Mixing
Some Questions On Signed Measures [closed]
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I am studying measure theory and I am beginner and looking for some counter examples or guides to prove these questions:
Let ${mu_n}$ be a sequence of positive measures on $(X, M)$ which converges to $mu$.
1-This sequence can be ascending or descending or an arbitrary sequence, in which case(s) $mu$ is a measure?
2- If $mu$ is a measure and all the $mu_n$ are a complete measure, then $mu$ is complete?
I think if this sequence be ascending or descending,$mu$ is a measure, but I have no idea to start the proof. and also I do not have any intuition about completeness.
any help would be great thanks.
real-analysis measure-theory
edited Feb 1 at 12:57
Clayton
19.6k33288
asked Feb 1 at 12:51
f.j1995f.j1995
153
$endgroup$
closed as off-topic by RRL, José Carlos Santos, mrtaurho, max_zorn, YiFan Feb 2 at 21:53
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – RRL, José Carlos Santos, mrtaurho, max_zorn, YiFan
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
$begingroup$
I am studying measure theory and I am beginner and looking for some counter examples or guides to prove these questions:
Let ${mu_n}$ be a sequence of positive measures on $(X, M)$ which converges to $mu$.
1-This sequence can be ascending or descending or an arbitrary sequence, in which case(s) $mu$ is a measure?
2- If $mu$ is a measure and all the $mu_n$ are a complete measure, then $mu$ is complete?
I think if this sequence be ascending or descending,$mu$ is a measure, but I have no idea to start the proof. and also I do not have any intuition about completeness.
any help would be great thanks.
real-analysis measure-theory
edited Feb 1 at 12:57
Clayton
19.6k33288
asked Feb 1 at 12:51
f.j1995f.j1995
153
$endgroup$
closed as off-topic by RRL, José Carlos Santos, mrtaurho, max_zorn, YiFan Feb 2 at 21:53
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – RRL, José Carlos Santos, mrtaurho, max_zorn, YiFan
If this question can be reworded to fit the rules in the help center, please edit the question.
$begingroup$
Exactly what do you mean by "$mu_n$ converges to $mu$"?
$endgroup$
– David C. Ullrich
Feb 1 at 14:38
$begingroup$
It means that the sequence of measures, converges pointwise to the mu
$endgroup$
– f.j1995
Feb 1 at 16:50
add a comment |
$begingroup$
I am studying measure theory and I am beginner and looking for some counter examples or guides to prove these questions:
Let ${mu_n}$ be a sequence of positive measures on $(X, M)$ which converges to $mu$.
1-This sequence can be ascending or descending or an arbitrary sequence, in which case(s) $mu$ is a measure?
2- If $mu$ is a measure and all the $mu_n$ are a complete measure, then $mu$ is complete?
I think if this sequence be ascending or descending,$mu$ is a measure, but I have no idea to start the proof. and also I do not have any intuition about completeness.
any help would be great thanks.
real-analysis measure-theory
edited Feb 1 at 12:57
Clayton
19.6k33288
asked Feb 1 at 12:51
f.j1995f.j1995
153
$endgroup$
I am studying measure theory and I am beginner and looking for some counter examples or guides to prove these questions:
Let ${mu_n}$ be a sequence of positive measures on $(X, M)$ which converges to $mu$.
1-This sequence can be ascending or descending or an arbitrary sequence, in which case(s) $mu$ is a measure?
2- If $mu$ is a measure and all the $mu_n$ are a complete measure, then $mu$ is complete?
I think if this sequence be ascending or descending,$mu$ is a measure, but I have no idea to start the proof. and also I do not have any intuition about completeness.
any help would be great thanks.
real-analysis measure-theory
real-analysis measure-theory
edited Feb 1 at 12:57
Clayton
19.6k33288
asked Feb 1 at 12:51
f.j1995f.j1995
153
edited Feb 1 at 12:57
Clayton
19.6k33288
asked Feb 1 at 12:51
f.j1995f.j1995
153
edited Feb 1 at 12:57
Clayton
19.6k33288
edited Feb 1 at 12:57
Clayton
19.6k33288
edited Feb 1 at 12:57
Clayton
19.6k33288
19.6k33288
asked Feb 1 at 12:51
f.j1995f.j1995
153
asked Feb 1 at 12:51
f.j1995f.j1995
153
asked Feb 1 at 12:51
f.j1995f.j1995
153
153
closed as off-topic by RRL, José Carlos Santos, mrtaurho, max_zorn, YiFan Feb 2 at 21:53
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – RRL, José Carlos Santos, mrtaurho, max_zorn, YiFan
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by RRL, José Carlos Santos, mrtaurho, max_zorn, YiFan Feb 2 at 21:53
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – RRL, José Carlos Santos, mrtaurho, max_zorn, YiFan
If this question can be reworded to fit the rules in the help center, please edit the question.
$begingroup$
Exactly what do you mean by "$mu_n$ converges to $mu$"?
