Mixing
If $mu_n rightarrow mu$, then $int f dmu_n$ does not converge to $int f dmu$
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Suppose that $mu_n$ is a sequence of measures converging setwise to a measure $mu$.
I want to show that 1) if $mu_n$ is a decreasing sequence of complete measures then $mu$ is not necessarily complete, 2) if $mu_n$ is a decreasing sequence of measures, then $int f dmu_n$ does not necessarily converge to $int f dmu$, where $f$ is a nonnegative measurable function, and 3) in general, $lim_{nrightarrow infty} int f dmu_n$ does not exist (but I’m not sure if this is correct).
For 1) here is my counterexample. Let $mu_n$ be the complete Lebesgue-Stieltjes measure on $mathbb R$ associated to the nondecreasing function $F(x)=x/n$.Then $mu_n$ decreases to the zero measure $mu=0$ . In particular, $mu[0,1]=0$ but the interval $[0,1]$ contains non measurable sets, showing that $mu$ is not complete. Is this right? Any easier counterexample?
For 2) take $mu_n$ as in the case 1, and let $f$ be the function $chi_{mathbb R}$. Then $int dmu_n=infty$ but $int dmu=0$. Is this counterexample correct? Is there any easier counterexample?
For 3) I couldn’t find any counterexample. Any suggestion?
real-analysis measure-theory proof-verification
asked Feb 1 at 23:56
User12239User12239
364216
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add a comment |
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Suppose that $mu_n$ is a sequence of measures converging setwise to a measure $mu$.
I want to show that 1) if $mu_n$ is a decreasing sequence of complete measures then $mu$ is not necessarily complete, 2) if $mu_n$ is a decreasing sequence of measures, then $int f dmu_n$ does not necessarily converge to $int f dmu$, where $f$ is a nonnegative measurable function, and 3) in general, $lim_{nrightarrow infty} int f dmu_n$ does not exist (but I’m not sure if this is correct).
For 1) here is my counterexample. Let $mu_n$ be the complete Lebesgue-Stieltjes measure on $mathbb R$ associated to the nondecreasing function $F(x)=x/n$.Then $mu_n$ decreases to the zero measure $mu=0$ . In particular, $mu[0,1]=0$ but the interval $[0,1]$ contains non measurable sets, showing that $mu$ is not complete. Is this right? Any easier counterexample?
For 2) take $mu_n$ as in the case 1, and let $f$ be the function $chi_{mathbb R}$. Then $int dmu_n=infty$ but $int dmu=0$. Is this counterexample correct? Is there any easier counterexample?
For 3) I couldn’t find any counterexample. Any suggestion?
real-analysis measure-theory proof-verification
asked Feb 1 at 23:56
User12239User12239
364216
$endgroup$
add a comment |
$begingroup$
Suppose that $mu_n$ is a sequence of measures converging setwise to a measure $mu$.
I want to show that 1) if $mu_n$ is a decreasing sequence of complete measures then $mu$ is not necessarily complete, 2) if $mu_n$ is a decreasing sequence of measures, then $int f dmu_n$ does not necessarily converge to $int f dmu$, where $f$ is a nonnegative measurable function, and 3) in general, $lim_{nrightarrow infty} int f dmu_n$ does not exist (but I’m not sure if this is correct).
For 1) here is my counterexample. Let $mu_n$ be the complete Lebesgue-Stieltjes measure on $mathbb R$ associated to the nondecreasing function $F(x)=x/n$.Then $mu_n$ decreases to the zero measure $mu=0$ . In particular, $mu[0,1]=0$ but the interval $[0,1]$ contains non measurable sets, showing that $mu$ is not complete. Is this right? Any easier counterexample?
For 2) take $mu_n$ as in the case 1, and let $f$ be the function $chi_{mathbb R}$. Then $int dmu_n=infty$ but $int dmu=0$. Is this counterexample correct? Is there any easier counterexample?
For 3) I couldn’t find any counterexample. Any suggestion?
real-analysis measure-theory proof-verification
asked Feb 1 at 23:56
User12239User12239
364216
$endgroup$
Suppose that $mu_n$ is a sequence of measures converging setwise to a measure $mu$.
I want to show that 1) if $mu_n$ is a decreasing sequence of complete measures then $mu$ is not necessarily complete, 2) if $mu_n$ is a decreasing sequence of measures, then $int f dmu_n$ does not necessarily converge to $int f dmu$, where $f$ is a nonnegative measurable function, and 3) in general, $lim_{nrightarrow infty} int f dmu_n$ does not exist (but I’m not sure if this is correct).
