Prove there is no map $T$ from $L_0[0,1]$ to $mathbb R$ satisfies that $T(fn) to 0$ iff $fn to 0$ a.e. on...
$begingroup$
Prove there is no map $T$ from $L_{0}[0,1]$ to $mathbb R$ satisfies that $T(fn)to 0$ if and only if fn $to$ $0$
a.e. on $[0,1]$.
L0 means Lebesgue measurable.
Hint: consider the condition that $fn to 0$ in measure or integral while not $fn to 0 a.e. on [0,1]$
Thank you for any helpful idea.
real-analysis lebesgue-measure
asked Dec 13 '18 at 12:57
user10754650user10754650
162
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closed as off-topic by Davide Giraudo, José Carlos Santos, onurcanbektas, jameselmore, Rebellos Dec 13 '18 at 18:12
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." – Davide Giraudo, José Carlos Santos, onurcanbektas, jameselmore, Rebellos
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
$begingroup$
Prove there is no map $T$ from $L_{0}[0,1]$ to $mathbb R$ satisfies that $T(fn)to 0$ if and only if fn $to$ $0$
a.e. on $[0,1]$.
L0 means Lebesgue measurable.
Hint: consider the condition that $fn to 0$ in measure or integral while not $fn to 0 a.e. on [0,1]$
Thank you for any helpful idea.
real-analysis lebesgue-measure
asked Dec 13 '18 at 12:57
user10754650user10754650
162
$endgroup$
closed as off-topic by Davide Giraudo, José Carlos Santos, onurcanbektas, jameselmore, Rebellos Dec 13 '18 at 18:12
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." – Davide Giraudo, José Carlos Santos, onurcanbektas, jameselmore, Rebellos
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
$begingroup$
Prove there is no map $T$ from $L_{0}[0,1]$ to $mathbb R$ satisfies that $T(fn)to 0$ if and only if fn $to$ $0$
a.e. on $[0,1]$.
L0 means Lebesgue measurable.
Hint: consider the condition that $fn to 0$ in measure or integral while not $fn to 0 a.e. on [0,1]$
Thank you for any helpful idea.
real-analysis lebesgue-measure
asked Dec 13 '18 at 12:57
user10754650user10754650
162
$endgroup$
Prove there is no map $T$ from $L_{0}[0,1]$ to $mathbb R$ satisfies that $T(fn)to 0$ if and only if fn $to$ $0$
a.e. on $[0,1]$.
L0 means Lebesgue measurable.
Hint: consider the condition that $fn to 0$ in measure or integral while not $fn to 0 a.e. on [0,1]$
Thank you for any helpful idea.
real-analysis lebesgue-measure
real-analysis lebesgue-measure
asked Dec 13 '18 at 12:57
user10754650user10754650
162
asked Dec 13 '18 at 12:57
user10754650user10754650
162
edited Dec 13 '18 at 14:13
user10754650
asked Dec 13 '18 at 12:57
user10754650user10754650
162
asked Dec 13 '18 at 12:57
user10754650user10754650
162
asked Dec 13 '18 at 12:57
user10754650user10754650
162
162
closed as off-topic by Davide Giraudo, José Carlos Santos, onurcanbektas, jameselmore, Rebellos Dec 13 '18 at 18:12
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." – Davide Giraudo, José Carlos Santos, onurcanbektas, jameselmore, Rebellos
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by Davide Giraudo, José Carlos Santos, onurcanbektas, jameselmore, Rebellos Dec 13 '18 at 18:12
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." – Davide Giraudo, José Carlos Santos, onurcanbektas, jameselmore, Rebellos
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
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$L_0[0,1]$ is the same as $U[0,1]$ probability distribution. If there is a map $T$ such that $T(f_n)rightarrow 0 iff f_nrightarrow 0$ a.s. then consider a sequence of random variables $X_n$ and a random variable $X$ on $[0,1]$ such that $X_nrightarrow X$ in probability $L_0[0,1]$, but not $X_nrightarrow X$ almost surely. Then for any sub-sequence $n_k$ there is a further sub-sequence $n_{k_l}$ such that $X_{n_{k_l}}rightarrow X$ alomst surely. This implies for any sub-sequence $n_k$ there is a further sub-sequence $n_{k_l}$ such that $T(X_{n_{k_l}}-X)rightarrow 0$. So, we get that $T(X_n-X)rightarrow 0Rightarrow X_nrightarrow X$ almost surely, which is a contradiction.
answered Dec 13 '18 at 13:46
John_WickJohn_Wick
1,621111
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Your first statement is a bit confusing; I think what you mean to say is that $f_n to f$ in $L^0$ is the same as $f_n to f$ in measure.
