Mixing
Cauchy a.e. implies existence of a measurable function to which this sequence converges a.e
$begingroup$
I have seen a similar exercise in Royden (if I recall correctly), but the statement included Cauchy convergence in measure.
In this case, there are measurable functions $f_n:Xrightarrowmathbb{R}$ such that ${f_n}$ is Cauchy a.e.
I need to prove that there exists a measurable function $f$ for which $f_nto f$ a.e.
Every Cauchy sequence has a finite limit then, intuitively, it seems that we can find a function $f$ which is finite-valued so that $f_n to f$ a.e.
I can't, however, justify this in a rigorous way. Can you provide a hint, please?
real-analysis measure-theory lebesgue-measure
asked Jan 28 at 15:14
Don DraperDon Draper
87110
$endgroup$
add a comment |
$begingroup$
I have seen a similar exercise in Royden (if I recall correctly), but the statement included Cauchy convergence in measure.
In this case, there are measurable functions $f_n:Xrightarrowmathbb{R}$ such that ${f_n}$ is Cauchy a.e.
I need to prove that there exists a measurable function $f$ for which $f_nto f$ a.e.
Every Cauchy sequence has a finite limit then, intuitively, it seems that we can find a function $f$ which is finite-valued so that $f_n to f$ a.e.
I can't, however, justify this in a rigorous way. Can you provide a hint, please?
real-analysis measure-theory lebesgue-measure
asked Jan 28 at 15:14
Don DraperDon Draper
87110
$endgroup$
1
$begingroup$
For any point $x$ outside a set of measure zero $N,$ the sequence of real numbers $f_n(x)$ is fundamental, hence converging to some $f(x)$ and on $N$ we define $f = 0.$ Q.E.D.
$endgroup$
– Will M.
Feb 4 at 21:37
add a comment |
$begingroup$
I have seen a similar exercise in Royden (if I recall correctly), but the statement included Cauchy convergence in measure.
In this case, there are measurable functions $f_n:Xrightarrowmathbb{R}$ such that ${f_n}$ is Cauchy a.e.
I need to prove that there exists a measurable function $f$ for which $f_nto f$ a.e.
Every Cauchy sequence has a finite limit then, intuitively, it seems that we can find a function $f$ which is finite-valued so that $f_n to f$ a.e.
I can't, however, justify this in a rigorous way. Can you provide a hint, please?
real-analysis measure-theory lebesgue-measure
asked Jan 28 at 15:14
Don DraperDon Draper
87110
$endgroup$
I have seen a similar exercise in Royden (if I recall correctly), but the statement included Cauchy convergence in measure.
In this case, there are measurable functions $f_n:Xrightarrowmathbb{R}$ such that ${f_n}$ is Cauchy a.e.
I need to prove that there exists a measurable function $f$ for which $f_nto f$ a.e.
Every Cauchy sequence has a finite limit then, intuitively, it seems that we can find a function $f$ which is finite-valued so that $f_n to f$ a.e.
I can't, however, justify this in a rigorous way. Can you provide a hint, please?
real-analysis measure-theory lebesgue-measure
real-analysis measure-theory lebesgue-measure
asked Jan 28 at 15:14
Don DraperDon Draper
87110
asked Jan 28 at 15:14
Don DraperDon Draper
87110
asked Jan 28 at 15:14
Don DraperDon Draper
87110
asked Jan 28 at 15:14
Don DraperDon Draper
87110
asked Jan 28 at 15:14
Don DraperDon Draper
87110
87110
1
$begingroup$
For any point $x$ outside a set of measure zero $N,$ the sequence of real numbers $f_n(x)$ is fundamental, hence converging to some $f(x)$ and on $N$ we define $f = 0.$ Q.E.D.
$endgroup$
– Will M.
Feb 4 at 21:37
add a comment |
1
$begingroup$
For any point $x$ outside a set of measure zero $N,$ the sequence of real numbers $f_n(x)$ is fundamental, hence converging to some $f(x)$ and on $N$ we define $f = 0.$ Q.E.D.
$endgroup$
– Will M.
Feb 4 at 21:37
1
1
$begingroup$
For any point $x$ outside a set of measure zero $N,$ the sequence of real numbers $f_n(x)$ is fundamental, hence converging to some $f(x)$ and on $N$ we define $f = 0.$ Q.E.D.
$endgroup$
– Will M.
Feb 4 at 21:37
$begingroup$
For any point $x$ outside a set of measure zero $N,$ the sequence of real numbers $f_n(x)$ is fundamental, hence converging to some $f(x)$ and on $N$ we define $f = 0.$ Q.E.D.
$endgroup$
– Will M.
Feb 4 at 21:37
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Let $mu$ be the measure on $X,$ and let $A={xin X: f_n(x) text { is Cauchy }}.$ We are given $Xsetminus A$ is measurable, with $mu(Xsetminus A)=0.$ Hence $A$ is measurable. For $xin A,$ we have $f_n(x)$ Cauchy, hence convergent, hence $f_n(x)to liminf f_n(x).$ But $liminf f_n(x)$ is a measurable function on $A.$ Hence the function
$$f(x) = begin{cases} liminf f_n(x),&xin A\ 0,&xin Xsetminus Aend{cases}$$
is measurable on $X,$ and $f_n(x)to f(x)$ $mu$-a.e.
answered Feb 2 at 20:11
zhw.zhw.