$endgroup$
– David C. Ullrich
Feb 1 at 14:38
$begingroup$
It means that the sequence of measures, converges pointwise to the mu
$endgroup$
– f.j1995
Feb 1 at 16:50
add a comment |
$begingroup$
Exactly what do you mean by "$mu_n$ converges to $mu$"?
$endgroup$
– David C. Ullrich
Feb 1 at 14:38
$begingroup$
It means that the sequence of measures, converges pointwise to the mu
$endgroup$
– f.j1995
Feb 1 at 16:50
$begingroup$
Exactly what do you mean by "$mu_n$ converges to $mu$"?
$endgroup$
– David C. Ullrich
Feb 1 at 14:38
$begingroup$
Exactly what do you mean by "$mu_n$ converges to $mu$"?
$endgroup$
– David C. Ullrich
Feb 1 at 14:38
$begingroup$
It means that the sequence of measures, converges pointwise to the mu
$endgroup$
– f.j1995
Feb 1 at 16:50
$begingroup$
It means that the sequence of measures, converges pointwise to the mu
$endgroup$
– f.j1995
Feb 1 at 16:50
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
It seems that the convergence in question is $$mu(E)=lim_{ntoinfty}mu_n(E)quad(Ein M).$$
1.a. In general $mu$ need not be a measure. For example, let $X=Bbb R$, let $M$ be the algebra of Lebesgue measurable sets, and let $$mu_n(E)=m(Ecap[n,infty)),$$whhere $m$ is Lebesgue measure. If you figure out what this limit is for a few sets you can show that $$mu(Bbb R)nesum_{kinBbb Z}mu([k,k+1)).$$
1.b. Note that $mu_{n+1}lemu_n$ in that example. If $mu_{n+1}gemu_n$ then yes $mu$ is a measure. Hint for that: If the $E_k$ are disjoint and $E=bigcup E_k$ you can use Monotone Convergence to show that $$mu(E)=sum_kmu(E_k).$$
- No. Let $X,M$ and $m$ be as in 1.a above. Let $$mu_n=frac 1n m.$$ Then $mu_n$ is complete, but $mu$ is not. (Because $mu=0$, so...)
answered Feb 1 at 21:03
David C. UllrichDavid C. Ullrich
61.7k44095
$endgroup$
$begingroup$
thank you so much, but for 2, what about increasing measures?
$endgroup$
– f.j1995
Feb 1 at 22:50
$begingroup$
Hadn't thought about it. So think about it: It turns out (2) is trivial for an increasing sequence of measures! Because then $mu(E)=0$ implies that....
$endgroup$
– David C. Ullrich
Feb 2 at 0:23
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
It seems that the convergence in question is $$mu(E)=lim_{ntoinfty}mu_n(E)quad(Ein M).$$
1.a. In general $mu$ need not be a measure. For example, let $X=Bbb R$, let $M$ be the algebra of Lebesgue measurable sets, and let $$mu_n(E)=m(Ecap[n,infty)),$$whhere $m$ is Lebesgue measure. If you figure out what this limit is for a few sets you can show that $$mu(Bbb R)nesum_{kinBbb Z}mu([k,k+1)).$$
1.b. Note that $mu_{n+1}lemu_n$ in that example. If $mu_{n+1}gemu_n$ then yes $mu$ is a measure. Hint for that: If the $E_k$ are disjoint and $E=bigcup E_k$ you can use Monotone Convergence to show that $$mu(E)=sum_kmu(E_k).$$
- No. Let $X,M$ and $m$ be as in 1.a above. Let $$mu_n=frac 1n m.$$ Then $mu_n$ is complete, but $mu$ is not. (Because $mu=0$, so...)
answered Feb 1 at 21:03
David C. UllrichDavid C. Ullrich
61.7k44095
$endgroup$
$begingroup$
thank you so much, but for 2, what about increasing measures?
$endgroup$
– f.j1995
Feb 1 at 22:50
$begingroup$
Hadn't thought about it. So think about it: It turns out (2) is trivial for an increasing sequence of measures! Because then $mu(E)=0$ implies that....
$endgroup$
– David C. Ullrich
Feb 2 at 0:23
add a comment |
$begingroup$
It seems that the convergence in question is $$mu(E)=lim_{ntoinfty}mu_n(E)quad(Ein M).$$
1.a. In general $mu$ need not be a measure. For example, let $X=Bbb R$, let $M$ be the algebra of Lebesgue measurable sets, and let $$mu_n(E)=m(Ecap[n,infty)),$$whhere $m$ is Lebesgue measure. If you figure out what this limit is for a few sets you can show that $$mu(Bbb R)nesum_{kinBbb Z}mu([k,k+1)).$$
1.b. Note that $mu_{n+1}lemu_n$ in that example. If $mu_{n+1}gemu_n$ then yes $mu$ is a measure. Hint for that: If the $E_k$ are disjoint and $E=bigcup E_k$ you can use Monotone Convergence to show that $$mu(E)=sum_kmu(E_k).$$
- No. Let $X,M$ and $m$ be as in 1.a above. Let $$mu_n=frac 1n m.$$ Then $mu_n$ is complete, but $mu$ is not. (Because $mu=0$, so...)
answered Feb 1 at 21:03
David C. UllrichDavid C. Ullrich
61.7k44095
$endgroup$
$begingroup$
thank you so much, but for 2, what about increasing measures?
