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understanding the solution of a given Lebesgue integration.

A basic doubt on Lebesgue integrationUnderstanding Lebesgue IntegrationWhat does Lebesgue measure zero mean and what are the advantages of Lebesgue integration over Riemann?Need a help in understanding a solution of a third problem in Israel Gohberg.Clarification of a step in a solution.The proof of Theorem 3 on page 73 in Royden “fourth edition”.Elaborating some points in the proof of the countable additivity of integration.

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2

$begingroup$

The question and its solution are given below:

My questions are:

I am supposed to calculate the Lebesgue integration of the function $f$ on the interval $[0, pi/2]$, why the solution divides the interval $[0, pi/2]$ into what is shown in the picture? depending on what theorem? I am suspecting if this solution is correct, could anyone tell me the correct solution?

Thanks!

real-analysis analysis measure-theory proof-writing lebesgue-integral

share|cite|improve this question

edited Oct 1 at 8:22

mathcounterexamples.net

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asked Oct 1 at 8:05

SecretlySecretly

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  Your approach completely misunderstands the problem. $Bbb Q$ is the set of all rational numbers. How did you mistake it for the intervals $[0,pi/6] cup [pi/4,pi/2]$?
  $endgroup$
  – Paul Sinclair
  Oct 1 at 16:31
  
  

  
  
  
  

add a comment
 | 

2

$begingroup$

The question and its solution are given below:

My questions are:

I am supposed to calculate the Lebesgue integration of the function $f$ on the interval $[0, pi/2]$, why the solution divides the interval $[0, pi/2]$ into what is shown in the picture? depending on what theorem? I am suspecting if this solution is correct, could anyone tell me the correct solution?

Thanks!

real-analysis analysis measure-theory proof-writing lebesgue-integral

share|cite|improve this question

edited Oct 1 at 8:22

mathcounterexamples.net

33.6k22 gold badges2424 silver badges6262 bronze badges

asked Oct 1 at 8:05

SecretlySecretly

2,40244 silver badges1616 bronze badges

$endgroup$

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  Your approach completely misunderstands the problem. $Bbb Q$ is the set of all rational numbers. How did you mistake it for the intervals $[0,pi/6] cup [pi/4,pi/2]$?
  $endgroup$
  – Paul Sinclair
  Oct 1 at 16:31
  
  

  
  
  
  

add a comment
 | 

$begingroup$

The question and its solution are given below:

My questions are:

I am supposed to calculate the Lebesgue integration of the function $f$ on the interval $[0, pi/2]$, why the solution divides the interval $[0, pi/2]$ into what is shown in the picture? depending on what theorem? I am suspecting if this solution is correct, could anyone tell me the correct solution?

Thanks!

real-analysis analysis measure-theory proof-writing lebesgue-integral

share|cite|improve this question

edited Oct 1 at 8:22

mathcounterexamples.net

33.6k22 gold badges2424 silver badges6262 bronze badges

asked Oct 1 at 8:05

SecretlySecretly

2,40244 silver badges1616 bronze badges

$endgroup$

share|cite|improve this question

edited Oct 1 at 8:22

mathcounterexamples.net

33.6k22 gold badges2424 silver badges6262 bronze badges

asked Oct 1 at 8:05

SecretlySecretly

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share|cite|improve this question

edited Oct 1 at 8:22

mathcounterexamples.net

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asked Oct 1 at 8:05

SecretlySecretly

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edited Oct 1 at 8:22

mathcounterexamples.net

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edited Oct 1 at 8:22

mathcounterexamples.net

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asked Oct 1 at 8:05

SecretlySecretly

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asked Oct 1 at 8:05

SecretlySecretly

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3 Answers
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$begingroup$

I think that your solution is not correct.

You must use the fact that the integral of each integrable function $f$ on a set of measure equal to $0$ is $0$.

But $m(cos^-1(mathbbQ))=0$, so

$int_0^fracpi2fdx =int_[0,fracpi2]cap cos^-1(mathbbQ)fdx + int_[0,fracpi2]cap cos^-1(mathbbQ)^cfdx =$

$=0+ int_[0,fracpi2]cap cos^-1(mathbbQ)^cfdx=$

$= int_[0,fracpi2]cap cos^-1(mathbbQ)^csin^2(x)dx=$

$= int_[0,fracpi2]cap cos^-1(mathbbQ)sin^2(x)dx + int_[0,fracpi2]cap cos^-1(mathbbQ)^csin^2(x)dx=$

$= int_[0,fracpi2]sin^2(x)dx$

Now you can compute this integral because the function $sin^2(x)$ is smooth and the domain of integration is an interval, so the Lebesgue integral is the Riemann integral.

share|cite|improve this answer

edited Oct 1 at 15:05

answered Oct 1 at 8:21

Federico FalluccaFederico Fallucca

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• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  but in case one $cos (x) in mathbbQ$ not the $x in mathbbQ$?
  $endgroup$
  – Mathstupid
  Oct 1 at 14:37
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Smart thanks you very much
  $endgroup$
  – Federico Fallucca
  Oct 1 at 15:05
  

