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Proof of additivity of domain for definite integrals

A proof in Spivak on integration.(Riemann Integrals) Three proofs from elementary analysisProve that if [a,b] and [b,c] are integrable, then [a,c] is integrableProving the converse of the Cauchy criterion for integrationIf a bounded function is integrable on each interval $[a,1]$, then it is integrable on $[0,1]$.Show $sinfrac1x$ for $xnot= 0$ and $f(0)=0$ is integrable on $[-1,1]$Proof Verification: Comparison Property for IntegralsIs my proof correct? If $f$ has a finite number of discontinuities on $[a, b]$, then it is integrable on $[a, b]$Show that the function $f : [0, 1] to mathbbR$ defined by $f(x) = cx$ for some fixed $c in mathbbR$ is Riemann IntegrableAdditivity of integral

2

3

$begingroup$

I would like to prove the following theorem:

Theorem If $c in (a,b)$ and $f$ is integrable on $[a,c]$ and $[c,b]$, then $f$ is integrable on $[a,b]$ and
$$int_a^bf = int_a^cf + int_c^bf.$$

I will make us of the following lemma:

Lemma For two non-empty sets $A, B$ which are bounded above, $sup(A+B) = sup A + sup B$ and $inf(A+B) = inf A + inf B$.

I would appreciate it if someone could verify my proof below since it's different from the ones I usually see in textbooks.

Proof Let $P_1$ be a partition of $[a,c]$, $P_2$ a partition of $[c,b]$ and $P = P_1 cup P_2$. Then clearly $P$ is a partition of $[a,b]$, $L(f,P) = L(f,P_1) + L(f,P_2)$ and $U(f,P) = U(f,P_1) + U(f,P_2)$. Then by the lemma,
$$supL(f,P) = supL(f,P_1) + supL(f,P_2),$$
and
$$infU(f,P) = infU(f,P_1) + infU(f,P_2). $$
Because $f$ is integrable on the respective intervals,
$$int_a^cf = supL(f,P_1) = infU(f,P_1) $$ and
$$int_c^bf = supL(f,P_2) = infU(f,P_2), $$ so we get that
$$supL(f,P) = infU(f,P) = int_a^bf = int_a^cf + int_c^bf, $$
showing that $f$ is integrable on $[a,b]$ and that the equality holds.

calculus real-analysis integration

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asked Aug 4 '15 at 7:17

SimeonSimeon

482419

$endgroup$

bumped to the homepage by Community♦ 2 days ago

This question has answers that may be good or bad; the system has marked it active so that they can be reviewed.

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  looks right to me
  $endgroup$
  – user251257
  Aug 4 '15 at 8:28
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Your $P,$ is not arbitrary as a partition of $[a,b]$ because it contains $c dots$
  $endgroup$
  – Tony Piccolo
  Aug 4 '15 at 11:08
  
  

  
  
  
  

add a comment | 

2

3

$begingroup$

I would like to prove the following theorem:

Theorem If $c in (a,b)$ and $f$ is integrable on $[a,c]$ and $[c,b]$, then $f$ is integrable on $[a,b]$ and
$$int_a^bf = int_a^cf + int_c^bf.$$

I will make us of the following lemma:

Lemma For two non-empty sets $A, B$ which are bounded above, $sup(A+B) = sup A + sup B$ and $inf(A+B) = inf A + inf B$.

I would appreciate it if someone could verify my proof below since it's different from the ones I usually see in textbooks.

Proof Let $P_1$ be a partition of $[a,c]$, $P_2$ a partition of $[c,b]$ and $P = P_1 cup P_2$. Then clearly $P$ is a partition of $[a,b]$, $L(f,P) = L(f,P_1) + L(f,P_2)$ and $U(f,P) = U(f,P_1) + U(f,P_2)$. Then by the lemma,
$$supL(f,P) = supL(f,P_1) + supL(f,P_2),$$
and
$$infU(f,P) = infU(f,P_1) + infU(f,P_2). $$
Because $f$ is integrable on the respective intervals,
$$int_a^cf = supL(f,P_1) = infU(f,P_1) $$ and
$$int_c^bf = supL(f,P_2) = infU(f,P_2), $$ so we get that
$$supL(f,P) = infU(f,P) = int_a^bf = int_a^cf + int_c^bf, $$
showing that $f$ is integrable on $[a,b]$ and that the equality holds.

calculus real-analysis integration

share|cite|improve this question

asked Aug 4 '15 at 7:17

SimeonSimeon

482419

$endgroup$

bumped to the homepage by Community♦ 2 days ago

This question has answers that may be good or bad; the system has marked it active so that they can be reviewed.

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  looks right to me
  $endgroup$
  – user251257
  Aug 4 '15 at 8:28
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Your $P,$ is not arbitrary as a partition of $[a,b]$ because it contains $c dots$
  $endgroup$
  – Tony Piccolo
  Aug 4 '15 at 11:08
  
  

  
  
  
  

add a comment | 

$begingroup$

I would like to prove the following theorem:

Theorem If $c in (a,b)$ and $f$ is integrable on $[a,c]$ and $[c,b]$, then $f$ is integrable on $[a,b]$ and
$$int_a^bf = int_a^cf + int_c^bf.$$

I will make us of the following lemma:

Lemma For two non-empty sets $A, B$ which are bounded above, $sup(A+B) = sup A + sup B$ and $inf(A+B) = inf A + inf B$.

I would appreciate it if someone could verify my proof below since it's different from the ones I usually see in textbooks.