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– Ian
Dec 13 '18 at 13:55
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$U[0,1]$ is a uniform probability distribution over $[0,1].$
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– John_Wick
Dec 13 '18 at 14:38
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And I clearly mentioned when I said in probability convergence and when I said almost sure convergence.
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– John_Wick
Dec 13 '18 at 14:39
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I just meant the first sentence in isolation, I still don't know what exactly you are trying to say there. Perhaps that $L^0[0,1]$ can be interpreted as all measurable functions of a $U[0,1]$ random variable?
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– Ian
Dec 13 '18 at 17:08
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How could you conclude T(Xn-X)->0 from T(Xnkl -X)->0 ?
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– user10754650
Dec 18 '18 at 6:09
add a comment |
$begingroup$
Suppose $T$ exists and take $fin text{Ker}(T)setminus {0}$. As $f$ is non zero we have $|f|geq ccdot 1_A$ for some set $A$ of positive measure and some positive $c$. Then is easy to show that $ncdot fnot rightarrow 0$ even though $T(ncdot f)rightarrow T(0)$.
answered Dec 13 '18 at 14:58
Yesterday was dramaticYesterday was dramatic
960718
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Where did you get that $T$ would have nontrivial kernel?
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– Ian
Dec 13 '18 at 17:09
1
$begingroup$
Because of a dimension argument. If $T$ is injective you are including $L_0[0,1]$ as a vector subspace of $mathbb{R}$ of dimension 1.
$endgroup$
– Yesterday was dramatic
Dec 13 '18 at 17:24
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
$L_0[0,1]$ is the same as $U[0,1]$ probability distribution. If there is a map $T$ such that $T(f_n)rightarrow 0 iff f_nrightarrow 0$ a.s. then consider a sequence of random variables $X_n$ and a random variable $X$ on $[0,1]$ such that $X_nrightarrow X$ in probability $L_0[0,1]$, but not $X_nrightarrow X$ almost surely. Then for any sub-sequence $n_k$ there is a further sub-sequence $n_{k_l}$ such that $X_{n_{k_l}}rightarrow X$ alomst surely. This implies for any sub-sequence $n_k$ there is a further sub-sequence $n_{k_l}$ such that $T(X_{n_{k_l}}-X)rightarrow 0$. So, we get that $T(X_n-X)rightarrow 0Rightarrow X_nrightarrow X$ almost surely, which is a contradiction.
answered Dec 13 '18 at 13:46
John_WickJohn_Wick
1,621111
$endgroup$
$begingroup$
Your first statement is a bit confusing; I think what you mean to say is that $f_n to f$ in $L^0$ is the same as $f_n to f$ in measure.
$endgroup$
– Ian
Dec 13 '18 at 13:55
$begingroup$
$U[0,1]$ is a uniform probability distribution over $[0,1].$
$endgroup$
– John_Wick
Dec 13 '18 at 14:38
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And I clearly mentioned when I said in probability convergence and when I said almost sure convergence.
$endgroup$
– John_Wick
Dec 13 '18 at 14:39
$begingroup$
I just meant the first sentence in isolation, I still don't know what exactly you are trying to say there. Perhaps that $L^0[0,1]$ can be interpreted as all measurable functions of a $U[0,1]$ random variable?
$endgroup$
– Ian
Dec 13 '18 at 17:08
$begingroup$
How could you conclude T(Xn-X)->0 from T(Xnkl -X)->0 ?
