A)$ for any measurable set $A$. I want to check if $L^1(mu)=bigcap_n L^1(mu_n)$.
1) First suppose that $mu_n$ is an increasing sequence of measures. I think in this case the relation above does not hold. Here is my counterexample. Let $mu_n:= ngamma$ where $gamma$ is the counting measure on $mathbb R$. Let $phi:= chi_{{0}}$. Then $int phi dmu_n=mu_n({0})=n$ for all $n$ so that $phiin bigcap_n (L^1(mu_n))$, but $int phi dmu=infty$ so that $phinotin L^1(mu)$. Is this correct? Is there any better counterexample?
2) Now suppose that $mu_n$ is decreasing. I think for this case as well the relation above is wrong. Again take $u_n$ as in the case above. We have $0=int dmu_n$ for each $n$ but $infty= int dmu$. Hence $chi_{mathbb R}$ is a counterexample. Is this correct? What about any better counterexample?
real-analysis measure-theory proof-verification
asked Feb 1 at 22:25
User12239User12239
364216
$endgroup$
|
show 1 more comment
$begingroup$
Let $mu_n$ be a sequence of positive measures on a measurable space $(X,M)$ converging to a measure $mu$,i.e. $lim_{nrightarrow infty} mu_n(A)=mu(A)$ for any measurable set $A$. I want to check if $L^1(mu)=bigcap_n L^1(mu_n)$.
1) First suppose that $mu_n$ is an increasing sequence of measures. I think in this case the relation above does not hold. Here is my counterexample. Let $mu_n:= ngamma$ where $gamma$ is the counting measure on $mathbb R$. Let $phi:= chi_{{0}}$. Then $int phi dmu_n=mu_n({0})=n$ for all $n$ so that $phiin bigcap_n (L^1(mu_n))$, but $int phi dmu=infty$ so that $phinotin L^1(mu)$. Is this correct? Is there any better counterexample?
2) Now suppose that $mu_n$ is decreasing. I think for this case as well the relation above is wrong. Again take $u_n$ as in the case above. We have $0=int dmu_n$ for each $n$ but $infty= int dmu$. Hence $chi_{mathbb R}$ is a counterexample. Is this correct? What about any better counterexample?
real-analysis measure-theory proof-verification
asked Feb 1 at 22:25
User12239User12239
364216
$endgroup$
$begingroup$
@kimchilover they converge pointwise.
$endgroup$
– User12239
Feb 1 at 22:40
$begingroup$
@kimchilover yes exactly. I should have written it out
$endgroup$
– User12239
Feb 1 at 22:42
$begingroup$
I edited my question. Thanks for your reminding me @kimchilover
$endgroup$
– User12239
Feb 1 at 22:46
$begingroup$
In (1), what is $mu$, and why do you think $mu_n$ converges to it, given your calculation that $mu_n({0})=n$?
$endgroup$
– kimchi lover
Feb 1 at 22:49
$begingroup$
@kimchilover In (1) the measure $mu$ takes $0$ on the null set and takes $infty$ otherwise
$endgroup$
– User12239
Feb 1 at 23:01
|
show 1 more comment
$begingroup$
Let $mu_n$ be a sequence of positive measures on a measurable space $(X,M)$ converging to a measure $mu$,i.e. $lim_{nrightarrow infty} mu_n(A)=mu(A)$ for any measurable set $A$. I want to check if $L^1(mu)=bigcap_n L^1(mu_n)$.
1) First suppose that $mu_n$ is an increasing sequence of measures. I think in this case the relation above does not hold. Here is my counterexample. Let $mu_n:= ngamma$ where $gamma$ is the counting measure on $mathbb R$. Let $phi:= chi_{{0}}$. Then $int phi dmu_n=mu_n({0})=n$ for all $n$ so that $phiin bigcap_n (L^1(mu_n))$, but $int phi dmu=infty$ so that $phinotin L^1(mu)$. Is this correct? Is there any better counterexample?