Proof Let $P_1$ be a partition of $[a,c]$, $P_2$ a partition of $[c,b]$ and $P = P_1 cup P_2$. Then clearly $P$ is a partition of $[a,b]$, $L(f,P) = L(f,P_1) + L(f,P_2)$ and $U(f,P) = U(f,P_1) + U(f,P_2)$. Then by the lemma,
$$supL(f,P) = supL(f,P_1) + supL(f,P_2),$$
and
$$infU(f,P) = infU(f,P_1) + infU(f,P_2). $$
Because $f$ is integrable on the respective intervals,
$$int_a^cf = supL(f,P_1) = infU(f,P_1) $$ and
$$int_c^bf = supL(f,P_2) = infU(f,P_2), $$ so we get that
$$supL(f,P) = infU(f,P) = int_a^bf = int_a^cf + int_c^bf, $$
showing that $f$ is integrable on $[a,b]$ and that the equality holds.

calculus real-analysis integration

share|cite|improve this question

asked Aug 4 '15 at 7:17

SimeonSimeon

482419

$endgroup$

share|cite|improve this question

asked Aug 4 '15 at 7:17

SimeonSimeon

482419

share|cite|improve this question

asked Aug 4 '15 at 7:17

SimeonSimeon

482419

asked Aug 4 '15 at 7:17

SimeonSimeon

482419

asked Aug 4 '15 at 7:17

SimeonSimeon

482419

1 Answer
1

active

oldest

votes

0

$begingroup$

There is another way of proving the theorem, namely:

Say $I_1 = [a,c]$ and $I_2 = [c,b]$.Then

$$int_a^c f = int_a^b f chi_I_1 quad & quad int_c^b f = int_a^b fchi_I_2$$

So by additivity of the integral

$$int_a^b f chi_I_1 + int_a^b fchi_I_2 = int_a^b f(chi_I_1 + chi_I_2) - int_c^c f(chi_I_1 + chi_I_2) = int_a^b f(chi_I_1 + chi_I_2) = int_a^b f,$$

where $$chi_I (x) = begincases
1 & xin I \
0 & otherwise
endcases$$

Or

You can use the property that

$$chi_I_1 cup I_2 = chi_I_1 + chi_I_2 - chi_I_1 cap I_2.$$

share|cite|improve this answer

answered Feb 22 '18 at 4:05

onurcanbektasonurcanbektas

3,46211037

$endgroup$

add a comment | 

Your Answer

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0

$begingroup$

There is another way of proving the theorem, namely:

Say $I_1 = [a,c]$ and $I_2 = [c,b]$.Then

$$int_a^c f = int_a^b f chi_I_1 quad & quad int_c^b f = int_a^b fchi_I_2$$

So by additivity of the integral

$$int_a^b f chi_I_1 + int_a^b fchi_I_2 = int_a^b f(chi_I_1 + chi_I_2) - int_c^c f(chi_I_1 + chi_I_2) = int_a^b f(chi_I_1 + chi_I_2) = int_a^b f,$$

where $$chi_I (x) = begincases
1 & xin I \
0 & otherwise
endcases$$

Or

You can use the property that

$$chi_I_1 cup I_2 = chi_I_1 + chi_I_2 - chi_I_1 cap I_2.$$

share|cite|improve this answer

answered Feb 22 '18 at 4:05

onurcanbektasonurcanbektas

3,46211037

$endgroup$

add a comment | 

0

$begingroup$

There is another way of proving the theorem, namely:

Say $I_1 = [a,c]$ and $I_2 = [c,b]$.Then

$$int_a^c f = int_a^b f chi_I_1 quad & quad int_c^b f = int_a^b fchi_I_2$$

So by additivity of the integral

$$int_a^b f chi_I_1 + int_a^b fchi_I_2 = int_a^b f(chi_I_1 + chi_I_2) - int_c^c f(chi_I_1 + chi_I_2) = int_a^b f(chi_I_1 + chi_I_2) = int_a^b f,$$

where $$chi_I (x) = begincases
1 & xin I \
0 & otherwise
endcases$$

Or

You can use the property that

$$chi_I_1 cup I_2 = chi_I_1 + chi_I_2 - chi_I_1 cap I_2.$$

share|cite|improve this answer

answered Feb 22 '18 at 4:05

onurcanbektasonurcanbektas

3,46211037

$endgroup$

add a comment | 

$begingroup$

There is another way of proving the theorem, namely:

Say $I_1 = [a,c]$ and $I_2 = [c,b]$.Then

$$int_a^c f = int_a^b f chi_I_1 quad & quad int_c^b f = int_a^b fchi_I_2$$

So by additivity of the integral

$$int_a^b f chi_I_1 + int_a^b fchi_I_2 = int_a^b f(chi_I_1 + chi_I_2) - int_c^c f(chi_I_1 + chi_I_2) = int_a^b f(chi_I_1 + chi_I_2) = int_a^b f,$$

where $$chi_I (x) = begincases
1 & xin I \
0 & otherwise
endcases$$

Or

You can use the property that

$$chi_I_1 cup I_2 = chi_I_1 + chi_I_2 - chi_I_1 cap I_2.$$

share|cite|improve this answer

answered Feb 22 '18 at 4:05

onurcanbektasonurcanbektas

3,46211037

$endgroup$

share|cite|improve this answer

answered Feb 22 '18 at 4:05

onurcanbektasonurcanbektas

3,46211037

answered Feb 22 '18 at 4:05

onurcanbektasonurcanbektas

3,46211037

answered Feb 22 '18 at 4:05

onurcanbektasonurcanbektas

3,46211037