$lambda$-almost-everywhere convergence implied by lim$lambda (${$x in E : | f_n(x) - f(x) | > epsilon$}$)...
$begingroup$
Let $E subset mathbb{R}$ be $lambda$-measurable and let $f_n,f: E -> mathbb{R}$ $lambda$-integrable, so that for all $epsilon > 0$
$lambda(${$xin E: |f_n(x)-f(x)| > epsilon$}$)->0$ as $n -> infty$
Proof that $f_n$ converges to $f$ $lambda$-almost-everywhere.
My attempt:
Assume $f_n$ does not converge almost everywhere to $f$ then there exists an $ epsilon$ for which for all $N in mathbb{N}$ there's a $n>=N$ with $|f_n(x)-f(x)|>epsilon$ for all $x in A$ with $A$ being a set that's measure isn't zero.
Then x is in infinitely many $H_n^epsilon:=${$x in E : |f_n(x)-f(x)|>epsilon$} and thus $sum_{n=1}^inftylambda(H_n^epsilon)=infty$.
On the other hand, since $f_n,f$ are integrable we get:
$infty>int_E|f_n(x)-f(x)|dlambda>int_{H_n^epsilon}|f_n(x)-f(x)|dlambda=sum_{n}int_{H_n^epsilon}|f_n(x)-f(x)|dlambda>sum_{n}epsilon lambda(H_n^epsilon)=infty$
What is a contradiction. And the assumption must be wrong.
I guess this is wrong as I didn't even use $lambda(${$xin E: |f_n(x)-f(x)| > epsilon$}$)->0$ as $n -> infty$. However, I do not find my mistake nor have I an idea on how to approach this problem.
I would really appreciate any help.
measure-theory lebesgue-measure almost-everywhere
asked Dec 4 '18 at 19:26
KekksKekks
418
$endgroup$
|
show 1 more comment
$begingroup$
Let $E subset mathbb{R}$ be $lambda$-measurable and let $f_n,f: E -> mathbb{R}$ $lambda$-integrable, so that for all $epsilon > 0$
$lambda(${$xin E: |f_n(x)-f(x)| > epsilon$}$)->0$ as $n -> infty$
Proof that $f_n$ converges to $f$ $lambda$-almost-everywhere.
My attempt:
Assume $f_n$ does not converge almost everywhere to $f$ then there exists an $ epsilon$ for which for all $N in mathbb{N}$ there's a $n>=N$ with $|f_n(x)-f(x)|>epsilon$ for all $x in A$ with $A$ being a set that's measure isn't zero.
Then x is in infinitely many $H_n^epsilon:=${$x in E : |f_n(x)-f(x)|>epsilon$} and thus $sum_{n=1}^inftylambda(H_n^epsilon)=infty$.
On the other hand, since $f_n,f$ are integrable we get:
$infty>int_E|f_n(x)-f(x)|dlambda>int_{H_n^epsilon}|f_n(x)-f(x)|dlambda=sum_{n}int_{H_n^epsilon}|f_n(x)-f(x)|dlambda>sum_{n}epsilon lambda(H_n^epsilon)=infty$
What is a contradiction. And the assumption must be wrong.
I guess this is wrong as I didn't even use $lambda(${$xin E: |f_n(x)-f(x)| > epsilon$}$)->0$ as $n -> infty$. However, I do not find my mistake nor have I an idea on how to approach this problem.
I would really appreciate any help.
measure-theory lebesgue-measure almost-everywhere
asked Dec 4 '18 at 19:26
KekksKekks
418
$endgroup$
$begingroup$
I didn't read your attempt, but maybe you want to read en.wikipedia.org/wiki/Borel%E2%80%93Cantelli_lemma. This is the standard tool to prove it
$endgroup$
– Federico
Dec 4 '18 at 19:33
$begingroup$
I think your statement of what it means to not converge almost everywhere is not quite right. It seems like you're claiming: There exists $epsilon > 0$ and a set of positive measure $A$ such that, for any $N in mathbb{N}$ there is $n > N$ with $|f_n(x) - f(x)| > epsilon$ for all $x in A$. Rather I think it should be the following: There exists a set $A$ of positive measure, such that for all $x in A$ there exists an $epsilon > 0$ such that, for any $N in mathbb{N}$ there exists $n > N$ with $|f_n(x) - f(x)| geq epsilon$.
