Proving $f_nto f$a.e implies $f_nto f$ almost uniformly [duplicate]
$begingroup$
This question already has an answer here:
Omitting the hypotheses of finiteness of the measure in Egorov theorem
3 answers
Exercise: Let $(f)_{ninmathbb{N}}$ be a sequence of functions such that $f_nto f$a.e(almost everyehere) and there exists $g$ integrable such that $|f_n|leqslant g$a.e
for all $ninmathbb{N}$. Prove that $f_nto f$ almost uniformly.
I think I can apply the following theorem:
Ergoroff Theorem:
Consider $Einmathscr{F}$(sigma-algebra), and $EinOmega$ defined on a measure space $(Omega,mathscr{F},mu)$. Suppose $mu(E)<infty$, and ${f_n}$ is a sequence of measurable functions on $Etomathbb{R}$ which are finite almost everywhere and converge almost everywhere to a function $f:Etomathbb{R}$ which is also finite almost everywhere. Then $f_nto f$ almost uniformly in $E$.
I now that $f_nto f$ a.e so $lim_{ntoinfty}f_n(x)=f(x)forall xin E$, But to apply Ergoroff theorem I need to prove that $mu(E)<infty$ or $mu(Omega)<infty$.
I know by the Dominated convergence theorem that $lim_{ntoinfty}int |f_n-f| dmu=0$ but I cannot see how shall I prove from there that $mu(E)$ or $mu(Omega)$ are limited.
Question:
Can someone provide me any help?
Thanks in advance!
Note:$f_n$ does not necessarily converge to $f$ uniformly. So the question is not a duplicate.
functional-analysis measure-theory proof-writing
asked Jan 13 at 12:27
Pedro GomesPedro Gomes
2,0082822
$endgroup$
marked as duplicate by Xander Henderson, mrtaurho, zz20s, Cesareo, metamorphy Jan 14 at 9:47
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
|
show 4 more comments
$begingroup$
This question already has an answer here:
Omitting the hypotheses of finiteness of the measure in Egorov theorem
3 answers
Exercise: Let $(f)_{ninmathbb{N}}$ be a sequence of functions such that $f_nto f$a.e(almost everyehere) and there exists $g$ integrable such that $|f_n|leqslant g$a.e
for all $ninmathbb{N}$. Prove that $f_nto f$ almost uniformly.
I think I can apply the following theorem:
Ergoroff Theorem:
Consider $Einmathscr{F}$(sigma-algebra), and $EinOmega$ defined on a measure space $(Omega,mathscr{F},mu)$. Suppose $mu(E)<infty$, and ${f_n}$ is a sequence of measurable functions on $Etomathbb{R}$ which are finite almost everywhere and converge almost everywhere to a function $f:Etomathbb{R}$ which is also finite almost everywhere. Then $f_nto f$ almost uniformly in $E$.
I now that $f_nto f$ a.e so $lim_{ntoinfty}f_n(x)=f(x)forall xin E$, But to apply Ergoroff theorem I need to prove that $mu(E)<infty$ or $mu(Omega)<infty$.
I know by the Dominated convergence theorem that $lim_{ntoinfty}int |f_n-f| dmu=0$ but I cannot see how shall I prove from there that $mu(E)$ or $mu(Omega)$ are limited.
Question:
Can someone provide me any help?
Thanks in advance!
Note:$f_n$ does not necessarily converge to $f$ uniformly. So the question is not a duplicate.
functional-analysis measure-theory proof-writing
asked Jan 13 at 12:27
Pedro GomesPedro Gomes
2,0082822
$endgroup$
marked as duplicate by Xander Henderson, mrtaurho, zz20s, Cesareo, metamorphy Jan 14 at 9:47
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
$begingroup$
If $lambda>0$ then $mu({x:g(x)>lambda})<infty$.
$endgroup$
– David C. Ullrich
Jan 13 at 13:02
$begingroup$
@Jakobian My question is not a duplicate. If you read carefully the exercise you find out $f_n$ does not necessarily converge to $f$ uniformly. However the answer you provide assumes $f_n$ to converge uniformly so it is not answering this question.
$endgroup$
– Pedro Gomes
Jan 13 at 15:13
$begingroup$
@PedroGomes no, it doesn't. It's exactly the answer to your question
$endgroup$
– Jakobian
Jan 13 at 15:20
$begingroup$
@DavidC.Ullrich In order to apply Ergoroff I need to prove the measure of the domain where the function converges is finite. It is already assumed in the question when $f_nto f$ a.e that $mu({x:g(x)>lambda})=0$.