For 1) here is my counterexample. Let $mu_n$ be the complete Lebesgue-Stieltjes measure on $mathbb R$ associated to the nondecreasing function $F(x)=x/n$.Then $mu_n$ decreases to the zero measure $mu=0$ . In particular, $mu[0,1]=0$ but the interval $[0,1]$ contains non measurable sets, showing that $mu$ is not complete. Is this right? Any easier counterexample?
For 2) take $mu_n$ as in the case 1, and let $f$ be the function $chi_{mathbb R}$. Then $int dmu_n=infty$ but $int dmu=0$. Is this counterexample correct? Is there any easier counterexample?
For 3) I couldn’t find any counterexample. Any suggestion?
real-analysis measure-theory proof-verification
real-analysis measure-theory proof-verification
asked Feb 1 at 23:56
User12239User12239
364216
asked Feb 1 at 23:56
User12239User12239
364216
asked Feb 1 at 23:56
User12239User12239
364216
asked Feb 1 at 23:56
User12239User12239
364216
asked Feb 1 at 23:56
User12239User12239
364216
364216
add a comment |
add a comment |
1 Answer
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For 3) let $mu_n(A)=int_A frac 1 {nx} I_{(1,n+1)} (x), dx$ for $n$ even, $mu_n(A)=int_A frac 2 {nx} I_{(1,n+1)} (x), dx$ for $n$ odd. Then $mu_n(A) to 0$ for each $A$ but $int |x|dmu_{n}(x)=1$, for each $n$ even, $2$ for each $n$ odd .
answered Feb 2 at 0:07
Kavi Rama MurthyKavi Rama Murthy
74.3k53270
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Thanks. Are my counterexamples above correct?
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– User12239
Feb 2 at 0:08
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@User12239 Yes, your answers to 1) and 2) are correct.
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– Kavi Rama Murthy
Feb 2 at 0:09
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Thank you for your help
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– User12239
Feb 2 at 0:10
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@User12239 Your counterexamples? Your example for (1) is exactlly the example I gave you earlier today: math.stackexchange.com/questions/3096191/…
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– David C. Ullrich
Feb 2 at 0:38
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@DavidC.Ullrich are you sure? I hadn’t posted anything here for a week or so
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– User12239
Feb 2 at 0:40
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show 2 more comments
Your Answer
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1 Answer
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$begingroup$
For 3) let $mu_n(A)=int_A frac 1 {nx} I_{(1,n+1)} (x), dx$ for $n$ even, $mu_n(A)=int_A frac 2 {nx} I_{(1,n+1)} (x), dx$ for $n$ odd. Then $mu_n(A) to 0$ for each $A$ but $int |x|dmu_{n}(x)=1$, for each $n$ even, $2$ for each $n$ odd .
answered Feb 2 at 0:07
Kavi Rama MurthyKavi Rama Murthy
74.3k53270
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$begingroup$
Thanks. Are my counterexamples above correct?
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– User12239
Feb 2 at 0:08
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@User12239 Yes, your answers to 1) and 2) are correct.
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– Kavi Rama Murthy
Feb 2 at 0:09
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Thank you for your help
$endgroup$
– User12239
Feb 2 at 0:10
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@User12239 Your counterexamples? Your example for (1) is exactlly the example I gave you earlier today: math.stackexchange.com/questions/3096191/…
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– David C. Ullrich
Feb 2 at 0:38
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@DavidC.Ullrich are you sure? I hadn’t posted anything here for a week or so
$endgroup$
– User12239
Feb 2 at 0:40
|
show 2 more comments
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For 3) let $mu_n(A)=int_A frac 1 {nx} I_{(1,n+1)} (x), dx$ for $n$ even, $mu_n(A)=int_A frac 2 {nx} I_{(1,n+1)} (x), dx$ for $n$ odd. Then $mu_n(A) to 0$ for each $A$ but $int |x|dmu_{n}(x)=1$, for each $n$ even, $2$ for each $n$ odd .
answered Feb 2 at 0:07
Kavi Rama MurthyKavi Rama Murthy
74.3k53270
$endgroup$
$begingroup$
Thanks. Are my counterexamples above correct?
$endgroup$
– User12239
Feb 2 at 0:08
$begingroup$
@User12239 Yes, your answers to 1) and 2) are correct.
