Example of non negative function with finite L1 seminorm but not integrable The Next CEO of Stack OverflowA pathological example of a differentiable function whose derivative is not integrableCollection is uniformly integrable, but individual is not integrableHow to show $int_[0, +infty) frac21+x^2 dx$ Lebesgue integrable?How to prove an inequality of Lebesgue integral?An integrable function is finite a.e.Prove $int_mathcalQ_k^c vert f vert to_k 0$.conditions on integrable function with counting measureIs my function Lebesgue integrable but not Riemann?If $f_n rightarrow f$ uniformly and $f_n$ are measureable, integrable functions with $mu$ $sigma$-finite. Then $f$ has not to be integrableDefinition of Lebesgue Integral
How seriously should I take size and weight limits of hand luggage?
Small nick on power cord from an electric alarm clock, and copper wiring exposed but intact
How to pronounce fünf in 45
"Eavesdropping" vs "Listen in on"
logical reads on global temp table, but not on session-level temp table
Early programmable calculators with RS-232
Read/write a pipe-delimited file line by line with some simple text manipulation
Simplify trigonometric expression using trigonometric identities
What steps are necessary to read a Modern SSD in Medieval Europe?
Prodigo = pro + ago?
How can a day be of 24 hours?
Strange use of "whether ... than ..." in official text
Is it possible to make a 9x9 table fit within the default margins?
Why did early computer designers eschew integers?
Another proof that dividing by 0 does not exist -- is it right?
Does the Idaho Potato Commission associate potato skins with healthy eating?
Is it correct to say moon starry nights?
Masking layers by a vector polygon layer in QGIS
Gauss' Posthumous Publications?
Planeswalker Ability and Death Timing
What happens if you break a law in another country outside of that country?
Why do we say “un seul M” and not “une seule M” even though M is a “consonne”?
Man transported from Alternate World into ours by a Neutrino Detector
Incomplete cube
Example of non negative function with finite L1 seminorm but not integrable
The Next CEO of Stack OverflowA pathological example of a differentiable function whose derivative is not integrableCollection is uniformly integrable, but individual is not integrableHow to show $int_[0, +infty) frac21+x^2 dx$ Lebesgue integrable?How to prove an inequality of Lebesgue integral?An integrable function is finite a.e.Prove $int_mathcalQ_k^c vert f vert to_k 0$.conditions on integrable function with counting measureIs my function Lebesgue integrable but not Riemann?If $f_n rightarrow f$ uniformly and $f_n$ are measureable, integrable functions with $mu$ $sigma$-finite. Then $f$ has not to be integrableDefinition of Lebesgue Integral
$begingroup$
I was wondering if there is an easy example for a function
$$ f : mathbbR rightarrow mathbbR_ge0$$
which has $ ||f||_1 in mathbbR$ but not Lebesgue integrable. The reverse holds, i.e. any Lebesgue integrable non negative function has finite L1 seminorm.
Let me give the definitions I am working with (Königsberger Analysis 2)
Definition: A hullseries of a function is a series
$$ Phi=sum_k=1^infty c_k 1_Q_k$$
with
i) Q_k are open cuboid (in our case of 1D, open intervals)
ii) for any x:
$$ |f(x)| le Phi (x) $$
The volume of the hullseries is
$$ I(Phi):= sum_k=1^infty c_k v(Q_k) $$
(here v(Qk) is the length of the interval)
Definition: L1-seminorm is
$$ ||f||_1 :=rm inf Phi rm; is; a; hullseries; for; f$$
Definition: a function phi $Rrightarrow C$ is called stepfunction, is there is a finite number of pairwise disjoint cuboids $Q_s$ (here intervals), such that
i) $phi$ is constant on each $Q_s$
ii) $phi(x)=0$ for all $x in Rbackslash cup Q_k$
Definition: A function is Lebesgue integrable if there is a series of stepfunctions $phi_k$ such that $||f-phi_k||_1 rightarrow 0$ for $krightarrow infty$
lebesgue-integral
asked Mar 20 at 8:53
lalalalalala
230110
$endgroup$
|
show 1 more comment
$begingroup$
I was wondering if there is an easy example for a function
$$ f : mathbbR rightarrow mathbbR_ge0$$
which has $ ||f||_1 in mathbbR$ but not Lebesgue integrable. The reverse holds, i.e. any Lebesgue integrable non negative function has finite L1 seminorm.
