Mdthbs

What is the farthest a camera can see?

Stephen King and steam/diesel/cyber-punk

Shifting tenses in the middle of narration

Would Mirko Vosk, Mind Drinker trigger Waste Not?

Is there a way to proportionalize fixed costs in a MILP?

Cases with long math equation

Identifying My Main Water Shutoff Valve / Setup

Big number puzzle

How to remove ambiguity: "... lives in the city of H, the capital of the province of NS, WHERE the unemployment rate is ..."?

Installing Windows to flash UEFI/ BIOS, then reinstalling Ubuntu

Why won't the Republicans use a superdelegate system like the DNC in their nomination process?

Is it possible to arrive in the US without a C-1 visa for a transit flight

Swap (and hibernation) on SSD in 2019?

How much can I judge a company based on a phone screening?

Why is there a large performance impact when looping over an array over 240 elements?

"Table of Astronomy's" depiction of the solar system models

Luggage Storage at Szechenyi Baths

Are there examples in Tanach of 3 or more parties having an ongoing conversation?

How would armour (and combat) change if the fighter didn't need to actually wear it?

Word for an event that will likely never happen again

Go to last file in vim

Lípínguapua dopo Pêpê

Did DOS zero out the BSS area when it loaded a program?

How can God warn people of the upcoming rapture without disrupting society?

There is $f$ such that $int_0^1 f dx = lim_{c downarrow 0} int_c^1 fdx$, but for $|f|$ this limit does not exist. How is that possible?

Prove this inequality $lvert a - brvert < frac{1}{2}lvert b rvert implies lvert a rvert > frac{1}{2}lvert b rvert$A Riemann integral with a jump discontinuity at its lower limitProve if $f(0) = 0$ then $lim_{x to 0^+}xint_x^1 frac{f(t)}{t^2}dt = 0$ for regulated function $f$Prob. 7 (b), Chap. 6, in Baby Rudin: Example of a function such that $lim_{c to 0+} int_c^1 f(x) mathrm{d}x$ exists but . . .The old and modern definitions of total variation are actually equivalent?Principles of math analysis by Rudin, Chapter 6 Problem 7Find a function for which integral does not exist but converges as a limit of a sequence

.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty{ margin-bottom:0;
}

3

$begingroup$

If $f$ in integrable on some interval $[a,b]$ then we know that $lvert f rvert $ is also integrable on that same interval.

There is a problem in Rudin's Principles of Mathematical analysis such that we construct an $f$ where

$$int_0^1 f dx = lim_{c downarrow 0} int_c^1 fdx$$

exists and yet for $lvert f rvert$ this limit fails to exist.

How does this not contradict the implication above?

One such constuction is to set $f(x) = (-1)^{k+1}(k+1), forall x in (frac{1}{k+1},frac{1}{k}]$.

real-analysis integration improper-integrals

share|cite|improve this question

edited 15 hours ago

Asaf Karagila♦

315k3434 gold badges454454 silver badges787787 bronze badges

asked yesterday

all.overall.over

9377 bronze badges

$endgroup$

add a comment | 

3

$begingroup$

If $f$ in integrable on some interval $[a,b]$ then we know that $lvert f rvert $ is also integrable on that same interval.

There is a problem in Rudin's Principles of Mathematical analysis such that we construct an $f$ where

$$int_0^1 f dx = lim_{c downarrow 0} int_c^1 fdx$$

exists and yet for $lvert f rvert$ this limit fails to exist.

How does this not contradict the implication above?

One such constuction is to set $f(x) = (-1)^{k+1}(k+1), forall x in (frac{1}{k+1},frac{1}{k}]$.

real-analysis integration improper-integrals

share|cite|improve this question

edited 15 hours ago

Asaf Karagila♦

315k3434 gold badges454454 silver badges787787 bronze badges

asked yesterday

all.overall.over

9377 bronze badges

$endgroup$

add a comment | 

$begingroup$

If $f$ in integrable on some interval $[a,b]$ then we know that $lvert f rvert $ is also integrable on that same interval.

There is a problem in Rudin's Principles of Mathematical analysis such that we construct an $f$ where

$$int_0^1 f dx = lim_{c downarrow 0} int_c^1 fdx$$

exists and yet for $lvert f rvert$ this limit fails to exist.

How does this not contradict the implication above?

