Why is Lebesgue measure theory asymmetric?

Clash Royale CLAN TAG#URR8PPP

up vote
17
down vote

favorite

6

A set $Esubseteq mathbbR^d$ is said to be Jordan measurable if its inner measure $m_*(E)$ and outer measure $m^*(E)$ are equal.However, Lebesgue mesure theory is developed with only outer measure.

A function is Riemann integrable iff its upper integral and lower integral are equal.However, in Lebesgue integration theory, we rarely use upper Lebesgue integral.

Why are outer measure and lower integral more important than inner measure and upper integral?

measure-theory lebesgue-measure

share|cite|improve this question

edited Aug 22 at 14:36

asked Aug 22 at 0:15

satoukibi

915

• 
  
  
  

  
  9
  

  
  
  
  
  
  

  
  Lebesgue measurability can also be defined via inner and outer measure, and I believe that was how it was originally defined, the "outer measure only" definition by Caratheodory coming later. That said, I can't speak to why the Caratheodory definition seems to be presented more frequently.
  – Noah Schweber
  Aug 22 at 2:41
  
  

  
  
  
  


• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  I do not think it is asymetric.. It may be true that for a set $Esubseteq mathbbR$, lebesgue outer measure is defined by covering it from out side by open intervals... But, you call a set $E$ Lebesgue measurable only if for any $Asubseteq mathbbR$ you have $mu(A)=mu(Acap E)+mu(Acap E^c)$.. This can be seen as taking care of outer approximation (in case when you consider $Acap E$) and inner approximation (when you consider $Acap E^C$).. Does it make some sense.. ?? It is vague but I can not make it any better :D
  – Praphulla Koushik
  Aug 22 at 2:55
  
  

  
  
  
  


• 
  
  
  

  
  4
  

  
  
  
  
  
  

  
  I agree with Noah and Praphulla — it’s rather an expository choice. As further evidence, see e.g. Bourbaki, Chap. IV, §4, exercises 6b) and 7b) or Godement (1948, p. 4, fourth displayed equation).
  – Francois Ziegler
  Aug 22 at 8:52
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  Also “Historical note” in Chap. V, pp. 126–127: In Lebesgue’s thesis, “imitating the Peano–Jordan method, the ‘outer measure’ of a bounded set $Asubsetmathbf R$ is defined... then, if $I$ is an interval containing $A$, the ‘inner measure’ of $A$ is the difference of the outer measures of $I$ and $I - A$; one thus obtains a notion of ‘measurable set’...”
  – Francois Ziegler
  Aug 22 at 9:25
  

  
  
  
  

add a comment | 

up vote
17
down vote

favorite

6

A set $Esubseteq mathbbR^d$ is said to be Jordan measurable if its inner measure $m_*(E)$ and outer measure $m^*(E)$ are equal.However, Lebesgue mesure theory is developed with only outer measure.

A function is Riemann integrable iff its upper integral and lower integral are equal.However, in Lebesgue integration theory, we rarely use upper Lebesgue integral.

Why are outer measure and lower integral more important than inner measure and upper integral?

measure-theory lebesgue-measure

share|cite|improve this question

edited Aug 22 at 14:36

asked Aug 22 at 0:15

satoukibi

915

• 
  
  
  

  
  9
  

  
  
  
  
  
  

  
  Lebesgue measurability can also be defined via inner and outer measure, and I believe that was how it was originally defined, the "outer measure only" definition by Caratheodory coming later. That said, I can't speak to why the Caratheodory definition seems to be presented more frequently.
  – Noah Schweber
  Aug 22 at 2:41
  
  

  
  
  
  


• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  I do not think it is asymetric.. It may be true that for a set $Esubseteq mathbbR$, lebesgue outer measure is defined by covering it from out side by open intervals... But, you call a set $E$ Lebesgue measurable only if for any $Asubseteq mathbbR$ you have $mu(A)=mu(Acap E)+mu(Acap E^c)$.. This can be seen as taking care of outer approximation (in case when you consider $Acap E$) and inner approximation (when you consider $Acap E^C$).. Does it make some sense.. ?? It is vague but I can not make it any better :D
  – Praphulla Koushik
  Aug 22 at 2:55
  