$endgroup$
– Pedro Gomes
Jan 13 at 15:21
1
$begingroup$
How do I know that you are pointing me in the right direction if you do not prove you are? Does it not sound like an authority fallacy?
$endgroup$
– Pedro Gomes
Jan 13 at 15:32
|
show 4 more comments
$begingroup$
This question already has an answer here:
Omitting the hypotheses of finiteness of the measure in Egorov theorem
3 answers
Exercise: Let $(f)_{ninmathbb{N}}$ be a sequence of functions such that $f_nto f$a.e(almost everyehere) and there exists $g$ integrable such that $|f_n|leqslant g$a.e
for all $ninmathbb{N}$. Prove that $f_nto f$ almost uniformly.
I think I can apply the following theorem:
Ergoroff Theorem:
Consider $Einmathscr{F}$(sigma-algebra), and $EinOmega$ defined on a measure space $(Omega,mathscr{F},mu)$. Suppose $mu(E)<infty$, and ${f_n}$ is a sequence of measurable functions on $Etomathbb{R}$ which are finite almost everywhere and converge almost everywhere to a function $f:Etomathbb{R}$ which is also finite almost everywhere. Then $f_nto f$ almost uniformly in $E$.
I now that $f_nto f$ a.e so $lim_{ntoinfty}f_n(x)=f(x)forall xin E$, But to apply Ergoroff theorem I need to prove that $mu(E)<infty$ or $mu(Omega)<infty$.
I know by the Dominated convergence theorem that $lim_{ntoinfty}int |f_n-f| dmu=0$ but I cannot see how shall I prove from there that $mu(E)$ or $mu(Omega)$ are limited.
Question:
Can someone provide me any help?
Thanks in advance!
Note:$f_n$ does not necessarily converge to $f$ uniformly. So the question is not a duplicate.
functional-analysis measure-theory proof-writing
asked Jan 13 at 12:27
Pedro GomesPedro Gomes
2,0082822
$endgroup$
This question already has an answer here:
Omitting the hypotheses of finiteness of the measure in Egorov theorem
3 answers
Exercise: Let $(f)_{ninmathbb{N}}$ be a sequence of functions such that $f_nto f$a.e(almost everyehere) and there exists $g$ integrable such that $|f_n|leqslant g$a.e
for all $ninmathbb{N}$. Prove that $f_nto f$ almost uniformly.
I think I can apply the following theorem:
Ergoroff Theorem:
Consider $Einmathscr{F}$(sigma-algebra), and $EinOmega$ defined on a measure space $(Omega,mathscr{F},mu)$. Suppose $mu(E)<infty$, and ${f_n}$ is a sequence of measurable functions on $Etomathbb{R}$ which are finite almost everywhere and converge almost everywhere to a function $f:Etomathbb{R}$ which is also finite almost everywhere. Then $f_nto f$ almost uniformly in $E$.
I now that $f_nto f$ a.e so $lim_{ntoinfty}f_n(x)=f(x)forall xin E$, But to apply Ergoroff theorem I need to prove that $mu(E)<infty$ or $mu(Omega)<infty$.
I know by the Dominated convergence theorem that $lim_{ntoinfty}int |f_n-f| dmu=0$ but I cannot see how shall I prove from there that $mu(E)$ or $mu(Omega)$ are limited.
Question:
Can someone provide me any help?
Thanks in advance!
Note:$f_n$ does not necessarily converge to $f$ uniformly. So the question is not a duplicate.
This question already has an answer here:
Omitting the hypotheses of finiteness of the measure in Egorov theorem
3 answers
functional-analysis measure-theory proof-writing
functional-analysis measure-theory proof-writing
asked Jan 13 at 12:27
Pedro GomesPedro Gomes
2,0082822
asked Jan 13 at 12:27
Pedro GomesPedro Gomes
2,0082822
edited Jan 13 at 15:14
Pedro Gomes
asked Jan 13 at 12:27
Pedro GomesPedro Gomes
2,0082822
asked Jan 13 at 12:27
Pedro GomesPedro Gomes
2,0082822
asked Jan 13 at 12:27
Pedro GomesPedro Gomes
2,0082822
2,0082822
marked as duplicate by Xander Henderson, mrtaurho, zz20s, Cesareo, metamorphy Jan 14 at 9:47
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
marked as duplicate by Xander Henderson, mrtaurho, zz20s, Cesareo, metamorphy Jan 14 at 9:47
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
$begingroup$
If $lambda>0$ then $mu({x:g(x)>lambda})<infty$.