Let me give the definitions I am working with (Königsberger Analysis 2)
Definition: A hullseries of a function is a series
$$ Phi=sum_k=1^infty c_k 1_Q_k$$
with
i) Q_k are open cuboid (in our case of 1D, open intervals)
ii) for any x:
$$ |f(x)| le Phi (x) $$
The volume of the hullseries is
$$ I(Phi):= sum_k=1^infty c_k v(Q_k) $$
(here v(Qk) is the length of the interval)
Definition: L1-seminorm is
$$ ||f||_1 :=rm inf Phi rm; is; a; hullseries; for; f$$
Definition: a function phi $Rrightarrow C$ is called stepfunction, is there is a finite number of pairwise disjoint cuboids $Q_s$ (here intervals), such that
i) $phi$ is constant on each $Q_s$
ii) $phi(x)=0$ for all $x in Rbackslash cup Q_k$
Definition: A function is Lebesgue integrable if there is a series of stepfunctions $phi_k$ such that $||f-phi_k||_1 rightarrow 0$ for $krightarrow infty$
lebesgue-integral
asked Mar 20 at 8:53
lalalalalala
230110
$endgroup$
1
$begingroup$
If $f$ is nonnegative then $f=|f|$, so the questions becomes trivial. Am I missing something?
$endgroup$
– uniquesolution
Mar 20 at 9:03
$begingroup$
@uniquesolution maybe I am missing something. Yes f=|f|. You would need to show that there is a sequence of stepfunctions converging to f in the one norm.
$endgroup$
– lalala
Mar 20 at 9:10
$begingroup$
Yes, you are missing something, because if you assume $|f|_1inmathbbR$ you are saying that the Lebesgue integral of $|f|$ exists, and it is a basic theorem that if $|f|$ is Lebesgue integrable, then so is $f$, but in your case we do not even need that, because $|f|=f$. Are you looking specifically for the proof of this theorem?
$endgroup$
– uniquesolution
Mar 20 at 9:16
$begingroup$
@uniquesolution When I say ||f||_1 is finite I am not implying that it is Lebesgue integrabel (in the framework of the definitions given by Königsberger)
$endgroup$
– lalala
Mar 20 at 9:21
$begingroup$
It is misleading to use a conventional notation for the norm in $L_1$ and later edit your post to define it to mean something else.
$endgroup$
– uniquesolution
Mar 20 at 13:14
|
show 1 more comment
$begingroup$
I was wondering if there is an easy example for a function
$$ f : mathbbR rightarrow mathbbR_ge0$$
which has $ ||f||_1 in mathbbR$ but not Lebesgue integrable. The reverse holds, i.e. any Lebesgue integrable non negative function has finite L1 seminorm.
Let me give the definitions I am working with (Königsberger Analysis 2)
Definition: A hullseries of a function is a series
$$ Phi=sum_k=1^infty c_k 1_Q_k$$
with
i) Q_k are open cuboid (in our case of 1D, open intervals)
ii) for any x:
$$ |f(x)| le Phi (x) $$
The volume of the hullseries is
$$ I(Phi):= sum_k=1^infty c_k v(Q_k) $$
(here v(Qk) is the length of the interval)
Definition: L1-seminorm is
$$ ||f||_1 :=rm inf Phi rm; is; a; hullseries; for; f$$
Definition: a function phi $Rrightarrow C$ is called stepfunction, is there is a finite number of pairwise disjoint cuboids $Q_s$ (here intervals), such that
i) $phi$ is constant on each $Q_s$
ii) $phi(x)=0$ for all $x in Rbackslash cup Q_k$
Definition: A function is Lebesgue integrable if there is a series of stepfunctions $phi_k$ such that $||f-phi_k||_1 rightarrow 0$ for $krightarrow infty$
lebesgue-integral
asked Mar 20 at 8:53
lalalalalala
230110
$endgroup$
I was wondering if there is an easy example for a function
$$ f : mathbbR rightarrow mathbbR_ge0$$
which has $ ||f||_1 in mathbbR$ but not Lebesgue integrable. The reverse holds, i.e. any Lebesgue integrable non negative function has finite L1 seminorm.