$endgroup$
– user10754650
Dec 18 '18 at 6:09
add a comment |
$begingroup$
$L_0[0,1]$ is the same as $U[0,1]$ probability distribution. If there is a map $T$ such that $T(f_n)rightarrow 0 iff f_nrightarrow 0$ a.s. then consider a sequence of random variables $X_n$ and a random variable $X$ on $[0,1]$ such that $X_nrightarrow X$ in probability $L_0[0,1]$, but not $X_nrightarrow X$ almost surely. Then for any sub-sequence $n_k$ there is a further sub-sequence $n_{k_l}$ such that $X_{n_{k_l}}rightarrow X$ alomst surely. This implies for any sub-sequence $n_k$ there is a further sub-sequence $n_{k_l}$ such that $T(X_{n_{k_l}}-X)rightarrow 0$. So, we get that $T(X_n-X)rightarrow 0Rightarrow X_nrightarrow X$ almost surely, which is a contradiction.
answered Dec 13 '18 at 13:46
John_WickJohn_Wick
1,621111
$endgroup$
$begingroup$
Your first statement is a bit confusing; I think what you mean to say is that $f_n to f$ in $L^0$ is the same as $f_n to f$ in measure.
$endgroup$
– Ian
Dec 13 '18 at 13:55
$begingroup$
$U[0,1]$ is a uniform probability distribution over $[0,1].$
$endgroup$
– John_Wick
Dec 13 '18 at 14:38
$begingroup$
And I clearly mentioned when I said in probability convergence and when I said almost sure convergence.
$endgroup$
– John_Wick
Dec 13 '18 at 14:39
$begingroup$
I just meant the first sentence in isolation, I still don't know what exactly you are trying to say there. Perhaps that $L^0[0,1]$ can be interpreted as all measurable functions of a $U[0,1]$ random variable?
$endgroup$
– Ian
Dec 13 '18 at 17:08
$begingroup$
How could you conclude T(Xn-X)->0 from T(Xnkl -X)->0 ?
$endgroup$
– user10754650
Dec 18 '18 at 6:09
add a comment |
$begingroup$
$L_0[0,1]$ is the same as $U[0,1]$ probability distribution. If there is a map $T$ such that $T(f_n)rightarrow 0 iff f_nrightarrow 0$ a.s. then consider a sequence of random variables $X_n$ and a random variable $X$ on $[0,1]$ such that $X_nrightarrow X$ in probability $L_0[0,1]$, but not $X_nrightarrow X$ almost surely. Then for any sub-sequence $n_k$ there is a further sub-sequence $n_{k_l}$ such that $X_{n_{k_l}}rightarrow X$ alomst surely. This implies for any sub-sequence $n_k$ there is a further sub-sequence $n_{k_l}$ such that $T(X_{n_{k_l}}-X)rightarrow 0$. So, we get that $T(X_n-X)rightarrow 0Rightarrow X_nrightarrow X$ almost surely, which is a contradiction.
answered Dec 13 '18 at 13:46
John_WickJohn_Wick
1,621111
$endgroup$
$L_0[0,1]$ is the same as $U[0,1]$ probability distribution. If there is a map $T$ such that $T(f_n)rightarrow 0 iff f_nrightarrow 0$ a.s. then consider a sequence of random variables $X_n$ and a random variable $X$ on $[0,1]$ such that $X_nrightarrow X$ in probability $L_0[0,1]$, but not $X_nrightarrow X$ almost surely. Then for any sub-sequence $n_k$ there is a further sub-sequence $n_{k_l}$ such that $X_{n_{k_l}}rightarrow X$ alomst surely. This implies for any sub-sequence $n_k$ there is a further sub-sequence $n_{k_l}$ such that $T(X_{n_{k_l}}-X)rightarrow 0$. So, we get that $T(X_n-X)rightarrow 0Rightarrow X_nrightarrow X$ almost surely, which is a contradiction.
answered Dec 13 '18 at 13:46
John_WickJohn_Wick
1,621111
answered Dec 13 '18 at 13:46
John_WickJohn_Wick
1,621111
answered Dec 13 '18 at 13:46
John_WickJohn_Wick
1,621111
answered Dec 13 '18 at 13:46
John_WickJohn_Wick
1,621111
1,621111
$begingroup$
Your first statement is a bit confusing; I think what you mean to say is that $f_n to f$ in $L^0$ is the same as $f_n to f$ in measure.