$endgroup$
– David C. Ullrich
Jan 13 at 13:02
$begingroup$
@Jakobian My question is not a duplicate. If you read carefully the exercise you find out $f_n$ does not necessarily converge to $f$ uniformly. However the answer you provide assumes $f_n$ to converge uniformly so it is not answering this question.
$endgroup$
– Pedro Gomes
Jan 13 at 15:13
$begingroup$
@PedroGomes no, it doesn't. It's exactly the answer to your question
$endgroup$
– Jakobian
Jan 13 at 15:20
$begingroup$
@DavidC.Ullrich In order to apply Ergoroff I need to prove the measure of the domain where the function converges is finite. It is already assumed in the question when $f_nto f$ a.e that $mu({x:g(x)>lambda})=0$.
$endgroup$
– Pedro Gomes
Jan 13 at 15:21
1
$begingroup$
How do I know that you are pointing me in the right direction if you do not prove you are? Does it not sound like an authority fallacy?
$endgroup$
– Pedro Gomes
Jan 13 at 15:32
|
show 4 more comments
$begingroup$
If $lambda>0$ then $mu({x:g(x)>lambda})<infty$.
$endgroup$
– David C. Ullrich
Jan 13 at 13:02
$begingroup$
@Jakobian My question is not a duplicate. If you read carefully the exercise you find out $f_n$ does not necessarily converge to $f$ uniformly. However the answer you provide assumes $f_n$ to converge uniformly so it is not answering this question.
$endgroup$
– Pedro Gomes
Jan 13 at 15:13
$begingroup$
@PedroGomes no, it doesn't. It's exactly the answer to your question
$endgroup$
– Jakobian
Jan 13 at 15:20
$begingroup$
@DavidC.Ullrich In order to apply Ergoroff I need to prove the measure of the domain where the function converges is finite. It is already assumed in the question when $f_nto f$ a.e that $mu({x:g(x)>lambda})=0$.
$endgroup$
– Pedro Gomes
Jan 13 at 15:21
1
$begingroup$
How do I know that you are pointing me in the right direction if you do not prove you are? Does it not sound like an authority fallacy?
$endgroup$
– Pedro Gomes
Jan 13 at 15:32
$begingroup$
If $lambda>0$ then $mu({x:g(x)>lambda})<infty$.
$endgroup$
– David C. Ullrich
Jan 13 at 13:02
$begingroup$
If $lambda>0$ then $mu({x:g(x)>lambda})<infty$.
$endgroup$
– David C. Ullrich
Jan 13 at 13:02
$begingroup$
@Jakobian My question is not a duplicate. If you read carefully the exercise you find out $f_n$ does not necessarily converge to $f$ uniformly. However the answer you provide assumes $f_n$ to converge uniformly so it is not answering this question.
$endgroup$
– Pedro Gomes
Jan 13 at 15:13
$begingroup$
@Jakobian My question is not a duplicate. If you read carefully the exercise you find out $f_n$ does not necessarily converge to $f$ uniformly. However the answer you provide assumes $f_n$ to converge uniformly so it is not answering this question.
$endgroup$
– Pedro Gomes
Jan 13 at 15:13
$begingroup$
@PedroGomes no, it doesn't. It's exactly the answer to your question
$endgroup$
– Jakobian
Jan 13 at 15:20
$begingroup$
@PedroGomes no, it doesn't. It's exactly the answer to your question
$endgroup$
– Jakobian
Jan 13 at 15:20
$begingroup$
@DavidC.Ullrich In order to apply Ergoroff I need to prove the measure of the domain where the function converges is finite. It is already assumed in the question when $f_nto f$ a.e that $mu({x:g(x)>lambda})=0$.
$endgroup$
– Pedro Gomes
Jan 13 at 15:21
$begingroup$
@DavidC.Ullrich In order to apply Ergoroff I need to prove the measure of the domain where the function converges is finite. It is already assumed in the question when $f_nto f$ a.e that $mu({x:g(x)>lambda})=0$.
$endgroup$
– Pedro Gomes
Jan 13 at 15:21
1
1
$begingroup$
How do I know that you are pointing me in the right direction if you do not prove you are? Does it not sound like an authority fallacy?