$endgroup$
– Sean Haight
Dec 4 '18 at 19:53
$begingroup$
@SeanHaight you're right.
$endgroup$
– Kekks
Dec 4 '18 at 20:06
$begingroup$
@Federico so i use the integrability of $f_n,f$ to shot that $sum_n lambda(H_n^{epsilon}) < infty$ so i can use Borel-Cantelli which implies the convergence a.e.?
$endgroup$
– Kekks
Dec 4 '18 at 20:06
$begingroup$
You write the set of non-convergence as an intersection of a union, like in the statement of Borel-Cantelli lemma, then use it to show that it has measure $0$.
$endgroup$
– Federico
Dec 4 '18 at 20:09
|
show 1 more comment
$begingroup$
Let $E subset mathbb{R}$ be $lambda$-measurable and let $f_n,f: E -> mathbb{R}$ $lambda$-integrable, so that for all $epsilon > 0$
$lambda(${$xin E: |f_n(x)-f(x)| > epsilon$}$)->0$ as $n -> infty$
Proof that $f_n$ converges to $f$ $lambda$-almost-everywhere.
My attempt:
Assume $f_n$ does not converge almost everywhere to $f$ then there exists an $ epsilon$ for which for all $N in mathbb{N}$ there's a $n>=N$ with $|f_n(x)-f(x)|>epsilon$ for all $x in A$ with $A$ being a set that's measure isn't zero.
Then x is in infinitely many $H_n^epsilon:=${$x in E : |f_n(x)-f(x)|>epsilon$} and thus $sum_{n=1}^inftylambda(H_n^epsilon)=infty$.
On the other hand, since $f_n,f$ are integrable we get:
$infty>int_E|f_n(x)-f(x)|dlambda>int_{H_n^epsilon}|f_n(x)-f(x)|dlambda=sum_{n}int_{H_n^epsilon}|f_n(x)-f(x)|dlambda>sum_{n}epsilon lambda(H_n^epsilon)=infty$
What is a contradiction. And the assumption must be wrong.
I guess this is wrong as I didn't even use $lambda(${$xin E: |f_n(x)-f(x)| > epsilon$}$)->0$ as $n -> infty$. However, I do not find my mistake nor have I an idea on how to approach this problem.
I would really appreciate any help.
measure-theory lebesgue-measure almost-everywhere
asked Dec 4 '18 at 19:26
KekksKekks
418
$endgroup$
Let $E subset mathbb{R}$ be $lambda$-measurable and let $f_n,f: E -> mathbb{R}$ $lambda$-integrable, so that for all $epsilon > 0$
$lambda(${$xin E: |f_n(x)-f(x)| > epsilon$}$)->0$ as $n -> infty$
Proof that $f_n$ converges to $f$ $lambda$-almost-everywhere.
My attempt:
Assume $f_n$ does not converge almost everywhere to $f$ then there exists an $ epsilon$ for which for all $N in mathbb{N}$ there's a $n>=N$ with $|f_n(x)-f(x)|>epsilon$ for all $x in A$ with $A$ being a set that's measure isn't zero.
Then x is in infinitely many $H_n^epsilon:=${$x in E : |f_n(x)-f(x)|>epsilon$} and thus $sum_{n=1}^inftylambda(H_n^epsilon)=infty$.
On the other hand, since $f_n,f$ are integrable we get:
$infty>int_E|f_n(x)-f(x)|dlambda>int_{H_n^epsilon}|f_n(x)-f(x)|dlambda=sum_{n}int_{H_n^epsilon}|f_n(x)-f(x)|dlambda>sum_{n}epsilon lambda(H_n^epsilon)=infty$
What is a contradiction. And the assumption must be wrong.
I guess this is wrong as I didn't even use $lambda(${$xin E: |f_n(x)-f(x)| > epsilon$}$)->0$ as $n -> infty$. However, I do not find my mistake nor have I an idea on how to approach this problem.