Let me give the definitions I am working with (Königsberger Analysis 2)
Definition: A hullseries of a function is a series
$$ Phi=sum_k=1^infty c_k 1_Q_k$$
with
i) Q_k are open cuboid (in our case of 1D, open intervals)
ii) for any x:
$$ |f(x)| le Phi (x) $$
The volume of the hullseries is
$$ I(Phi):= sum_k=1^infty c_k v(Q_k) $$
(here v(Qk) is the length of the interval)
Definition: L1-seminorm is
$$ ||f||_1 :=rm inf Phi rm; is; a; hullseries; for; f$$
Definition: a function phi $Rrightarrow C$ is called stepfunction, is there is a finite number of pairwise disjoint cuboids $Q_s$ (here intervals), such that
i) $phi$ is constant on each $Q_s$
ii) $phi(x)=0$ for all $x in Rbackslash cup Q_k$
Definition: A function is Lebesgue integrable if there is a series of stepfunctions $phi_k$ such that $||f-phi_k||_1 rightarrow 0$ for $krightarrow infty$
lebesgue-integral
lebesgue-integral
asked Mar 20 at 8:53
lalalalalala
230110
asked Mar 20 at 8:53
lalalalalala
230110
edited Mar 20 at 9:22
lalala
asked Mar 20 at 8:53
lalalalalala
230110
asked Mar 20 at 8:53
lalalalalala
230110
asked Mar 20 at 8:53
lalalalalala
230110
230110
1
$begingroup$
If $f$ is nonnegative then $f=|f|$, so the questions becomes trivial. Am I missing something?
$endgroup$
– uniquesolution
Mar 20 at 9:03
$begingroup$
@uniquesolution maybe I am missing something. Yes f=|f|. You would need to show that there is a sequence of stepfunctions converging to f in the one norm.
$endgroup$
– lalala
Mar 20 at 9:10
$begingroup$
Yes, you are missing something, because if you assume $|f|_1inmathbbR$ you are saying that the Lebesgue integral of $|f|$ exists, and it is a basic theorem that if $|f|$ is Lebesgue integrable, then so is $f$, but in your case we do not even need that, because $|f|=f$. Are you looking specifically for the proof of this theorem?
$endgroup$
– uniquesolution
Mar 20 at 9:16
$begingroup$
@uniquesolution When I say ||f||_1 is finite I am not implying that it is Lebesgue integrabel (in the framework of the definitions given by Königsberger)
$endgroup$
– lalala
Mar 20 at 9:21
$begingroup$
It is misleading to use a conventional notation for the norm in $L_1$ and later edit your post to define it to mean something else.
$endgroup$
– uniquesolution
Mar 20 at 13:14
|
show 1 more comment
1
$begingroup$
If $f$ is nonnegative then $f=|f|$, so the questions becomes trivial. Am I missing something?
$endgroup$
– uniquesolution
Mar 20 at 9:03
$begingroup$
@uniquesolution maybe I am missing something. Yes f=|f|. You would need to show that there is a sequence of stepfunctions converging to f in the one norm.
$endgroup$
– lalala
Mar 20 at 9:10
$begingroup$
Yes, you are missing something, because if you assume $|f|_1inmathbbR$ you are saying that the Lebesgue integral of $|f|$ exists, and it is a basic theorem that if $|f|$ is Lebesgue integrable, then so is $f$, but in your case we do not even need that, because $|f|=f$. Are you looking specifically for the proof of this theorem?
$endgroup$
– uniquesolution
Mar 20 at 9:16
$begingroup$
@uniquesolution When I say ||f||_1 is finite I am not implying that it is Lebesgue integrabel (in the framework of the definitions given by Königsberger)
$endgroup$
– lalala
Mar 20 at 9:21
$begingroup$
It is misleading to use a conventional notation for the norm in $L_1$ and later edit your post to define it to mean something else.