2) Now suppose that $mu_n$ is decreasing. I think for this case as well the relation above is wrong. Again take $u_n$ as in the case above. We have $0=int dmu_n$ for each $n$ but $infty= int dmu$. Hence $chi_{mathbb R}$ is a counterexample. Is this correct? What about any better counterexample?
real-analysis measure-theory proof-verification
asked Feb 1 at 22:25
User12239User12239
364216
$endgroup$
Let $mu_n$ be a sequence of positive measures on a measurable space $(X,M)$ converging to a measure $mu$,i.e. $lim_{nrightarrow infty} mu_n(A)=mu(A)$ for any measurable set $A$. I want to check if $L^1(mu)=bigcap_n L^1(mu_n)$.
1) First suppose that $mu_n$ is an increasing sequence of measures. I think in this case the relation above does not hold. Here is my counterexample. Let $mu_n:= ngamma$ where $gamma$ is the counting measure on $mathbb R$. Let $phi:= chi_{{0}}$. Then $int phi dmu_n=mu_n({0})=n$ for all $n$ so that $phiin bigcap_n (L^1(mu_n))$, but $int phi dmu=infty$ so that $phinotin L^1(mu)$. Is this correct? Is there any better counterexample?
2) Now suppose that $mu_n$ is decreasing. I think for this case as well the relation above is wrong. Again take $u_n$ as in the case above. We have $0=int dmu_n$ for each $n$ but $infty= int dmu$. Hence $chi_{mathbb R}$ is a counterexample. Is this correct? What about any better counterexample?
real-analysis measure-theory proof-verification
real-analysis measure-theory proof-verification
asked Feb 1 at 22:25
User12239User12239
364216
asked Feb 1 at 22:25
User12239User12239
364216
edited Feb 1 at 22:45
User12239
asked Feb 1 at 22:25
User12239User12239
364216
asked Feb 1 at 22:25
User12239User12239
364216
asked Feb 1 at 22:25
User12239User12239
364216
364216
$begingroup$
@kimchilover they converge pointwise.
$endgroup$
– User12239
Feb 1 at 22:40
$begingroup$
@kimchilover yes exactly. I should have written it out
$endgroup$
– User12239
Feb 1 at 22:42
$begingroup$
I edited my question. Thanks for your reminding me @kimchilover
$endgroup$
– User12239
Feb 1 at 22:46
$begingroup$
In (1), what is $mu$, and why do you think $mu_n$ converges to it, given your calculation that $mu_n({0})=n$?
$endgroup$
– kimchi lover
Feb 1 at 22:49
$begingroup$
@kimchilover In (1) the measure $mu$ takes $0$ on the null set and takes $infty$ otherwise
$endgroup$
– User12239
Feb 1 at 23:01
|
show 1 more comment
$begingroup$
@kimchilover they converge pointwise.
$endgroup$
– User12239
Feb 1 at 22:40
$begingroup$
@kimchilover yes exactly. I should have written it out
$endgroup$
– User12239
Feb 1 at 22:42
$begingroup$
I edited my question. Thanks for your reminding me @kimchilover
$endgroup$
– User12239
Feb 1 at 22:46
$begingroup$
In (1), what is $mu$, and why do you think $mu_n$ converges to it, given your calculation that $mu_n({0})=n$?
$endgroup$
– kimchi lover
Feb 1 at 22:49
$begingroup$
@kimchilover In (1) the measure $mu$ takes $0$ on the null set and takes $infty$ otherwise
$endgroup$
– User12239
Feb 1 at 23:01
$begingroup$
@kimchilover they converge pointwise.
$endgroup$
– User12239
Feb 1 at 22:40
$begingroup$
@kimchilover they converge pointwise.
$endgroup$
– User12239
Feb 1 at 22:40
$begingroup$
@kimchilover yes exactly. I should have written it out
$endgroup$
– User12239
Feb 1 at 22:42
$begingroup$
@kimchilover yes exactly. I should have written it out
$endgroup$
– User12239
Feb 1 at 22:42
$begingroup$
I edited my question. Thanks for your reminding me @kimchilover
$endgroup$
– User12239
Feb 1 at 22:46
$begingroup$
I edited my question. Thanks for your reminding me @kimchilover
$endgroup$
– User12239
Feb 1 at 22:46
$begingroup$
In (1), what is $mu$, and why do you think $mu_n$ converges to it, given your calculation that $mu_n({0})=n$?