  

  
  
  
  


• 
  
  
  

  
  4
  

  
  
  
  
  
  

  
  I agree with Noah and Praphulla — it’s rather an expository choice. As further evidence, see e.g. Bourbaki, Chap. IV, §4, exercises 6b) and 7b) or Godement (1948, p. 4, fourth displayed equation).
  – Francois Ziegler
  Aug 22 at 8:52
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  Also “Historical note” in Chap. V, pp. 126–127: In Lebesgue’s thesis, “imitating the Peano–Jordan method, the ‘outer measure’ of a bounded set $Asubsetmathbf R$ is defined... then, if $I$ is an interval containing $A$, the ‘inner measure’ of $A$ is the difference of the outer measures of $I$ and $I - A$; one thus obtains a notion of ‘measurable set’...”
  – Francois Ziegler
  Aug 22 at 9:25
  

  
  
  
  

add a comment | 

up vote
17
down vote

favorite

6

up vote
17
down vote

favorite

6

6

A set $Esubseteq mathbbR^d$ is said to be Jordan measurable if its inner measure $m_*(E)$ and outer measure $m^*(E)$ are equal.However, Lebesgue mesure theory is developed with only outer measure.

A function is Riemann integrable iff its upper integral and lower integral are equal.However, in Lebesgue integration theory, we rarely use upper Lebesgue integral.

Why are outer measure and lower integral more important than inner measure and upper integral?

measure-theory lebesgue-measure

share|cite|improve this question

edited Aug 22 at 14:36

asked Aug 22 at 0:15

satoukibi

915

A set $Esubseteq mathbbR^d$ is said to be Jordan measurable if its inner measure $m_*(E)$ and outer measure $m^*(E)$ are equal.However, Lebesgue mesure theory is developed with only outer measure.

A function is Riemann integrable iff its upper integral and lower integral are equal.However, in Lebesgue integration theory, we rarely use upper Lebesgue integral.

Why are outer measure and lower integral more important than inner measure and upper integral?

measure-theory lebesgue-measure

share|cite|improve this question

edited Aug 22 at 14:36

asked Aug 22 at 0:15

satoukibi

915

share|cite|improve this question

share|cite|improve this question

edited Aug 22 at 14:36

edited Aug 22 at 14:36

edited Aug 22 at 14:36

asked Aug 22 at 0:15

satoukibi

915

asked Aug 22 at 0:15

satoukibi

915

asked Aug 22 at 0:15

satoukibi

915

915

• 
  
  
  

  
  9
  

  
  
  
  
  
  

  
  Lebesgue measurability can also be defined via inner and outer measure, and I believe that was how it was originally defined, the "outer measure only" definition by Caratheodory coming later. That said, I can't speak to why the Caratheodory definition seems to be presented more frequently.
  – Noah Schweber
  Aug 22 at 2:41
  
  

  
  
  
  


• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  I do not think it is asymetric.. It may be true that for a set $Esubseteq mathbbR$, lebesgue outer measure is defined by covering it from out side by open intervals... But, you call a set $E$ Lebesgue measurable only if for any $Asubseteq mathbbR$ you have $mu(A)=mu(Acap E)+mu(Acap E^c)$.. This can be seen as taking care of outer approximation (in case when you consider $Acap E$) and inner approximation (when you consider $Acap E^C$).. Does it make some sense.. ?? It is vague but I can not make it any better :D
  – Praphulla Koushik
  Aug 22 at 2:55
  
  

  
  
  
  


• 
  
  
  

  
  4
  

  
  
  
  
  
  

  
  I agree with Noah and Praphulla — it’s rather an expository choice. As further evidence, see e.g. Bourbaki, Chap. IV, §4, exercises 6b) and 7b) or Godement (1948, p. 4, fourth displayed equation).
  – Francois Ziegler
  Aug 22 at 8:52
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  Also “Historical note” in Chap. V, pp. 126–127: In Lebesgue’s thesis, “imitating the Peano–Jordan method, the ‘outer measure’ of a bounded set $Asubsetmathbf R$ is defined... then, if $I$ is an interval containing $A$, the ‘inner measure’ of $A$ is the difference of the outer measures of $I$ and $I - A$; one thus obtains a notion of ‘measurable set’...”
  – Francois Ziegler
  Aug 22 at 9:25
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  9
  