$endgroup$
– Pedro Gomes
Jan 13 at 15:32
$begingroup$
How do I know that you are pointing me in the right direction if you do not prove you are? Does it not sound like an authority fallacy?
$endgroup$
– Pedro Gomes
Jan 13 at 15:32
|
show 4 more comments
1 Answer
1
active
oldest
votes
$begingroup$
Ok, a bigger hint. Let $$E_k={x:g(x)>1/k}.$$Since $mu(E_k)<infty$, Egoroff shows that there exists $S_ksubset E_k$ such that $f_nto f$ uniformly on $E_ksetminus S_k$ and $$mu(S_k)<epsilon/2^k.$$
So if $S=bigcup_{k=1}^infty S_k$ then $mu(S)<epsilon$. And it's possible to prove that $f_nto f$ uniformly on $Xsetminus S$. (There's still something to be proved in that last sentence, it's not quite just trivial by definition. Hint: So far we haven't used the fact that $|f_n|le g$.)
answered Jan 13 at 16:27
David C. UllrichDavid C. Ullrich
61.8k44095
$endgroup$
$begingroup$
Is it not $E_k={x:g(x)<1/k}$. $S_k$ is subset of the set where $f_n$ converges to $f$? Or is there something I am missing?
$endgroup$
– Pedro Gomes
Jan 13 at 17:50
$begingroup$
$E_k$ is not what you say it is. So I don't know whether you're asking about $E_k$ or about ${x:g(x)<1/k}$. But in either case: So what?
$endgroup$
– David C. Ullrich
Jan 13 at 23:09
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Ok, a bigger hint. Let $$E_k={x:g(x)>1/k}.$$Since $mu(E_k)<infty$, Egoroff shows that there exists $S_ksubset E_k$ such that $f_nto f$ uniformly on $E_ksetminus S_k$ and $$mu(S_k)<epsilon/2^k.$$
So if $S=bigcup_{k=1}^infty S_k$ then $mu(S)<epsilon$. And it's possible to prove that $f_nto f$ uniformly on $Xsetminus S$. (There's still something to be proved in that last sentence, it's not quite just trivial by definition. Hint: So far we haven't used the fact that $|f_n|le g$.)
answered Jan 13 at 16:27
David C. UllrichDavid C. Ullrich
61.8k44095
$endgroup$
$begingroup$
Is it not $E_k={x:g(x)<1/k}$. $S_k$ is subset of the set where $f_n$ converges to $f$? Or is there something I am missing?
$endgroup$
– Pedro Gomes
Jan 13 at 17:50
$begingroup$
$E_k$ is not what you say it is. So I don't know whether you're asking about $E_k$ or about ${x:g(x)<1/k}$. But in either case: So what?
$endgroup$
– David C. Ullrich
Jan 13 at 23:09
add a comment |
$begingroup$
Ok, a bigger hint. Let $$E_k={x:g(x)>1/k}.$$Since $mu(E_k)<infty$, Egoroff shows that there exists $S_ksubset E_k$ such that $f_nto f$ uniformly on $E_ksetminus S_k$ and $$mu(S_k)<epsilon/2^k.$$
So if $S=bigcup_{k=1}^infty S_k$ then $mu(S)<epsilon$. And it's possible to prove that $f_nto f$ uniformly on $Xsetminus S$. (There's still something to be proved in that last sentence, it's not quite just trivial by definition. Hint: So far we haven't used the fact that $|f_n|le g$.)
answered Jan 13 at 16:27
David C. UllrichDavid C. Ullrich
61.8k44095
$endgroup$
$begingroup$
Is it not $E_k={x:g(x)<1/k}$. $S_k$ is subset of the set where $f_n$ converges to $f$? Or is there something I am missing?
$endgroup$
– Pedro Gomes
Jan 13 at 17:50
$begingroup$
$E_k$ is not what you say it is. So I don't know whether you're asking about $E_k$ or about ${x:g(x)<1/k}$. But in either case: So what?