$endgroup$
– Kavi Rama Murthy
Feb 2 at 0:09
$begingroup$
Thank you for your help
$endgroup$
– User12239
Feb 2 at 0:10
$begingroup$
@User12239 Your counterexamples? Your example for (1) is exactlly the example I gave you earlier today: math.stackexchange.com/questions/3096191/…
$endgroup$
– David C. Ullrich
Feb 2 at 0:38
$begingroup$
@DavidC.Ullrich are you sure? I hadn’t posted anything here for a week or so
$endgroup$
– User12239
Feb 2 at 0:40
|
show 2 more comments
$begingroup$
For 3) let $mu_n(A)=int_A frac 1 {nx} I_{(1,n+1)} (x), dx$ for $n$ even, $mu_n(A)=int_A frac 2 {nx} I_{(1,n+1)} (x), dx$ for $n$ odd. Then $mu_n(A) to 0$ for each $A$ but $int |x|dmu_{n}(x)=1$, for each $n$ even, $2$ for each $n$ odd .
answered Feb 2 at 0:07
Kavi Rama MurthyKavi Rama Murthy
74.3k53270
$endgroup$
For 3) let $mu_n(A)=int_A frac 1 {nx} I_{(1,n+1)} (x), dx$ for $n$ even, $mu_n(A)=int_A frac 2 {nx} I_{(1,n+1)} (x), dx$ for $n$ odd. Then $mu_n(A) to 0$ for each $A$ but $int |x|dmu_{n}(x)=1$, for each $n$ even, $2$ for each $n$ odd .
answered Feb 2 at 0:07
Kavi Rama MurthyKavi Rama Murthy
74.3k53270
edited Feb 2 at 0:19
answered Feb 2 at 0:07
Kavi Rama MurthyKavi Rama Murthy
74.3k53270
answered Feb 2 at 0:07
Kavi Rama MurthyKavi Rama Murthy
74.3k53270
answered Feb 2 at 0:07
Kavi Rama MurthyKavi Rama Murthy
74.3k53270
74.3k53270
$begingroup$
Thanks. Are my counterexamples above correct?
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– User12239
Feb 2 at 0:08
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@User12239 Yes, your answers to 1) and 2) are correct.
$endgroup$
– Kavi Rama Murthy
Feb 2 at 0:09
$begingroup$
Thank you for your help
$endgroup$
– User12239
Feb 2 at 0:10
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@User12239 Your counterexamples? Your example for (1) is exactlly the example I gave you earlier today: math.stackexchange.com/questions/3096191/…
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– David C. Ullrich
Feb 2 at 0:38
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@DavidC.Ullrich are you sure? I hadn’t posted anything here for a week or so
$endgroup$
– User12239
Feb 2 at 0:40
|
show 2 more comments
$begingroup$
Thanks. Are my counterexamples above correct?
$endgroup$
– User12239
Feb 2 at 0:08
$begingroup$
@User12239 Yes, your answers to 1) and 2) are correct.
$endgroup$
– Kavi Rama Murthy
Feb 2 at 0:09
$begingroup$
Thank you for your help
$endgroup$
– User12239
Feb 2 at 0:10
$begingroup$
@User12239 Your counterexamples? Your example for (1) is exactlly the example I gave you earlier today: math.stackexchange.com/questions/3096191/…
$endgroup$
– David C. Ullrich
Feb 2 at 0:38
$begingroup$
@DavidC.Ullrich are you sure? I hadn’t posted anything here for a week or so
$endgroup$
– User12239
Feb 2 at 0:40
$begingroup$
Thanks. Are my counterexamples above correct?
$endgroup$
– User12239
Feb 2 at 0:08
$begingroup$
Thanks. Are my counterexamples above correct?
$endgroup$
– User12239
Feb 2 at 0:08
$begingroup$
@User12239 Yes, your answers to 1) and 2) are correct.
$endgroup$
– Kavi Rama Murthy
Feb 2 at 0:09
$begingroup$
@User12239 Yes, your answers to 1) and 2) are correct.
$endgroup$
– Kavi Rama Murthy
Feb 2 at 0:09
$begingroup$
Thank you for your help
$endgroup$
– User12239
Feb 2 at 0:10
$begingroup$
Thank you for your help
$endgroup$
– User12239
Feb 2 at 0:10
$begingroup$
@User12239 Your counterexamples? Your example for (1) is exactlly the example I gave you earlier today: math.stackexchange.com/questions/3096191/…
$endgroup$
– David C. Ullrich
Feb 2 at 0:38
$begingroup$
@User12239 Your counterexamples? Your example for (1) is exactlly the example I gave you earlier today: math.stackexchange.com/questions/3096191/…
$endgroup$
– David C. Ullrich
Feb 2 at 0:38
$begingroup$
@DavidC.Ullrich are you sure? I hadn’t posted anything here for a week or so
$endgroup$
– User12239
Feb 2 at 0:40
$begingroup$
@DavidC.Ullrich are you sure? I hadn’t posted anything here for a week or so
$endgroup$
– User12239
Feb 2 at 0:40
|
show 2 more comments
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