  
  
  
  
  
  

  
  Lebesgue measurability can also be defined via inner and outer measure, and I believe that was how it was originally defined, the "outer measure only" definition by Caratheodory coming later. That said, I can't speak to why the Caratheodory definition seems to be presented more frequently.
  – Noah Schweber
  Aug 22 at 2:41
  
  

  
  
  
  


• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  I do not think it is asymetric.. It may be true that for a set $Esubseteq mathbbR$, lebesgue outer measure is defined by covering it from out side by open intervals... But, you call a set $E$ Lebesgue measurable only if for any $Asubseteq mathbbR$ you have $mu(A)=mu(Acap E)+mu(Acap E^c)$.. This can be seen as taking care of outer approximation (in case when you consider $Acap E$) and inner approximation (when you consider $Acap E^C$).. Does it make some sense.. ?? It is vague but I can not make it any better :D
  – Praphulla Koushik
  Aug 22 at 2:55
  
  

  
  
  
  


• 
  
  
  

  
  4
  

  
  
  
  
  
  

  
  I agree with Noah and Praphulla — it’s rather an expository choice. As further evidence, see e.g. Bourbaki, Chap. IV, §4, exercises 6b) and 7b) or Godement (1948, p. 4, fourth displayed equation).
  – Francois Ziegler
  Aug 22 at 8:52
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  Also “Historical note” in Chap. V, pp. 126–127: In Lebesgue’s thesis, “imitating the Peano–Jordan method, the ‘outer measure’ of a bounded set $Asubsetmathbf R$ is defined... then, if $I$ is an interval containing $A$, the ‘inner measure’ of $A$ is the difference of the outer measures of $I$ and $I - A$; one thus obtains a notion of ‘measurable set’...”
  – Francois Ziegler
  Aug 22 at 9:25
  

  
  
  
  

9

9

Lebesgue measurability can also be defined via inner and outer measure, and I believe that was how it was originally defined, the "outer measure only" definition by Caratheodory coming later. That said, I can't speak to why the Caratheodory definition seems to be presented more frequently.
– Noah Schweber
Aug 22 at 2:41

Lebesgue measurability can also be defined via inner and outer measure, and I believe that was how it was originally defined, the "outer measure only" definition by Caratheodory coming later. That said, I can't speak to why the Caratheodory definition seems to be presented more frequently.
– Noah Schweber
Aug 22 at 2:41

3

3

I do not think it is asymetric.. It may be true that for a set $Esubseteq mathbbR$, lebesgue outer measure is defined by covering it from out side by open intervals... But, you call a set $E$ Lebesgue measurable only if for any $Asubseteq mathbbR$ you have $mu(A)=mu(Acap E)+mu(Acap E^c)$.. This can be seen as taking care of outer approximation (in case when you consider $Acap E$) and inner approximation (when you consider $Acap E^C$).. Does it make some sense.. ?? It is vague but I can not make it any better :D
– Praphulla Koushik
Aug 22 at 2:55

I do not think it is asymetric.. It may be true that for a set $Esubseteq mathbbR$, lebesgue outer measure is defined by covering it from out side by open intervals... But, you call a set $E$ Lebesgue measurable only if for any $Asubseteq mathbbR$ you have $mu(A)=mu(Acap E)+mu(Acap E^c)$.. This can be seen as taking care of outer approximation (in case when you consider $Acap E$) and inner approximation (when you consider $Acap E^C$).. Does it make some sense.. ?? It is vague but I can not make it any better :D
– Praphulla Koushik
Aug 22 at 2:55

4

4

I agree with Noah and Praphulla — it’s rather an expository choice. As further evidence, see e.g. Bourbaki, Chap. IV, §4, exercises 6b) and 7b) or Godement (1948, p. 4, fourth displayed equation).
– Francois Ziegler
Aug 22 at 8:52

