If ${f_n }$ is a sequence of integrable function on $R$ , and ${ f_n } $ converges to a measurable function...
$begingroup$
If ${f_n}_{n=1}^{infty}$ is a sequence of integrable function on $R$ (a function $g$ is integrable means that $int_R g ,dx lt infty$). And $f_n $ converges uniformly to a measurable function $f$ , means that for any $epsilon$ ,there exist $N$ , and when $n ge N$ , we have $$|f_n(x)-f(x)|lt epsilon$$ for every $x in R $ . If we imposed some other properties of $f$ or $f_n$ (not the condition of Fatou or dominated convergence theorem) , can we prove or disprove that $$lim_{n to infty} int_R f_n(x) , dx = int_R f(x) , dx ,,,,,,,,,,,, text{ (1)}$$
My attempt :
a) If we do not assume the uniform convegence , then (1) might not hold , just consider the the characteristic funtion of $(n,+infty)$
b) If we assume the uniform convergence , and also assume that $f$ is integrable , then $(1)$ might not hold , consider the function $f_n=frac 1n $ whenever $x in (0,n)$ and $f_n=0$ otherwise . Then $f_n$ converges to $0$ uniformly but LHS of (1) is equal to $1$ while the RHS of (1) is equal to $0$ .
c) If we assume the uniform convergence , and also assume $f,f_n$ are non-negtive function , $f$ is not integrable. Then by Fatou's theorem we can check (1) is right for this condition .
My question: does (1) hold for the following two conditions?
d) The similar condition of c) , but we do not assume $f_n $ is non-negtive .
e) The similar condition of c) , but we do not assume $f_n , f$ are non-negtive
real-analysis
asked Jan 8 at 19:28
J.GuoJ.Guo
2799
$endgroup$
add a comment |
$begingroup$
If ${f_n}_{n=1}^{infty}$ is a sequence of integrable function on $R$ (a function $g$ is integrable means that $int_R g ,dx lt infty$). And $f_n $ converges uniformly to a measurable function $f$ , means that for any $epsilon$ ,there exist $N$ , and when $n ge N$ , we have $$|f_n(x)-f(x)|lt epsilon$$ for every $x in R $ . If we imposed some other properties of $f$ or $f_n$ (not the condition of Fatou or dominated convergence theorem) , can we prove or disprove that $$lim_{n to infty} int_R f_n(x) , dx = int_R f(x) , dx ,,,,,,,,,,,, text{ (1)}$$
My attempt :
a) If we do not assume the uniform convegence , then (1) might not hold , just consider the the characteristic funtion of $(n,+infty)$
b) If we assume the uniform convergence , and also assume that $f$ is integrable , then $(1)$ might not hold , consider the function $f_n=frac 1n $ whenever $x in (0,n)$ and $f_n=0$ otherwise . Then $f_n$ converges to $0$ uniformly but LHS of (1) is equal to $1$ while the RHS of (1) is equal to $0$ .
c) If we assume the uniform convergence , and also assume $f,f_n$ are non-negtive function , $f$ is not integrable. Then by Fatou's theorem we can check (1) is right for this condition .
My question: does (1) hold for the following two conditions?
d) The similar condition of c) , but we do not assume $f_n $ is non-negtive .
e) The similar condition of c) , but we do not assume $f_n , f$ are non-negtive
real-analysis
asked Jan 8 at 19:28
J.GuoJ.Guo
2799
$endgroup$
2
$begingroup$
I am pretty sure $d$ will not hold. Take $f(x)=1/x$ on $[1,infty)$, and $f_n(x)=1/x$ on $[1,n]$, is $-1/n$ on $[n,n+nln(n)]$ and $0$ afterwards.
$endgroup$
– Mindlack
Jan 8 at 19:39
$begingroup$
I see . Thank you for your help .
$endgroup$
– J.Guo
Jan 8 at 19:48
add a comment |
$begingroup$
If ${f_n}_{n=1}^{infty}$ is a sequence of integrable function on $R$ (a function $g$ is integrable means that $int_R g ,dx lt infty$). And $f_n $ converges uniformly to a measurable function $f$ , means that for any $epsilon$ ,there exist $N$ , and when $n ge N$ , we have $$|f_n(x)-f(x)|lt epsilon$$ for every $x in R $ . If we imposed some other properties of $f$ or $f_n$ (not the condition of Fatou or dominated convergence theorem) , can we prove or disprove that $$lim_{n to infty} int_R f_n(x) , dx = int_R f(x) , dx ,,,,,,,,,,,, text{ (1)}$$
My attempt :
a) If we do not assume the uniform convegence , then (1) might not hold , just consider the the characteristic funtion of $(n,+infty)$
b) If we assume the uniform convergence , and also assume that $f$ is integrable , then $(1)$ might not hold , consider the function $f_n=frac 1n $ whenever $x in (0,n)$ and $f_n=0$ otherwise . Then $f_n$ converges to $0$ uniformly but LHS of (1) is equal to $1$ while the RHS of (1) is equal to $0$ .
c) If we assume the uniform convergence , and also assume $f,f_n$ are non-negtive function , $f$ is not integrable. Then by Fatou's theorem we can check (1) is right for this condition .