$endgroup$
– David C. Ullrich
Jan 13 at 23:09
add a comment |
$begingroup$
Ok, a bigger hint. Let $$E_k={x:g(x)>1/k}.$$Since $mu(E_k)<infty$, Egoroff shows that there exists $S_ksubset E_k$ such that $f_nto f$ uniformly on $E_ksetminus S_k$ and $$mu(S_k)<epsilon/2^k.$$
So if $S=bigcup_{k=1}^infty S_k$ then $mu(S)<epsilon$. And it's possible to prove that $f_nto f$ uniformly on $Xsetminus S$. (There's still something to be proved in that last sentence, it's not quite just trivial by definition. Hint: So far we haven't used the fact that $|f_n|le g$.)
answered Jan 13 at 16:27
David C. UllrichDavid C. Ullrich
61.8k44095
$endgroup$
Ok, a bigger hint. Let $$E_k={x:g(x)>1/k}.$$Since $mu(E_k)<infty$, Egoroff shows that there exists $S_ksubset E_k$ such that $f_nto f$ uniformly on $E_ksetminus S_k$ and $$mu(S_k)<epsilon/2^k.$$
So if $S=bigcup_{k=1}^infty S_k$ then $mu(S)<epsilon$. And it's possible to prove that $f_nto f$ uniformly on $Xsetminus S$. (There's still something to be proved in that last sentence, it's not quite just trivial by definition. Hint: So far we haven't used the fact that $|f_n|le g$.)
answered Jan 13 at 16:27
David C. UllrichDavid C. Ullrich
61.8k44095
answered Jan 13 at 16:27
David C. UllrichDavid C. Ullrich
61.8k44095
answered Jan 13 at 16:27
David C. UllrichDavid C. Ullrich
61.8k44095
answered Jan 13 at 16:27
David C. UllrichDavid C. Ullrich
61.8k44095
61.8k44095
$begingroup$
Is it not $E_k={x:g(x)<1/k}$. $S_k$ is subset of the set where $f_n$ converges to $f$? Or is there something I am missing?
$endgroup$
– Pedro Gomes
Jan 13 at 17:50
$begingroup$
$E_k$ is not what you say it is. So I don't know whether you're asking about $E_k$ or about ${x:g(x)<1/k}$. But in either case: So what?
$endgroup$
– David C. Ullrich
Jan 13 at 23:09
add a comment |
$begingroup$
Is it not $E_k={x:g(x)<1/k}$. $S_k$ is subset of the set where $f_n$ converges to $f$? Or is there something I am missing?
$endgroup$
– Pedro Gomes
Jan 13 at 17:50
$begingroup$
$E_k$ is not what you say it is. So I don't know whether you're asking about $E_k$ or about ${x:g(x)<1/k}$. But in either case: So what?
$endgroup$
– David C. Ullrich
Jan 13 at 23:09
$begingroup$
Is it not $E_k={x:g(x)<1/k}$. $S_k$ is subset of the set where $f_n$ converges to $f$? Or is there something I am missing?
$endgroup$
– Pedro Gomes
Jan 13 at 17:50
$begingroup$
Is it not $E_k={x:g(x)<1/k}$. $S_k$ is subset of the set where $f_n$ converges to $f$? Or is there something I am missing?
$endgroup$
– Pedro Gomes
Jan 13 at 17:50
$begingroup$
$E_k$ is not what you say it is. So I don't know whether you're asking about $E_k$ or about ${x:g(x)<1/k}$. But in either case: So what?
$endgroup$
– David C. Ullrich
Jan 13 at 23:09
$begingroup$
$E_k$ is not what you say it is. So I don't know whether you're asking about $E_k$ or about ${x:g(x)<1/k}$. But in either case: So what?
$endgroup$
– David C. Ullrich
Jan 13 at 23:09
add a comment |
$begingroup$
If $lambda>0$ then $mu({x:g(x)>lambda})<infty$.
$endgroup$
– David C. Ullrich
Jan 13 at 13:02
$begingroup$
@Jakobian My question is not a duplicate. If you read carefully the exercise you find out $f_n$ does not necessarily converge to $f$ uniformly. However the answer you provide assumes $f_n$ to converge uniformly so it is not answering this question.
$endgroup$
– Pedro Gomes
Jan 13 at 15:13
$begingroup$
@PedroGomes no, it doesn't. It's exactly the answer to your question
$endgroup$
– Jakobian
Jan 13 at 15:20
$begingroup$
@DavidC.Ullrich In order to apply Ergoroff I need to prove the measure of the domain where the function converges is finite. It is already assumed in the question when $f_nto f$ a.e that $mu({x:g(x)>lambda})=0$.
$endgroup$
– Pedro Gomes
Jan 13 at 15:21
1
$begingroup$
How do I know that you are pointing me in the right direction if you do not prove you are? Does it not sound like an authority fallacy?
$endgroup$
– Pedro Gomes
Jan 13 at 15:32