I would really appreciate any help.
measure-theory lebesgue-measure almost-everywhere
measure-theory lebesgue-measure almost-everywhere
asked Dec 4 '18 at 19:26
KekksKekks
418
asked Dec 4 '18 at 19:26
KekksKekks
418
asked Dec 4 '18 at 19:26
KekksKekks
418
asked Dec 4 '18 at 19:26
KekksKekks
418
asked Dec 4 '18 at 19:26
KekksKekks
418
418
$begingroup$
I didn't read your attempt, but maybe you want to read en.wikipedia.org/wiki/Borel%E2%80%93Cantelli_lemma. This is the standard tool to prove it
$endgroup$
– Federico
Dec 4 '18 at 19:33
$begingroup$
I think your statement of what it means to not converge almost everywhere is not quite right. It seems like you're claiming: There exists $epsilon > 0$ and a set of positive measure $A$ such that, for any $N in mathbb{N}$ there is $n > N$ with $|f_n(x) - f(x)| > epsilon$ for all $x in A$. Rather I think it should be the following: There exists a set $A$ of positive measure, such that for all $x in A$ there exists an $epsilon > 0$ such that, for any $N in mathbb{N}$ there exists $n > N$ with $|f_n(x) - f(x)| geq epsilon$.
$endgroup$
– Sean Haight
Dec 4 '18 at 19:53
$begingroup$
@SeanHaight you're right.
$endgroup$
– Kekks
Dec 4 '18 at 20:06
$begingroup$
@Federico so i use the integrability of $f_n,f$ to shot that $sum_n lambda(H_n^{epsilon}) < infty$ so i can use Borel-Cantelli which implies the convergence a.e.?
$endgroup$
– Kekks
Dec 4 '18 at 20:06
$begingroup$
You write the set of non-convergence as an intersection of a union, like in the statement of Borel-Cantelli lemma, then use it to show that it has measure $0$.
$endgroup$
– Federico
Dec 4 '18 at 20:09
|
show 1 more comment
$begingroup$
I didn't read your attempt, but maybe you want to read en.wikipedia.org/wiki/Borel%E2%80%93Cantelli_lemma. This is the standard tool to prove it
$endgroup$
– Federico
Dec 4 '18 at 19:33
$begingroup$
I think your statement of what it means to not converge almost everywhere is not quite right. It seems like you're claiming: There exists $epsilon > 0$ and a set of positive measure $A$ such that, for any $N in mathbb{N}$ there is $n > N$ with $|f_n(x) - f(x)| > epsilon$ for all $x in A$. Rather I think it should be the following: There exists a set $A$ of positive measure, such that for all $x in A$ there exists an $epsilon > 0$ such that, for any $N in mathbb{N}$ there exists $n > N$ with $|f_n(x) - f(x)| geq epsilon$.
$endgroup$
– Sean Haight
Dec 4 '18 at 19:53
$begingroup$
@SeanHaight you're right.
$endgroup$
– Kekks
Dec 4 '18 at 20:06
$begingroup$
@Federico so i use the integrability of $f_n,f$ to shot that $sum_n lambda(H_n^{epsilon}) < infty$ so i can use Borel-Cantelli which implies the convergence a.e.?
$endgroup$
– Kekks
Dec 4 '18 at 20:06
$begingroup$
You write the set of non-convergence as an intersection of a union, like in the statement of Borel-Cantelli lemma, then use it to show that it has measure $0$.
$endgroup$
– Federico
Dec 4 '18 at 20:09
$begingroup$
I didn't read your attempt, but maybe you want to read en.wikipedia.org/wiki/Borel%E2%80%93Cantelli_lemma. This is the standard tool to prove it
$endgroup$
– Federico
Dec 4 '18 at 19:33
$begingroup$
I didn't read your attempt, but maybe you want to read en.wikipedia.org/wiki/Borel%E2%80%93Cantelli_lemma. This is the standard tool to prove it
$endgroup$
– Federico
Dec 4 '18 at 19:33
$begingroup$
I think your statement of what it means to not converge almost everywhere is not quite right. It seems like you're claiming: There exists $epsilon > 0$ and a set of positive measure $A$ such that, for any $N in mathbb{N}$ there is $n > N$ with $|f_n(x) - f(x)| > epsilon$ for all $x in A$. Rather I think it should be the following: There exists a set $A$ of positive measure, such that for all $x in A$ there exists an $epsilon > 0$ such that, for any $N in mathbb{N}$ there exists $n > N$ with $|f_n(x) - f(x)| geq epsilon$.