$endgroup$
– Ian
Dec 13 '18 at 13:55
$begingroup$
$U[0,1]$ is a uniform probability distribution over $[0,1].$
$endgroup$
– John_Wick
Dec 13 '18 at 14:38
$begingroup$
And I clearly mentioned when I said in probability convergence and when I said almost sure convergence.
$endgroup$
– John_Wick
Dec 13 '18 at 14:39
$begingroup$
I just meant the first sentence in isolation, I still don't know what exactly you are trying to say there. Perhaps that $L^0[0,1]$ can be interpreted as all measurable functions of a $U[0,1]$ random variable?
$endgroup$
– Ian
Dec 13 '18 at 17:08
$begingroup$
How could you conclude T(Xn-X)->0 from T(Xnkl -X)->0 ?
$endgroup$
– user10754650
Dec 18 '18 at 6:09
add a comment |
$begingroup$
Your first statement is a bit confusing; I think what you mean to say is that $f_n to f$ in $L^0$ is the same as $f_n to f$ in measure.
$endgroup$
– Ian
Dec 13 '18 at 13:55
$begingroup$
$U[0,1]$ is a uniform probability distribution over $[0,1].$
$endgroup$
– John_Wick
Dec 13 '18 at 14:38
$begingroup$
And I clearly mentioned when I said in probability convergence and when I said almost sure convergence.
$endgroup$
– John_Wick
Dec 13 '18 at 14:39
$begingroup$
I just meant the first sentence in isolation, I still don't know what exactly you are trying to say there. Perhaps that $L^0[0,1]$ can be interpreted as all measurable functions of a $U[0,1]$ random variable?
$endgroup$
– Ian
Dec 13 '18 at 17:08
$begingroup$
How could you conclude T(Xn-X)->0 from T(Xnkl -X)->0 ?
$endgroup$
– user10754650
Dec 18 '18 at 6:09
$begingroup$
Your first statement is a bit confusing; I think what you mean to say is that $f_n to f$ in $L^0$ is the same as $f_n to f$ in measure.
$endgroup$
– Ian
Dec 13 '18 at 13:55
$begingroup$
Your first statement is a bit confusing; I think what you mean to say is that $f_n to f$ in $L^0$ is the same as $f_n to f$ in measure.
$endgroup$
– Ian
Dec 13 '18 at 13:55
$begingroup$
$U[0,1]$ is a uniform probability distribution over $[0,1].$
$endgroup$
– John_Wick
Dec 13 '18 at 14:38
$begingroup$
$U[0,1]$ is a uniform probability distribution over $[0,1].$
$endgroup$
– John_Wick
Dec 13 '18 at 14:38
$begingroup$
And I clearly mentioned when I said in probability convergence and when I said almost sure convergence.
$endgroup$
– John_Wick
Dec 13 '18 at 14:39
$begingroup$
And I clearly mentioned when I said in probability convergence and when I said almost sure convergence.
$endgroup$
– John_Wick
Dec 13 '18 at 14:39
$begingroup$
I just meant the first sentence in isolation, I still don't know what exactly you are trying to say there. Perhaps that $L^0[0,1]$ can be interpreted as all measurable functions of a $U[0,1]$ random variable?
$endgroup$
– Ian
Dec 13 '18 at 17:08
$begingroup$
I just meant the first sentence in isolation, I still don't know what exactly you are trying to say there. Perhaps that $L^0[0,1]$ can be interpreted as all measurable functions of a $U[0,1]$ random variable?
$endgroup$
– Ian
Dec 13 '18 at 17:08
$begingroup$
How could you conclude T(Xn-X)->0 from T(Xnkl -X)->0 ?
$endgroup$
– user10754650
Dec 18 '18 at 6:09
$begingroup$
How could you conclude T(Xn-X)->0 from T(Xnkl -X)->0 ?
$endgroup$
– user10754650
Dec 18 '18 at 6:09
add a comment |
$begingroup$
Suppose $T$ exists and take $fin text{Ker}(T)setminus {0}$. As $f$ is non zero we have $|f|geq ccdot 1_A$ for some set $A$ of positive measure and some positive $c$. Then is easy to show that $ncdot fnot rightarrow 0$ even though $T(ncdot f)rightarrow T(0)$.
answered Dec 13 '18 at 14:58
Yesterday was dramaticYesterday was dramatic
960718
$endgroup$
$begingroup$
Where did you get that $T$ would have nontrivial kernel?