  
  
  
  

add a comment
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4

$begingroup$

Your solution is not correct. Since $$mxmid cos(x)in mathbb Q=0,$$
you have $$int_0^pi/2f(x),mathrm d x=int_0^pi/2sin^2(x),mathrm d x.$$

share|cite|improve this answer

answered Oct 1 at 8:16

SurbSurb

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add a comment
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2

$begingroup$

A simpler approach: $cos x$ takes any particular value only in a countbale set. Hence $x: cos x in mathbb Q$ is a countable set. But countable sets have measure $0$. Hence the given integral is same as $int_0^pi /2 sin^2x dx=frac pi 4$.

share|cite|improve this answer

answered Oct 1 at 8:15

Kabo MurphyKabo Murphy

145k99 gold badges4949 silver badges106106 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  is this because the integration of $- cos(x)$ is $sin (x)$ ..... so we are always trying to change any Lebesgue integration to Riemann integration?
  $endgroup$
  – Mathstupid
  Oct 1 at 14:39
  
  

  
  
  
  

add a comment
 | 

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2

$begingroup$

I think that your solution is not correct.

You must use the fact that the integral of each integrable function $f$ on a set of measure equal to $0$ is $0$.

But $m(cos^-1(mathbbQ))=0$, so

$int_0^fracpi2fdx =int_[0,fracpi2]cap cos^-1(mathbbQ)fdx + int_[0,fracpi2]cap cos^-1(mathbbQ)^cfdx =$

$=0+ int_[0,fracpi2]cap cos^-1(mathbbQ)^cfdx=$

$= int_[0,fracpi2]cap cos^-1(mathbbQ)^csin^2(x)dx=$

$= int_[0,fracpi2]cap cos^-1(mathbbQ)sin^2(x)dx + int_[0,fracpi2]cap cos^-1(mathbbQ)^csin^2(x)dx=$

$= int_[0,fracpi2]sin^2(x)dx$

Now you can compute this integral because the function $sin^2(x)$ is smooth and the domain of integration is an interval, so the Lebesgue integral is the Riemann integral.

share|cite|improve this answer

edited Oct 1 at 15:05

answered Oct 1 at 8:21

Federico FalluccaFederico Fallucca

4,17733 silver badges1212 bronze badges

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  but in case one $cos (x) in mathbbQ$ not the $x in mathbbQ$?
  $endgroup$
  – Mathstupid
  Oct 1 at 14:37
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Smart thanks you very much
  $endgroup$
  – Federico Fallucca
  Oct 1 at 15:05
  

  
  
  
  

add a comment
 | 

2

$begingroup$

I think that your solution is not correct.

You must use the fact that the integral of each integrable function $f$ on a set of measure equal to $0$ is $0$.

But $m(cos^-1(mathbbQ))=0$, so

$int_0^fracpi2fdx =int_[0,fracpi2]cap cos^-1(mathbbQ)fdx + int_[0,fracpi2]cap cos^-1(mathbbQ)^cfdx =$

$=0+ int_[0,fracpi2]cap cos^-1(mathbbQ)^cfdx=$

$= int_[0,fracpi2]cap cos^-1(mathbbQ)^csin^2(x)dx=$

$= int_[0,fracpi2]cap cos^-1(mathbbQ)sin^2(x)dx + int_[0,fracpi2]cap cos^-1(mathbbQ)^csin^2(x)dx=$

$= int_[0,fracpi2]sin^2(x)dx$

Now you can compute this integral because the function $sin^2(x)$ is smooth and the domain of integration is an interval, so the Lebesgue integral is the Riemann integral.

share|cite|improve this answer

edited Oct 1 at 15:05

answered Oct 1 at 8:21

Federico FalluccaFederico Fallucca

4,17733 silver badges1212 bronze badges

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  but in case one $cos (x) in mathbbQ$ not the $x in mathbbQ$?
  $endgroup$
  – Mathstupid
  Oct 1 at 14:37
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Smart thanks you very much
  $endgroup$
  – Federico Fallucca
  Oct 1 at 15:05
  

  
  
  
  

add a comment
 | 

$begingroup$

I think that your solution is not correct.

You must use the fact that the integral of each integrable function $f$ on a set of measure equal to $0$ is $0$.