$endgroup$
– Sean Haight
Dec 4 '18 at 19:53
$begingroup$
I think your statement of what it means to not converge almost everywhere is not quite right. It seems like you're claiming: There exists $epsilon > 0$ and a set of positive measure $A$ such that, for any $N in mathbb{N}$ there is $n > N$ with $|f_n(x) - f(x)| > epsilon$ for all $x in A$. Rather I think it should be the following: There exists a set $A$ of positive measure, such that for all $x in A$ there exists an $epsilon > 0$ such that, for any $N in mathbb{N}$ there exists $n > N$ with $|f_n(x) - f(x)| geq epsilon$.
$endgroup$
– Sean Haight
Dec 4 '18 at 19:53
$begingroup$
@SeanHaight you're right.
$endgroup$
– Kekks
Dec 4 '18 at 20:06
$begingroup$
@SeanHaight you're right.
$endgroup$
– Kekks
Dec 4 '18 at 20:06
$begingroup$
@Federico so i use the integrability of $f_n,f$ to shot that $sum_n lambda(H_n^{epsilon}) < infty$ so i can use Borel-Cantelli which implies the convergence a.e.?
$endgroup$
– Kekks
Dec 4 '18 at 20:06
$begingroup$
@Federico so i use the integrability of $f_n,f$ to shot that $sum_n lambda(H_n^{epsilon}) < infty$ so i can use Borel-Cantelli which implies the convergence a.e.?
$endgroup$
– Kekks
Dec 4 '18 at 20:06
$begingroup$
You write the set of non-convergence as an intersection of a union, like in the statement of Borel-Cantelli lemma, then use it to show that it has measure $0$.
$endgroup$
– Federico
Dec 4 '18 at 20:09
$begingroup$
You write the set of non-convergence as an intersection of a union, like in the statement of Borel-Cantelli lemma, then use it to show that it has measure $0$.
$endgroup$
– Federico
Dec 4 '18 at 20:09
|
show 1 more comment
1 Answer
1
active
oldest
votes
$begingroup$
Fix $k$ and pick $n_k$ such that $lambda E_k le {1 over 2^k}$ where $E_k ={ x | |f_{n_k}(x)-f(x)| > { 1over k} }$.
This is the Borel Cantelli lemma:
Note that $int sum_k 1_{E_k} = sum_k int 1_{E_k} = sum_{k} lambda E_k < infty$
and so $sum_k 1_{E_k}$ is finite ae. Hence for ae. $x$, $x$ is in at most a finite number of $E_k$ and so $f_{n_k}(x) to f(x)$.
Addendum:
The convergence cannot be everywhere. Here is an example that is easier to describe than to write an explicit expression. Choose the space $[0,1]$, let $f = 0$
and define $f_n$ as follows: $f_1 = 1_{[0,{1 over 2}]}, f_2 = 1_{[{1 over 2},1]}$, $f_3 = 1_{[0,{1 over 4}]}$, $f_4 = 1_{[{1 over 4}, {1 over 2}]}$, $f_5 = 1_{[{1 over 2}, {3 over 4}]}$, etc. It should be clear that $lambda { x | |f_n(x)| > epsilon} to 0$, but for every $x in [0,1)$, we have $f_n(x) = 1$ infinitely often and $f_n(x) = 0$ and infinitely often.
However, if we pick the subsequence corresponding to the functions $1_{[0,{1 over 2^n}]}$ then we see that this subsequence converges pointwise to $0$ for $x in (0,1]$.
answered Dec 4 '18 at 21:16
copper.hatcopper.hat
126k559160
$endgroup$
$begingroup$
This only proves that a subsequence converges, isnt it?
$endgroup$
– Kekks
Dec 4 '18 at 21:24
$begingroup$
Correct. ${}{}{}$
$endgroup$
– copper.hat
Dec 4 '18 at 21:25
add a comment |
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$begingroup$
Fix $k$ and pick $n_k$ such that $lambda E_k le {1 over 2^k}$ where $E_k ={ x | |f_{n_k}(x)-f(x)| > { 1over k} }$.
This is the Borel Cantelli lemma:
Note that $int sum_k 1_{E_k} = sum_k int 1_{E_k} = sum_{k} lambda E_k < infty$
and so $sum_k 1_{E_k}$ is finite ae. Hence for ae. $x$, $x$ is in at most a finite number of $E_k$ and so $f_{n_k}(x) to f(x)$.