$endgroup$
– kimchi lover
Feb 1 at 22:49
$begingroup$
In (1), what is $mu$, and why do you think $mu_n$ converges to it, given your calculation that $mu_n({0})=n$?
$endgroup$
– kimchi lover
Feb 1 at 22:49
$begingroup$
@kimchilover In (1) the measure $mu$ takes $0$ on the null set and takes $infty$ otherwise
$endgroup$
– User12239
Feb 1 at 23:01
$begingroup$
@kimchilover In (1) the measure $mu$ takes $0$ on the null set and takes $infty$ otherwise
$endgroup$
– User12239
Feb 1 at 23:01
|
show 1 more comment
1 Answer
1
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oldest
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The answer is clearly no in general.
To see this, pick an arbitrary measure $nu$ such that $L^1(nu) subsetneq L^1(mu)$. Now replace $mu_1$ by $nu$.
answered Feb 1 at 22:52
N. S.N. S.
105k7115210
$endgroup$
$begingroup$
Thank you for for this general fact
$endgroup$
– User12239
Feb 1 at 23:30
add a comment |
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1 Answer
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1 Answer
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$begingroup$
The answer is clearly no in general.
To see this, pick an arbitrary measure $nu$ such that $L^1(nu) subsetneq L^1(mu)$. Now replace $mu_1$ by $nu$.
answered Feb 1 at 22:52
N. S.N. S.
105k7115210
$endgroup$
$begingroup$
Thank you for for this general fact
$endgroup$
– User12239
Feb 1 at 23:30
add a comment |
$begingroup$
The answer is clearly no in general.
To see this, pick an arbitrary measure $nu$ such that $L^1(nu) subsetneq L^1(mu)$. Now replace $mu_1$ by $nu$.
answered Feb 1 at 22:52
N. S.N. S.
105k7115210
$endgroup$
$begingroup$
Thank you for for this general fact
$endgroup$
– User12239
Feb 1 at 23:30
add a comment |
$begingroup$
The answer is clearly no in general.
To see this, pick an arbitrary measure $nu$ such that $L^1(nu) subsetneq L^1(mu)$. Now replace $mu_1$ by $nu$.
answered Feb 1 at 22:52
N. S.N. S.
105k7115210
$endgroup$
The answer is clearly no in general.
To see this, pick an arbitrary measure $nu$ such that $L^1(nu) subsetneq L^1(mu)$. Now replace $mu_1$ by $nu$.
answered Feb 1 at 22:52
N. S.N. S.
105k7115210
answered Feb 1 at 22:52
N. S.N. S.
105k7115210
answered Feb 1 at 22:52
N. S.N. S.
105k7115210
answered Feb 1 at 22:52
N. S.N. S.
105k7115210
105k7115210
$begingroup$
Thank you for for this general fact
$endgroup$
– User12239
Feb 1 at 23:30
add a comment |
$begingroup$
Thank you for for this general fact
$endgroup$
– User12239
Feb 1 at 23:30
$begingroup$
Thank you for for this general fact
$endgroup$
– User12239
Feb 1 at 23:30
$begingroup$
Thank you for for this general fact
$endgroup$
– User12239
Feb 1 at 23:30
add a comment |
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$begingroup$
@kimchilover they converge pointwise.
$endgroup$
– User12239
Feb 1 at 22:40
$begingroup$
@kimchilover yes exactly. I should have written it out
$endgroup$
– User12239
Feb 1 at 22:42
$begingroup$
I edited my question. Thanks for your reminding me @kimchilover
$endgroup$
– User12239
Feb 1 at 22:46
$begingroup$
In (1), what is $mu$, and why do you think $mu_n$ converges to it, given your calculation that $mu_n({0})=n$?
$endgroup$
– kimchi lover
Feb 1 at 22:49
$begingroup$
@kimchilover In (1) the measure $mu$ takes $0$ on the null set and takes $infty$ otherwise
$endgroup$
– User12239
Feb 1 at 23:01