$endgroup$
– Ian
Dec 13 '18 at 17:09
1
$begingroup$
Because of a dimension argument. If $T$ is injective you are including $L_0[0,1]$ as a vector subspace of $mathbb{R}$ of dimension 1.
$endgroup$
– Yesterday was dramatic
Dec 13 '18 at 17:24
add a comment |
$begingroup$
Suppose $T$ exists and take $fin text{Ker}(T)setminus {0}$. As $f$ is non zero we have $|f|geq ccdot 1_A$ for some set $A$ of positive measure and some positive $c$. Then is easy to show that $ncdot fnot rightarrow 0$ even though $T(ncdot f)rightarrow T(0)$.
answered Dec 13 '18 at 14:58
Yesterday was dramaticYesterday was dramatic
960718
$endgroup$
$begingroup$
Where did you get that $T$ would have nontrivial kernel?
$endgroup$
– Ian
Dec 13 '18 at 17:09
1
$begingroup$
Because of a dimension argument. If $T$ is injective you are including $L_0[0,1]$ as a vector subspace of $mathbb{R}$ of dimension 1.
$endgroup$
– Yesterday was dramatic
Dec 13 '18 at 17:24
add a comment |
$begingroup$
Suppose $T$ exists and take $fin text{Ker}(T)setminus {0}$. As $f$ is non zero we have $|f|geq ccdot 1_A$ for some set $A$ of positive measure and some positive $c$. Then is easy to show that $ncdot fnot rightarrow 0$ even though $T(ncdot f)rightarrow T(0)$.
answered Dec 13 '18 at 14:58
Yesterday was dramaticYesterday was dramatic
960718
$endgroup$
Suppose $T$ exists and take $fin text{Ker}(T)setminus {0}$. As $f$ is non zero we have $|f|geq ccdot 1_A$ for some set $A$ of positive measure and some positive $c$. Then is easy to show that $ncdot fnot rightarrow 0$ even though $T(ncdot f)rightarrow T(0)$.
answered Dec 13 '18 at 14:58
Yesterday was dramaticYesterday was dramatic
960718
answered Dec 13 '18 at 14:58
Yesterday was dramaticYesterday was dramatic
960718
answered Dec 13 '18 at 14:58
Yesterday was dramaticYesterday was dramatic
960718
answered Dec 13 '18 at 14:58
Yesterday was dramaticYesterday was dramatic
960718
960718
$begingroup$
Where did you get that $T$ would have nontrivial kernel?
$endgroup$
– Ian
Dec 13 '18 at 17:09
1
$begingroup$
Because of a dimension argument. If $T$ is injective you are including $L_0[0,1]$ as a vector subspace of $mathbb{R}$ of dimension 1.
$endgroup$
– Yesterday was dramatic
Dec 13 '18 at 17:24
add a comment |
$begingroup$
Where did you get that $T$ would have nontrivial kernel?
$endgroup$
– Ian
Dec 13 '18 at 17:09
1
$begingroup$
Because of a dimension argument. If $T$ is injective you are including $L_0[0,1]$ as a vector subspace of $mathbb{R}$ of dimension 1.
$endgroup$
– Yesterday was dramatic
Dec 13 '18 at 17:24
$begingroup$
Where did you get that $T$ would have nontrivial kernel?
$endgroup$
– Ian
Dec 13 '18 at 17:09
$begingroup$
Where did you get that $T$ would have nontrivial kernel?
$endgroup$
– Ian
Dec 13 '18 at 17:09
1
1
$begingroup$
Because of a dimension argument. If $T$ is injective you are including $L_0[0,1]$ as a vector subspace of $mathbb{R}$ of dimension 1.
$endgroup$
– Yesterday was dramatic
Dec 13 '18 at 17:24
$begingroup$
Because of a dimension argument. If $T$ is injective you are including $L_0[0,1]$ as a vector subspace of $mathbb{R}$ of dimension 1.
$endgroup$
– Yesterday was dramatic
Dec 13 '18 at 17:24
add a comment |