I agree with Noah and Praphulla — it’s rather an expository choice. As further evidence, see e.g. Bourbaki, Chap. IV, §4, exercises 6b) and 7b) or Godement (1948, p. 4, fourth displayed equation).
– Francois Ziegler
Aug 22 at 8:52

2

2

Also “Historical note” in Chap. V, pp. 126–127: In Lebesgue’s thesis, “imitating the Peano–Jordan method, the ‘outer measure’ of a bounded set $Asubsetmathbf R$ is defined... then, if $I$ is an interval containing $A$, the ‘inner measure’ of $A$ is the difference of the outer measures of $I$ and $I - A$; one thus obtains a notion of ‘measurable set’...”
– Francois Ziegler
Aug 22 at 9:25

Also “Historical note” in Chap. V, pp. 126–127: In Lebesgue’s thesis, “imitating the Peano–Jordan method, the ‘outer measure’ of a bounded set $Asubsetmathbf R$ is defined... then, if $I$ is an interval containing $A$, the ‘inner measure’ of $A$ is the difference of the outer measures of $I$ and $I - A$; one thus obtains a notion of ‘measurable set’...”
– Francois Ziegler
Aug 22 at 9:25

add a comment | 

5 Answers
5

active

oldest

votes

up vote
15
down vote



accepted

I have a (possibly idiosyncratic) view that the natural form of measure theory is for finite measure spaces and bounded functions. Other cases are obviously very important, but we have to work harder to get them. You can see this is many of the proofs, where the finite case is easier, and we have to work a bit more to generalize it. For example, the usual proof of the Radon-Nikodym theorem works that way.

In the finite measure space case, with bounded functions, everything can be made symmetric. The symmetry is broken in the general case, because allowing infinity breaks it. In integration, this asymmetry shows up in the way we have to have separate theorems for the non-negative measurable functions and the integrable functions. For bounded functions on finite measure spaces you don't need to impose any extra conditions.

share|cite|improve this answer

answered Aug 22 at 9:26

arsmath

4,19412340

• 
  
  
  

  
  5
  

  
  
  
  
  
  

  
  One very explicit way in which the symmetry is broken is by declaring $0 cdot infty = infty cdot 0$ to equal $0$ rather than $infty$. On the non-negative extended reals $[0,+infty]$, this makes multiplication continuous from below, but not from above: if $a_n,b_n in [0,+infty]$ increase to $a,b$ respectively, then $a_n b_n$ increases to $ab$, but the same is not true for decreasing sequences. I think this already explains much of the asymmetry in Lebesgue measure theory.
  – Terry Tao
  Aug 22 at 14:56
  
  

  
  
  
  


• 
  
  
  

  
  5
  

  
  
  
  
  
  

  
  If one adopted the opposite convention $0 cdot infty = infty cdot 0 = infty$, and reversed the roles of upper and lower integrals, I think one also gets a self-consistent theory, but one that is completely degenerate as soon as one allows $infty$ as a possible value of a function or as a possible value of a measure: all integrals would be infinite or divergent, because all functions are infinite on the empty set.
  – Terry Tao
  Aug 22 at 15:03
  
  

  
  
  
  

add a comment | 

up vote
10
down vote

I think your statement about Jordan is actually wrong. If $m_*(E) = infty$ and $m^*(E) = infty$, then $E$ need not be Jordan measurable. If you talk only about bounded sets $E$, then your characterization is correct. But it is also correct for Lebesgue measure (using Lebesgue inner and outer measure).

The reason for Caratheodory's criterion is to define measurability when even bounded sets could have infinite measure, so that restricting to bounded sets no longer helps. One of Caratheorory's examples was an "arc length" measure for sets in $mathbb R^n$. In that case, there is no obvious way to define inner measure. But we still can define outer measure. And then we need a criterion for measurability that uses only outer measure.

More recent mathematicians have developed a way to start only with an "inner measure" and go from there.

share|cite|improve this answer

answered Aug 22 at 12:57

Gerald Edgar

27k269153

add a comment | 

up vote
6
down vote

I think, the reason is that if the ground space has infinite measure, you can not define the measurable sets as those for which inner measure equals the outer measure: it may happen that both are infinite, while the set is still not measurable.