My question: does (1) hold for the following two conditions?
d) The similar condition of c) , but we do not assume $f_n $ is non-negtive .
e) The similar condition of c) , but we do not assume $f_n , f$ are non-negtive
real-analysis
asked Jan 8 at 19:28
J.GuoJ.Guo
2799
$endgroup$
If ${f_n}_{n=1}^{infty}$ is a sequence of integrable function on $R$ (a function $g$ is integrable means that $int_R g ,dx lt infty$). And $f_n $ converges uniformly to a measurable function $f$ , means that for any $epsilon$ ,there exist $N$ , and when $n ge N$ , we have $$|f_n(x)-f(x)|lt epsilon$$ for every $x in R $ . If we imposed some other properties of $f$ or $f_n$ (not the condition of Fatou or dominated convergence theorem) , can we prove or disprove that $$lim_{n to infty} int_R f_n(x) , dx = int_R f(x) , dx ,,,,,,,,,,,, text{ (1)}$$
My attempt :
a) If we do not assume the uniform convegence , then (1) might not hold , just consider the the characteristic funtion of $(n,+infty)$
b) If we assume the uniform convergence , and also assume that $f$ is integrable , then $(1)$ might not hold , consider the function $f_n=frac 1n $ whenever $x in (0,n)$ and $f_n=0$ otherwise . Then $f_n$ converges to $0$ uniformly but LHS of (1) is equal to $1$ while the RHS of (1) is equal to $0$ .
c) If we assume the uniform convergence , and also assume $f,f_n$ are non-negtive function , $f$ is not integrable. Then by Fatou's theorem we can check (1) is right for this condition .
My question: does (1) hold for the following two conditions?
d) The similar condition of c) , but we do not assume $f_n $ is non-negtive .
e) The similar condition of c) , but we do not assume $f_n , f$ are non-negtive
real-analysis
real-analysis
asked Jan 8 at 19:28
J.GuoJ.Guo
2799
asked Jan 8 at 19:28
J.GuoJ.Guo
2799
asked Jan 8 at 19:28
J.GuoJ.Guo
2799
asked Jan 8 at 19:28
J.GuoJ.Guo
2799
asked Jan 8 at 19:28
J.GuoJ.Guo
2799
2799
2
$begingroup$
I am pretty sure $d$ will not hold. Take $f(x)=1/x$ on $[1,infty)$, and $f_n(x)=1/x$ on $[1,n]$, is $-1/n$ on $[n,n+nln(n)]$ and $0$ afterwards.
$endgroup$
– Mindlack
Jan 8 at 19:39
$begingroup$
I see . Thank you for your help .
$endgroup$
– J.Guo
Jan 8 at 19:48
add a comment |
2
$begingroup$
I am pretty sure $d$ will not hold. Take $f(x)=1/x$ on $[1,infty)$, and $f_n(x)=1/x$ on $[1,n]$, is $-1/n$ on $[n,n+nln(n)]$ and $0$ afterwards.
$endgroup$
– Mindlack
Jan 8 at 19:39
$begingroup$
I see . Thank you for your help .
$endgroup$
– J.Guo
Jan 8 at 19:48
2
2
$begingroup$
I am pretty sure $d$ will not hold. Take $f(x)=1/x$ on $[1,infty)$, and $f_n(x)=1/x$ on $[1,n]$, is $-1/n$ on $[n,n+nln(n)]$ and $0$ afterwards.
$endgroup$
– Mindlack
Jan 8 at 19:39
$begingroup$
I am pretty sure $d$ will not hold. Take $f(x)=1/x$ on $[1,infty)$, and $f_n(x)=1/x$ on $[1,n]$, is $-1/n$ on $[n,n+nln(n)]$ and $0$ afterwards.
$endgroup$
– Mindlack
Jan 8 at 19:39
$begingroup$
I see . Thank you for your help .
$endgroup$
– J.Guo
Jan 8 at 19:48
$begingroup$
I see . Thank you for your help .
$endgroup$
– J.Guo
Jan 8 at 19:48
add a comment |
0
active
oldest
votes
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%2f3066614%2fif-f-n-is-a-sequence-of-integrable-function-on-r-and-f-n-con%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
0
active
oldest
votes
0
active
oldest
votes
active
oldest
votes
active
oldest
votes
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%2f3066614%2fif-f-n-is-a-sequence-of-integrable-function-on-r-and-f-n-con%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
2
$begingroup$
I am pretty sure $d$ will not hold. Take $f(x)=1/x$ on $[1,infty)$, and $f_n(x)=1/x$ on $[1,n]$, is $-1/n$ on $[n,n+nln(n)]$ and $0$ afterwards.
$endgroup$
– Mindlack
Jan 8 at 19:39
$begingroup$
I see . Thank you for your help .
$endgroup$
– J.Guo
Jan 8 at 19:48