Addendum:
The convergence cannot be everywhere. Here is an example that is easier to describe than to write an explicit expression. Choose the space $[0,1]$, let $f = 0$
and define $f_n$ as follows: $f_1 = 1_{[0,{1 over 2}]}, f_2 = 1_{[{1 over 2},1]}$, $f_3 = 1_{[0,{1 over 4}]}$, $f_4 = 1_{[{1 over 4}, {1 over 2}]}$, $f_5 = 1_{[{1 over 2}, {3 over 4}]}$, etc. It should be clear that $lambda { x | |f_n(x)| > epsilon} to 0$, but for every $x in [0,1)$, we have $f_n(x) = 1$ infinitely often and $f_n(x) = 0$ and infinitely often.
However, if we pick the subsequence corresponding to the functions $1_{[0,{1 over 2^n}]}$ then we see that this subsequence converges pointwise to $0$ for $x in (0,1]$.
answered Dec 4 '18 at 21:16
copper.hatcopper.hat
126k559160
$endgroup$
$begingroup$
This only proves that a subsequence converges, isnt it?
$endgroup$
– Kekks
Dec 4 '18 at 21:24
$begingroup$
Correct. ${}{}{}$
$endgroup$
– copper.hat
Dec 4 '18 at 21:25
add a comment |
$begingroup$
Fix $k$ and pick $n_k$ such that $lambda E_k le {1 over 2^k}$ where $E_k ={ x | |f_{n_k}(x)-f(x)| > { 1over k} }$.
This is the Borel Cantelli lemma:
Note that $int sum_k 1_{E_k} = sum_k int 1_{E_k} = sum_{k} lambda E_k < infty$
and so $sum_k 1_{E_k}$ is finite ae. Hence for ae. $x$, $x$ is in at most a finite number of $E_k$ and so $f_{n_k}(x) to f(x)$.
Addendum:
The convergence cannot be everywhere. Here is an example that is easier to describe than to write an explicit expression. Choose the space $[0,1]$, let $f = 0$
and define $f_n$ as follows: $f_1 = 1_{[0,{1 over 2}]}, f_2 = 1_{[{1 over 2},1]}$, $f_3 = 1_{[0,{1 over 4}]}$, $f_4 = 1_{[{1 over 4}, {1 over 2}]}$, $f_5 = 1_{[{1 over 2}, {3 over 4}]}$, etc. It should be clear that $lambda { x | |f_n(x)| > epsilon} to 0$, but for every $x in [0,1)$, we have $f_n(x) = 1$ infinitely often and $f_n(x) = 0$ and infinitely often.
However, if we pick the subsequence corresponding to the functions $1_{[0,{1 over 2^n}]}$ then we see that this subsequence converges pointwise to $0$ for $x in (0,1]$.
answered Dec 4 '18 at 21:16
copper.hatcopper.hat
126k559160
$endgroup$
$begingroup$
This only proves that a subsequence converges, isnt it?
$endgroup$
– Kekks
Dec 4 '18 at 21:24
$begingroup$
Correct. ${}{}{}$
$endgroup$
– copper.hat
Dec 4 '18 at 21:25
add a comment |
$begingroup$
Fix $k$ and pick $n_k$ such that $lambda E_k le {1 over 2^k}$ where $E_k ={ x | |f_{n_k}(x)-f(x)| > { 1over k} }$.
This is the Borel Cantelli lemma:
Note that $int sum_k 1_{E_k} = sum_k int 1_{E_k} = sum_{k} lambda E_k < infty$
and so $sum_k 1_{E_k}$ is finite ae. Hence for ae. $x$, $x$ is in at most a finite number of $E_k$ and so $f_{n_k}(x) to f(x)$.
Addendum:
The convergence cannot be everywhere. Here is an example that is easier to describe than to write an explicit expression. Choose the space $[0,1]$, let $f = 0$
and define $f_n$ as follows: $f_1 = 1_{[0,{1 over 2}]}, f_2 = 1_{[{1 over 2},1]}$, $f_3 = 1_{[0,{1 over 4}]}$, $f_4 = 1_{[{1 over 4}, {1 over 2}]}$, $f_5 = 1_{[{1 over 2}, {3 over 4}]}$, etc. It should be clear that $lambda { x | |f_n(x)| > epsilon} to 0$, but for every $x in [0,1)$, we have $f_n(x) = 1$ infinitely often and $f_n(x) = 0$ and infinitely often.