$endgroup$
– uniquesolution
Mar 20 at 13:14
1
1
$begingroup$
If $f$ is nonnegative then $f=|f|$, so the questions becomes trivial. Am I missing something?
$endgroup$
– uniquesolution
Mar 20 at 9:03
$begingroup$
If $f$ is nonnegative then $f=|f|$, so the questions becomes trivial. Am I missing something?
$endgroup$
– uniquesolution
Mar 20 at 9:03
$begingroup$
@uniquesolution maybe I am missing something. Yes f=|f|. You would need to show that there is a sequence of stepfunctions converging to f in the one norm.
$endgroup$
– lalala
Mar 20 at 9:10
$begingroup$
@uniquesolution maybe I am missing something. Yes f=|f|. You would need to show that there is a sequence of stepfunctions converging to f in the one norm.
$endgroup$
– lalala
Mar 20 at 9:10
$begingroup$
Yes, you are missing something, because if you assume $|f|_1inmathbbR$ you are saying that the Lebesgue integral of $|f|$ exists, and it is a basic theorem that if $|f|$ is Lebesgue integrable, then so is $f$, but in your case we do not even need that, because $|f|=f$. Are you looking specifically for the proof of this theorem?
$endgroup$
– uniquesolution
Mar 20 at 9:16
$begingroup$
Yes, you are missing something, because if you assume $|f|_1inmathbbR$ you are saying that the Lebesgue integral of $|f|$ exists, and it is a basic theorem that if $|f|$ is Lebesgue integrable, then so is $f$, but in your case we do not even need that, because $|f|=f$. Are you looking specifically for the proof of this theorem?
$endgroup$
– uniquesolution
Mar 20 at 9:16
$begingroup$
@uniquesolution When I say ||f||_1 is finite I am not implying that it is Lebesgue integrabel (in the framework of the definitions given by Königsberger)
$endgroup$
– lalala
Mar 20 at 9:21
$begingroup$
@uniquesolution When I say ||f||_1 is finite I am not implying that it is Lebesgue integrabel (in the framework of the definitions given by Königsberger)
$endgroup$
– lalala
Mar 20 at 9:21
$begingroup$
It is misleading to use a conventional notation for the norm in $L_1$ and later edit your post to define it to mean something else.
$endgroup$
– uniquesolution
Mar 20 at 13:14
$begingroup$
It is misleading to use a conventional notation for the norm in $L_1$ and later edit your post to define it to mean something else.
$endgroup$
– uniquesolution
Mar 20 at 13:14
|
show 1 more comment
0
active
oldest
votes
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%2f3155178%2fexample-of-non-negative-function-with-finite-l1-seminorm-but-not-integrable%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%2f3155178%2fexample-of-non-negative-function-with-finite-l1-seminorm-but-not-integrable%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
$begingroup$
If $f$ is nonnegative then $f=|f|$, so the questions becomes trivial. Am I missing something?
$endgroup$
– uniquesolution
Mar 20 at 9:03
$begingroup$
@uniquesolution maybe I am missing something. Yes f=|f|. You would need to show that there is a sequence of stepfunctions converging to f in the one norm.
$endgroup$
– lalala
Mar 20 at 9:10
$begingroup$
Yes, you are missing something, because if you assume $|f|_1inmathbbR$ you are saying that the Lebesgue integral of $|f|$ exists, and it is a basic theorem that if $|f|$ is Lebesgue integrable, then so is $f$, but in your case we do not even need that, because $|f|=f$. Are you looking specifically for the proof of this theorem?
$endgroup$
– uniquesolution
Mar 20 at 9:16
$begingroup$
@uniquesolution When I say ||f||_1 is finite I am not implying that it is Lebesgue integrabel (in the framework of the definitions given by Königsberger)
$endgroup$
– lalala
Mar 20 at 9:21
$begingroup$
It is misleading to use a conventional notation for the norm in $L_1$ and later edit your post to define it to mean something else.
$endgroup$
– uniquesolution
Mar 20 at 13:14