But $m(cos^-1(mathbbQ))=0$, so

$int_0^fracpi2fdx =int_[0,fracpi2]cap cos^-1(mathbbQ)fdx + int_[0,fracpi2]cap cos^-1(mathbbQ)^cfdx =$

$=0+ int_[0,fracpi2]cap cos^-1(mathbbQ)^cfdx=$

$= int_[0,fracpi2]cap cos^-1(mathbbQ)^csin^2(x)dx=$

$= int_[0,fracpi2]cap cos^-1(mathbbQ)sin^2(x)dx + int_[0,fracpi2]cap cos^-1(mathbbQ)^csin^2(x)dx=$

$= int_[0,fracpi2]sin^2(x)dx$

Now you can compute this integral because the function $sin^2(x)$ is smooth and the domain of integration is an interval, so the Lebesgue integral is the Riemann integral.

share|cite|improve this answer

edited Oct 1 at 15:05

answered Oct 1 at 8:21

Federico FalluccaFederico Fallucca

4,17733 silver badges1212 bronze badges

$endgroup$

share|cite|improve this answer

edited Oct 1 at 15:05

answered Oct 1 at 8:21

Federico FalluccaFederico Fallucca

4,17733 silver badges1212 bronze badges

answered Oct 1 at 8:21

Federico FalluccaFederico Fallucca

4,17733 silver badges1212 bronze badges

answered Oct 1 at 8:21

Federico FalluccaFederico Fallucca

4,17733 silver badges1212 bronze badges

4

$begingroup$

Your solution is not correct. Since $$mxmid cos(x)in mathbb Q=0,$$
you have $$int_0^pi/2f(x),mathrm d x=int_0^pi/2sin^2(x),mathrm d x.$$

share|cite|improve this answer

answered Oct 1 at 8:16

SurbSurb

43.9k99 gold badges4848 silver badges8888 bronze badges

$endgroup$

add a comment
 | 

4

$begingroup$

Your solution is not correct. Since $$mxmid cos(x)in mathbb Q=0,$$
you have $$int_0^pi/2f(x),mathrm d x=int_0^pi/2sin^2(x),mathrm d x.$$

share|cite|improve this answer

answered Oct 1 at 8:16

SurbSurb

43.9k99 gold badges4848 silver badges8888 bronze badges

$endgroup$

add a comment
 | 

$begingroup$

Your solution is not correct. Since $$mxmid cos(x)in mathbb Q=0,$$
you have $$int_0^pi/2f(x),mathrm d x=int_0^pi/2sin^2(x),mathrm d x.$$

share|cite|improve this answer

answered Oct 1 at 8:16

SurbSurb

43.9k99 gold badges4848 silver badges8888 bronze badges

$endgroup$

share|cite|improve this answer

answered Oct 1 at 8:16

SurbSurb

43.9k99 gold badges4848 silver badges8888 bronze badges

answered Oct 1 at 8:16

SurbSurb

43.9k99 gold badges4848 silver badges8888 bronze badges

answered Oct 1 at 8:16

SurbSurb

43.9k99 gold badges4848 silver badges8888 bronze badges

2

$begingroup$

A simpler approach: $cos x$ takes any particular value only in a countbale set. Hence $x: cos x in mathbb Q$ is a countable set. But countable sets have measure $0$. Hence the given integral is same as $int_0^pi /2 sin^2x dx=frac pi 4$.

share|cite|improve this answer

answered Oct 1 at 8:15

Kabo MurphyKabo Murphy

145k99 gold badges4949 silver badges106106 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  is this because the integration of $- cos(x)$ is $sin (x)$ ..... so we are always trying to change any Lebesgue integration to Riemann integration?
  $endgroup$
  – Mathstupid
  Oct 1 at 14:39
  
  

  
  
  
  

add a comment
 | 

2

$begingroup$

A simpler approach: $cos x$ takes any particular value only in a countbale set. Hence $x: cos x in mathbb Q$ is a countable set. But countable sets have measure $0$. Hence the given integral is same as $int_0^pi /2 sin^2x dx=frac pi 4$.

share|cite|improve this answer

answered Oct 1 at 8:15

Kabo MurphyKabo Murphy

145k99 gold badges4949 silver badges106106 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  is this because the integration of $- cos(x)$ is $sin (x)$ ..... so we are always trying to change any Lebesgue integration to Riemann integration?
  $endgroup$
  – Mathstupid
  Oct 1 at 14:39
  
  

  
  
  
  

add a comment
 | 

$begingroup$

A simpler approach: $cos x$ takes any particular value only in a countbale set. Hence $x: cos x in mathbb Q$ is a countable set. But countable sets have measure $0$. Hence the given integral is same as $int_0^pi /2 sin^2x dx=frac pi 4$.

share|cite|improve this answer

answered Oct 1 at 8:15

Kabo MurphyKabo Murphy

145k99 gold badges4949 silver badges106106 bronze badges

$endgroup$

share|cite|improve this answer

answered Oct 1 at 8:15

Kabo MurphyKabo Murphy

145k99 gold badges4949 silver badges106106 bronze badges

answered Oct 1 at 8:15

Kabo MurphyKabo Murphy

145k99 gold badges4949 silver badges106106 bronze badges

answered Oct 1 at 8:15

Kabo MurphyKabo Murphy

145k99 gold badges4949 silver badges106106 bronze badges