$endgroup$
– f.j1995
Feb 1 at 22:50
$begingroup$
Hadn't thought about it. So think about it: It turns out (2) is trivial for an increasing sequence of measures! Because then $mu(E)=0$ implies that....
$endgroup$
– David C. Ullrich
Feb 2 at 0:23
add a comment |
$begingroup$
It seems that the convergence in question is $$mu(E)=lim_{ntoinfty}mu_n(E)quad(Ein M).$$
1.a. In general $mu$ need not be a measure. For example, let $X=Bbb R$, let $M$ be the algebra of Lebesgue measurable sets, and let $$mu_n(E)=m(Ecap[n,infty)),$$whhere $m$ is Lebesgue measure. If you figure out what this limit is for a few sets you can show that $$mu(Bbb R)nesum_{kinBbb Z}mu([k,k+1)).$$
1.b. Note that $mu_{n+1}lemu_n$ in that example. If $mu_{n+1}gemu_n$ then yes $mu$ is a measure. Hint for that: If the $E_k$ are disjoint and $E=bigcup E_k$ you can use Monotone Convergence to show that $$mu(E)=sum_kmu(E_k).$$
- No. Let $X,M$ and $m$ be as in 1.a above. Let $$mu_n=frac 1n m.$$ Then $mu_n$ is complete, but $mu$ is not. (Because $mu=0$, so...)
answered Feb 1 at 21:03
David C. UllrichDavid C. Ullrich
61.7k44095
$endgroup$
It seems that the convergence in question is $$mu(E)=lim_{ntoinfty}mu_n(E)quad(Ein M).$$
1.a. In general $mu$ need not be a measure. For example, let $X=Bbb R$, let $M$ be the algebra of Lebesgue measurable sets, and let $$mu_n(E)=m(Ecap[n,infty)),$$whhere $m$ is Lebesgue measure. If you figure out what this limit is for a few sets you can show that $$mu(Bbb R)nesum_{kinBbb Z}mu([k,k+1)).$$
1.b. Note that $mu_{n+1}lemu_n$ in that example. If $mu_{n+1}gemu_n$ then yes $mu$ is a measure. Hint for that: If the $E_k$ are disjoint and $E=bigcup E_k$ you can use Monotone Convergence to show that $$mu(E)=sum_kmu(E_k).$$
- No. Let $X,M$ and $m$ be as in 1.a above. Let $$mu_n=frac 1n m.$$ Then $mu_n$ is complete, but $mu$ is not. (Because $mu=0$, so...)
answered Feb 1 at 21:03
David C. UllrichDavid C. Ullrich
61.7k44095
answered Feb 1 at 21:03
David C. UllrichDavid C. Ullrich
61.7k44095
answered Feb 1 at 21:03
David C. UllrichDavid C. Ullrich
61.7k44095
answered Feb 1 at 21:03
David C. UllrichDavid C. Ullrich
61.7k44095
61.7k44095
$begingroup$
thank you so much, but for 2, what about increasing measures?
$endgroup$
– f.j1995
Feb 1 at 22:50
$begingroup$
Hadn't thought about it. So think about it: It turns out (2) is trivial for an increasing sequence of measures! Because then $mu(E)=0$ implies that....
$endgroup$
– David C. Ullrich
Feb 2 at 0:23
add a comment |
$begingroup$
thank you so much, but for 2, what about increasing measures?
$endgroup$
– f.j1995
Feb 1 at 22:50
$begingroup$
Hadn't thought about it. So think about it: It turns out (2) is trivial for an increasing sequence of measures! Because then $mu(E)=0$ implies that....
$endgroup$
– David C. Ullrich
Feb 2 at 0:23
$begingroup$
thank you so much, but for 2, what about increasing measures?
$endgroup$
– f.j1995
Feb 1 at 22:50
$begingroup$
thank you so much, but for 2, what about increasing measures?
$endgroup$
– f.j1995
Feb 1 at 22:50
$begingroup$
Hadn't thought about it. So think about it: It turns out (2) is trivial for an increasing sequence of measures! Because then $mu(E)=0$ implies that....
$endgroup$
– David C. Ullrich
Feb 2 at 0:23
$begingroup$
Hadn't thought about it. So think about it: It turns out (2) is trivial for an increasing sequence of measures! Because then $mu(E)=0$ implies that....
$endgroup$
– David C. Ullrich
Feb 2 at 0:23
add a comment |
$begingroup$
Exactly what do you mean by "$mu_n$ converges to $mu$"?
$endgroup$
– David C. Ullrich
Feb 1 at 14:38
$begingroup$
It means that the sequence of measures, converges pointwise to the mu
$endgroup$
– f.j1995
Feb 1 at 16:50