74.8k43175
$endgroup$
add a comment |
$begingroup$
The set, on which $(f_n)_n$ is a Cauchy-sequence, can be written as
$$Omega = bigcap_{k=1}^infty bigcup_{N=1}^infty bigcap_{n,m ge N} { |f_n - f_m| < 1/k}.$$
Since $mathbb{R}$ is complete, for any $omega in Omega$ the limes $lim_{n rightarrow infty} f_n(omega)$ exists in $mathbb{R}$, Your assumptations says also that $mu(Omega^c) =0$, i.e. $Omega^c$ is a nullset. Next define $$f:= lim_{n rightarrow infty} f_n(omega) 1_Omega(omega).$$
Then $f_n rightarrow f$ a.e. and, since $f_n 1_{Omega}$ is measurable for any $n in mathbb{N}$, also $f$ is measurable (as a pointwise limes).
answered Jan 28 at 16:26
p4schp4sch
5,475318
$endgroup$
$begingroup$
Can you please add more details on the $Omega$ set? What does it stand for?
$endgroup$
– Don Draper
Feb 2 at 17:23
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3090985%2fcauchy-a-e-implies-existence-of-a-measurable-function-to-which-this-sequence-co%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Let $mu$ be the measure on $X,$ and let $A={xin X: f_n(x) text { is Cauchy }}.$ We are given $Xsetminus A$ is measurable, with $mu(Xsetminus A)=0.$ Hence $A$ is measurable. For $xin A,$ we have $f_n(x)$ Cauchy, hence convergent, hence $f_n(x)to liminf f_n(x).$ But $liminf f_n(x)$ is a measurable function on $A.$ Hence the function
$$f(x) = begin{cases} liminf f_n(x),&xin A\ 0,&xin Xsetminus Aend{cases}$$
is measurable on $X,$ and $f_n(x)to f(x)$ $mu$-a.e.
answered Feb 2 at 20:11
zhw.zhw.
74.8k43175
$endgroup$
add a comment |
$begingroup$
Let $mu$ be the measure on $X,$ and let $A={xin X: f_n(x) text { is Cauchy }}.$ We are given $Xsetminus A$ is measurable, with $mu(Xsetminus A)=0.$ Hence $A$ is measurable. For $xin A,$ we have $f_n(x)$ Cauchy, hence convergent, hence $f_n(x)to liminf f_n(x).$ But $liminf f_n(x)$ is a measurable function on $A.$ Hence the function
$$f(x) = begin{cases} liminf f_n(x),&xin A\ 0,&xin Xsetminus Aend{cases}$$
is measurable on $X,$ and $f_n(x)to f(x)$ $mu$-a.e.
answered Feb 2 at 20:11
zhw.zhw.
74.8k43175
$endgroup$
add a comment |
$begingroup$
Let $mu$ be the measure on $X,$ and let $A={xin X: f_n(x) text { is Cauchy }}.$ We are given $Xsetminus A$ is measurable, with $mu(Xsetminus A)=0.$ Hence $A$ is measurable. For $xin A,$ we have $f_n(x)$ Cauchy, hence convergent, hence $f_n(x)to liminf f_n(x).$ But $liminf f_n(x)$ is a measurable function on $A.$ Hence the function
$$f(x) = begin{cases} liminf f_n(x),&xin A\ 0,&xin Xsetminus Aend{cases}$$
is measurable on $X,$ and $f_n(x)to f(x)$ $mu$-a.e.
answered Feb 2 at 20:11
zhw.zhw.
74.8k43175
$endgroup$
Let $mu$ be the measure on $X,$ and let $A={xin X: f_n(x) text { is Cauchy }}.$ We are given $Xsetminus A$ is measurable, with $mu(Xsetminus A)=0.$ Hence $A$ is measurable. For $xin A,$ we have $f_n(x)$ Cauchy, hence convergent, hence $f_n(x)to liminf f_n(x).$ But $liminf f_n(x)$ is a measurable function on $A.$ Hence the function
$$f(x) = begin{cases} liminf f_n(x),&xin A\ 0,&xin Xsetminus Aend{cases}$$
is measurable on $X,$ and $f_n(x)to f(x)$ $mu$-a.e.
answered Feb 2 at 20:11
zhw.zhw.
74.8k43175
edited Feb 3 at 16:39
answered Feb 2 at 20:11
zhw.zhw.
74.8k43175
answered Feb 2 at 20:11
zhw.zhw.
74.8k43175
answered Feb 2 at 20:11
zhw.zhw.