One such constuction is to set $f(x) = (-1)^{k+1}(k+1), forall x in (frac{1}{k+1},frac{1}{k}]$.

real-analysis integration improper-integrals

share|cite|improve this question

edited 15 hours ago

Asaf Karagila♦

315k3434 gold badges454454 silver badges787787 bronze badges

asked yesterday

all.overall.over

9377 bronze badges

$endgroup$

share|cite|improve this question

edited 15 hours ago

Asaf Karagila♦

315k3434 gold badges454454 silver badges787787 bronze badges

asked yesterday

all.overall.over

9377 bronze badges

share|cite|improve this question

edited 15 hours ago

Asaf Karagila♦

315k3434 gold badges454454 silver badges787787 bronze badges

asked yesterday

all.overall.over

9377 bronze badges

edited 15 hours ago

Asaf Karagila♦

315k3434 gold badges454454 silver badges787787 bronze badges

edited 15 hours ago

Asaf Karagila♦

315k3434 gold badges454454 silver badges787787 bronze badges

asked yesterday

all.overall.over

9377 bronze badges

asked yesterday

all.overall.over

9377 bronze badges

2 Answers
2

active

oldest

votes

8

$begingroup$

There is no contradiction because $f$ is not Riemann-integrable on $[0,1]$. The fact that the limit $lim_{c to 0}int_c ^1f(x)dx$ exists does not mean that $f$ is Riemann-integrable on $[0,1]$. Note that Rudin says in the exercise that we can define the symbol $int_0^1f(x)dx$ to mean said limit in the case where we have a function on $(0,1]$. It does not mean that $f$ is Riemann-integrable on $[0,1]$. (Indeed, it isn't in your example. It isn't even bounded.) It is just an assignment of a value to a symbol.

PS: Note that part of the exercise is even to show that the two (a priori possibly conflicting) definitions of the symbol $int_ 0^1f(x)dx$ agree when $f$ is Riemann-integrable.

share|cite|improve this answer

edited yesterday

answered yesterday

Aloizio Macedo♦Aloizio Macedo

24.3k22 gold badges4040 silver badges8989 bronze badges

$endgroup$

add a comment | 

5

$begingroup$

The implication $f$ is integrable $Rightarrow$ $|f|$ is integrable is true for Lebesgue-integrable functions. But the function that is constructed in this example is not Lebesgue-integrable. Its improper Riemann integral exists. That is a different property that does not imply Lebesgue-integrability.

share|cite|improve this answer

answered yesterday

Hans EnglerHans Engler

11k11 gold badge2020 silver badges3636 bronze badges

$endgroup$

• 
  
  
  

  
  4
  

  
  
  
  
  
  

  
  $begingroup$
  This answer is misleading. Lebesgue integration is not even mentioned in the context of this exercise on Rudin, not to mention the fact that the very first sentence leaves itself quite open to the interpretation that the affirmation "$f$ integrable $implies$ $|f|$ integrable" only holds for Lebesgue integration, which is not true. It holds for Riemann integration as well. The issue is that OP is overloading a symbol to mean more than it does.
  $endgroup$
  – Aloizio Macedo♦
  yesterday
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Good point. Replace "Lebesgue" with "Riemann" as done in your answer.
  $endgroup$
  – Hans Engler
  8 hours ago
  

  
  
  
  

add a comment | 

Your Answer

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
});


}
});

draft saved

draft discarded

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

Name

Email

Required, but never shown

StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3322465%2fthere-is-f-such-that-int-01-f-dx-lim-c-downarrow-0-int-c1-fdx-but%23new-answer', 'question_page');
}
);

Post as a guest

Name

Email

Required, but never shown

8

$begingroup$

There is no contradiction because $f$ is not Riemann-integrable on $[0,1]$. The fact that the limit $lim_{c to 0}int_c ^1f(x)dx$ exists does not mean that $f$ is Riemann-integrable on $[0,1]$. Note that Rudin says in the exercise that we can define the symbol $int_0^1f(x)dx$ to mean said limit in the case where we have a function on $(0,1]$. It does not mean that $f$ is Riemann-integrable on $[0,1]$. (Indeed, it isn't in your example. It isn't even bounded.) It is just an assignment of a value to a symbol.

PS: Note that part of the exercise is even to show that the two (a priori possibly conflicting) definitions of the symbol $int_ 0^1f(x)dx$ agree when $f$ is Riemann-integrable.

share|cite|improve this answer

edited yesterday

answered yesterday

Aloizio Macedo♦Aloizio Macedo

24.3k22 gold badges4040 silver badges8989 bronze badges

$endgroup$

add a comment | 

8

$begingroup$

There is no contradiction because $f$ is not Riemann-integrable on $[0,1]$. The fact that the limit $lim_{c to 0}int_c ^1f(x)dx$ exists does not mean that $f$ is Riemann-integrable on $[0,1]$. Note that Rudin says in the exercise that we can define the symbol $int_0^1f(x)dx$ to mean said limit in the case where we have a function on $(0,1]$. It does not mean that $f$ is Riemann-integrable on $[0,1]$. (Indeed, it isn't in your example. It isn't even bounded.) It is just an assignment of a value to a symbol.