However, if we pick the subsequence corresponding to the functions $1_{[0,{1 over 2^n}]}$ then we see that this subsequence converges pointwise to $0$ for $x in (0,1]$.
answered Dec 4 '18 at 21:16
copper.hatcopper.hat
126k559160
$endgroup$
Fix $k$ and pick $n_k$ such that $lambda E_k le {1 over 2^k}$ where $E_k ={ x | |f_{n_k}(x)-f(x)| > { 1over k} }$.
This is the Borel Cantelli lemma:
Note that $int sum_k 1_{E_k} = sum_k int 1_{E_k} = sum_{k} lambda E_k < infty$
and so $sum_k 1_{E_k}$ is finite ae. Hence for ae. $x$, $x$ is in at most a finite number of $E_k$ and so $f_{n_k}(x) to f(x)$.
Addendum:
The convergence cannot be everywhere. Here is an example that is easier to describe than to write an explicit expression. Choose the space $[0,1]$, let $f = 0$
and define $f_n$ as follows: $f_1 = 1_{[0,{1 over 2}]}, f_2 = 1_{[{1 over 2},1]}$, $f_3 = 1_{[0,{1 over 4}]}$, $f_4 = 1_{[{1 over 4}, {1 over 2}]}$, $f_5 = 1_{[{1 over 2}, {3 over 4}]}$, etc. It should be clear that $lambda { x | |f_n(x)| > epsilon} to 0$, but for every $x in [0,1)$, we have $f_n(x) = 1$ infinitely often and $f_n(x) = 0$ and infinitely often.
However, if we pick the subsequence corresponding to the functions $1_{[0,{1 over 2^n}]}$ then we see that this subsequence converges pointwise to $0$ for $x in (0,1]$.
answered Dec 4 '18 at 21:16
copper.hatcopper.hat
126k559160
edited Dec 4 '18 at 21:45
answered Dec 4 '18 at 21:16
copper.hatcopper.hat
126k559160
answered Dec 4 '18 at 21:16
copper.hatcopper.hat
126k559160
answered Dec 4 '18 at 21:16
copper.hatcopper.hat
126k559160
126k559160
$begingroup$
This only proves that a subsequence converges, isnt it?
$endgroup$
– Kekks
Dec 4 '18 at 21:24
$begingroup$
Correct. ${}{}{}$
$endgroup$
– copper.hat
Dec 4 '18 at 21:25
add a comment |
$begingroup$
This only proves that a subsequence converges, isnt it?
$endgroup$
– Kekks
Dec 4 '18 at 21:24
$begingroup$
Correct. ${}{}{}$
$endgroup$
– copper.hat
Dec 4 '18 at 21:25
$begingroup$
This only proves that a subsequence converges, isnt it?
$endgroup$
– Kekks
Dec 4 '18 at 21:24
$begingroup$
This only proves that a subsequence converges, isnt it?
$endgroup$
– Kekks
Dec 4 '18 at 21:24
$begingroup$
Correct. ${}{}{}$
$endgroup$
– copper.hat
Dec 4 '18 at 21:25
$begingroup$
Correct. ${}{}{}$
$endgroup$
– copper.hat
Dec 4 '18 at 21:25
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I didn't read your attempt, but maybe you want to read en.wikipedia.org/wiki/Borel%E2%80%93Cantelli_lemma. This is the standard tool to prove it
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– Federico
Dec 4 '18 at 19:33
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I think your statement of what it means to not converge almost everywhere is not quite right. It seems like you're claiming: There exists $epsilon > 0$ and a set of positive measure $A$ such that, for any $N in mathbb{N}$ there is $n > N$ with $|f_n(x) - f(x)| > epsilon$ for all $x in A$. Rather I think it should be the following: There exists a set $A$ of positive measure, such that for all $x in A$ there exists an $epsilon > 0$ such that, for any $N in mathbb{N}$ there exists $n > N$ with $|f_n(x) - f(x)| geq epsilon$.
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– Sean Haight
Dec 4 '18 at 19:53
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@SeanHaight you're right.
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– Kekks
Dec 4 '18 at 20:06
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@Federico so i use the integrability of $f_n,f$ to shot that $sum_n lambda(H_n^{epsilon}) < infty$ so i can use Borel-Cantelli which implies the convergence a.e.?
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– Kekks
Dec 4 '18 at 20:06
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You write the set of non-convergence as an intersection of a union, like in the statement of Borel-Cantelli lemma, then use it to show that it has measure $0$.
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– Federico
Dec 4 '18 at 20:09