74.8k43175
74.8k43175
add a comment |
add a comment |
$begingroup$
The set, on which $(f_n)_n$ is a Cauchy-sequence, can be written as
$$Omega = bigcap_{k=1}^infty bigcup_{N=1}^infty bigcap_{n,m ge N} { |f_n - f_m| < 1/k}.$$
Since $mathbb{R}$ is complete, for any $omega in Omega$ the limes $lim_{n rightarrow infty} f_n(omega)$ exists in $mathbb{R}$, Your assumptations says also that $mu(Omega^c) =0$, i.e. $Omega^c$ is a nullset. Next define $$f:= lim_{n rightarrow infty} f_n(omega) 1_Omega(omega).$$
Then $f_n rightarrow f$ a.e. and, since $f_n 1_{Omega}$ is measurable for any $n in mathbb{N}$, also $f$ is measurable (as a pointwise limes).
answered Jan 28 at 16:26
p4schp4sch
5,475318
$endgroup$
$begingroup$
Can you please add more details on the $Omega$ set? What does it stand for?
$endgroup$
– Don Draper
Feb 2 at 17:23
add a comment |
$begingroup$
The set, on which $(f_n)_n$ is a Cauchy-sequence, can be written as
$$Omega = bigcap_{k=1}^infty bigcup_{N=1}^infty bigcap_{n,m ge N} { |f_n - f_m| < 1/k}.$$
Since $mathbb{R}$ is complete, for any $omega in Omega$ the limes $lim_{n rightarrow infty} f_n(omega)$ exists in $mathbb{R}$, Your assumptations says also that $mu(Omega^c) =0$, i.e. $Omega^c$ is a nullset. Next define $$f:= lim_{n rightarrow infty} f_n(omega) 1_Omega(omega).$$
Then $f_n rightarrow f$ a.e. and, since $f_n 1_{Omega}$ is measurable for any $n in mathbb{N}$, also $f$ is measurable (as a pointwise limes).
answered Jan 28 at 16:26
p4schp4sch
5,475318
$endgroup$
$begingroup$
Can you please add more details on the $Omega$ set? What does it stand for?
$endgroup$
– Don Draper
Feb 2 at 17:23
add a comment |
$begingroup$
The set, on which $(f_n)_n$ is a Cauchy-sequence, can be written as
$$Omega = bigcap_{k=1}^infty bigcup_{N=1}^infty bigcap_{n,m ge N} { |f_n - f_m| < 1/k}.$$
Since $mathbb{R}$ is complete, for any $omega in Omega$ the limes $lim_{n rightarrow infty} f_n(omega)$ exists in $mathbb{R}$, Your assumptations says also that $mu(Omega^c) =0$, i.e. $Omega^c$ is a nullset. Next define $$f:= lim_{n rightarrow infty} f_n(omega) 1_Omega(omega).$$
Then $f_n rightarrow f$ a.e. and, since $f_n 1_{Omega}$ is measurable for any $n in mathbb{N}$, also $f$ is measurable (as a pointwise limes).
answered Jan 28 at 16:26
p4schp4sch
5,475318
$endgroup$
The set, on which $(f_n)_n$ is a Cauchy-sequence, can be written as
$$Omega = bigcap_{k=1}^infty bigcup_{N=1}^infty bigcap_{n,m ge N} { |f_n - f_m| < 1/k}.$$
Since $mathbb{R}$ is complete, for any $omega in Omega$ the limes $lim_{n rightarrow infty} f_n(omega)$ exists in $mathbb{R}$, Your assumptations says also that $mu(Omega^c) =0$, i.e. $Omega^c$ is a nullset. Next define $$f:= lim_{n rightarrow infty} f_n(omega) 1_Omega(omega).$$
Then $f_n rightarrow f$ a.e. and, since $f_n 1_{Omega}$ is measurable for any $n in mathbb{N}$, also $f$ is measurable (as a pointwise limes).
answered Jan 28 at 16:26
p4schp4sch
5,475318
edited Feb 3 at 9:55
answered Jan 28 at 16:26
p4schp4sch
5,475318
answered Jan 28 at 16:26
p4schp4sch
5,475318
answered Jan 28 at 16:26
p4schp4sch
5,475318
5,475318
$begingroup$
Can you please add more details on the $Omega$ set? What does it stand for?
$endgroup$
– Don Draper
Feb 2 at 17:23
add a comment |
$begingroup$
Can you please add more details on the $Omega$ set? What does it stand for?
$endgroup$
– Don Draper
Feb 2 at 17:23
$begingroup$
Can you please add more details on the $Omega$ set? What does it stand for?
$endgroup$
– Don Draper
Feb 2 at 17:23
$begingroup$
Can you please add more details on the $Omega$ set? What does it stand for?
$endgroup$
– Don Draper
Feb 2 at 17:23
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3090985%2fcauchy-a-e-implies-existence-of-a-measurable-function-to-which-this-sequence-co%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
1
$begingroup$
For any point $x$ outside a set of measure zero $N,$ the sequence of real numbers $f_n(x)$ is fundamental, hence converging to some $f(x)$ and on $N$ we define $f = 0.$ Q.E.D.
$endgroup$
– Will M.
Feb 4 at 21:37