PS: Note that part of the exercise is even to show that the two (a priori possibly conflicting) definitions of the symbol $int_ 0^1f(x)dx$ agree when $f$ is Riemann-integrable.

share|cite|improve this answer

edited yesterday

answered yesterday

Aloizio Macedo♦Aloizio Macedo

24.3k22 gold badges4040 silver badges8989 bronze badges

$endgroup$

add a comment | 

$begingroup$

There is no contradiction because $f$ is not Riemann-integrable on $[0,1]$. The fact that the limit $lim_{c to 0}int_c ^1f(x)dx$ exists does not mean that $f$ is Riemann-integrable on $[0,1]$. Note that Rudin says in the exercise that we can define the symbol $int_0^1f(x)dx$ to mean said limit in the case where we have a function on $(0,1]$. It does not mean that $f$ is Riemann-integrable on $[0,1]$. (Indeed, it isn't in your example. It isn't even bounded.) It is just an assignment of a value to a symbol.

PS: Note that part of the exercise is even to show that the two (a priori possibly conflicting) definitions of the symbol $int_ 0^1f(x)dx$ agree when $f$ is Riemann-integrable.

share|cite|improve this answer

edited yesterday

answered yesterday

Aloizio Macedo♦Aloizio Macedo

24.3k22 gold badges4040 silver badges8989 bronze badges

$endgroup$

share|cite|improve this answer

edited yesterday

answered yesterday

Aloizio Macedo♦Aloizio Macedo

24.3k22 gold badges4040 silver badges8989 bronze badges

answered yesterday

Aloizio Macedo♦Aloizio Macedo

24.3k22 gold badges4040 silver badges8989 bronze badges

answered yesterday

Aloizio Macedo♦Aloizio Macedo

24.3k22 gold badges4040 silver badges8989 bronze badges

5

$begingroup$

The implication $f$ is integrable $Rightarrow$ $|f|$ is integrable is true for Lebesgue-integrable functions. But the function that is constructed in this example is not Lebesgue-integrable. Its improper Riemann integral exists. That is a different property that does not imply Lebesgue-integrability.

share|cite|improve this answer

answered yesterday

Hans EnglerHans Engler

11k11 gold badge2020 silver badges3636 bronze badges

$endgroup$

• 
  
  
  

  
  4
  

  
  
  
  
  
  

  
  $begingroup$
  This answer is misleading. Lebesgue integration is not even mentioned in the context of this exercise on Rudin, not to mention the fact that the very first sentence leaves itself quite open to the interpretation that the affirmation "$f$ integrable $implies$ $|f|$ integrable" only holds for Lebesgue integration, which is not true. It holds for Riemann integration as well. The issue is that OP is overloading a symbol to mean more than it does.
  $endgroup$
  – Aloizio Macedo♦
  yesterday
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Good point. Replace "Lebesgue" with "Riemann" as done in your answer.
  $endgroup$
  – Hans Engler
  8 hours ago
  

  
  
  
  

add a comment | 

5

$begingroup$

The implication $f$ is integrable $Rightarrow$ $|f|$ is integrable is true for Lebesgue-integrable functions. But the function that is constructed in this example is not Lebesgue-integrable. Its improper Riemann integral exists. That is a different property that does not imply Lebesgue-integrability.

share|cite|improve this answer

answered yesterday

Hans EnglerHans Engler

11k11 gold badge2020 silver badges3636 bronze badges

$endgroup$

• 
  
  
  

  
  4
  

  
  
  
  
  
  

  
  $begingroup$
  This answer is misleading. Lebesgue integration is not even mentioned in the context of this exercise on Rudin, not to mention the fact that the very first sentence leaves itself quite open to the interpretation that the affirmation "$f$ integrable $implies$ $|f|$ integrable" only holds for Lebesgue integration, which is not true. It holds for Riemann integration as well. The issue is that OP is overloading a symbol to mean more than it does.
  $endgroup$
  – Aloizio Macedo♦
  yesterday
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Good point. Replace "Lebesgue" with "Riemann" as done in your answer.
  $endgroup$
  – Hans Engler
  8 hours ago
  

  
  
  
  

add a comment | 

$begingroup$

The implication $f$ is integrable $Rightarrow$ $|f|$ is integrable is true for Lebesgue-integrable functions. But the function that is constructed in this example is not Lebesgue-integrable. Its improper Riemann integral exists. That is a different property that does not imply Lebesgue-integrability.

share|cite|improve this answer

answered yesterday

Hans EnglerHans Engler

11k11 gold badge2020 silver badges3636 bronze badges

$endgroup$

share|cite|improve this answer

answered yesterday

Hans EnglerHans Engler

11k11 gold badge2020 silver badges3636 bronze badges

answered yesterday

Hans EnglerHans Engler

11k11 gold badge2020 silver badges3636 bronze badges

answered yesterday

Hans EnglerHans Engler

11k11 gold badge2020 silver badges3636 bronze badges