Note also (this may be related) that outer and inner regarity behave differently in general. For example, the sigma-finite Borel measure on the Polish space is inner regular, but not always outer regular (example: counting measure of rational numbers as a measure on $mathbbR$.)

share|cite|improve this answer

answered Aug 22 at 7:17

Fedor Petrov

44.6k5107211

add a comment | 

up vote
1
down vote

Concerning Tao comment that the symmetry is broken by declaring 0⋅∞ = ∞⋅0 = 0, I would like to add that this is the reason why Lebesgue integral does not satisfy Newton-Leibniz formula. Namely, for Cantor-Lebesgue function f, f(1)–f(0) = 1 but ∫01f’ = 0 because f’ = ∞ on Cantor set C which has measure 0 (and f’ = 0 on its complement). But if we realize that the measure of C is 0 = (1, 2/3, 4/9, ... ) and f’ = ∞ = (1, 3/2, 9/4, ...) then we see that this particular 0⋅∞ is not 0 but exactly 1, as it should be by Newton-Leibniz formula. We have the similar problem with countable additivity of limiting frequencies, which is usually contradicted by an infinite lottery with tokens 1,2,3,4,…. This contradiction also depends on ∞⋅0 = 0 and disappears if we really calculate the relevant ∞⋅0 ( https://www.fsb.unizg.hr/matematika/download/ZS/clanci/ZS-a_note_on_probability_frequency.pdf )

share|cite|improve this answer

answered Aug 28 at 18:05

Zvonimir Sikic

111

add a comment | 

up vote
0
down vote

My two cents. Outer measures are sub-additive on countable coverings: $$Asubset cup _jinmathbbN A_jquad Rightarrow quad mu( A)le sum_jinmathbbNmu(A_j)$$
which is somehow a nicer and more practical property than the analogous dual property of super-additivity for inner measures (even in the case of finite measures). It gives a bound on the set $A$ in terms of the supposedly simpler sets $A_j$. Also, we like sub-additivity more than super-additivity, because it recalls norms.

share|cite|improve this answer

answered Aug 28 at 19:50

Pietro Majer

37.9k277180

add a comment | 

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: "504"
;
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',
convertImagesToLinks: true,
noModals: false,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
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

StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmathoverflow.net%2fquestions%2f308856%2fwhy-is-lebesgue-measure-theory-asymmetric%23new-answer', 'question_page');

);

Post as a guest

Name Email

5 Answers
5

active

oldest

votes

5 Answers
5

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
15
down vote



accepted

I have a (possibly idiosyncratic) view that the natural form of measure theory is for finite measure spaces and bounded functions. Other cases are obviously very important, but we have to work harder to get them. You can see this is many of the proofs, where the finite case is easier, and we have to work a bit more to generalize it. For example, the usual proof of the Radon-Nikodym theorem works that way.

In the finite measure space case, with bounded functions, everything can be made symmetric. The symmetry is broken in the general case, because allowing infinity breaks it. In integration, this asymmetry shows up in the way we have to have separate theorems for the non-negative measurable functions and the integrable functions. For bounded functions on finite measure spaces you don't need to impose any extra conditions.

share|cite|improve this answer

answered Aug 22 at 9:26

arsmath

4,19412340

• 
  
  
  

  
  5
  

  
  
  
  
  
  

  
  One very explicit way in which the symmetry is broken is by declaring $0 cdot infty = infty cdot 0$ to equal $0$ rather than $infty$. On the non-negative extended reals $[0,+infty]$, this makes multiplication continuous from below, but not from above: if $a_n,b_n in [0,+infty]$ increase to $a,b$ respectively, then $a_n b_n$ increases to $ab$, but the same is not true for decreasing sequences. I think this already explains much of the asymmetry in Lebesgue measure theory.
  – Terry Tao
  Aug 22 at 14:56
  
  

  
  
  
  


• 
  
  
  

  
  5
  

  
  
  
  
  
  

  
  If one adopted the opposite convention $0 cdot infty = infty cdot 0 = infty$, and reversed the roles of upper and lower integrals, I think one also gets a self-consistent theory, but one that is completely degenerate as soon as one allows $infty$ as a possible value of a function or as a possible value of a measure: all integrals would be infinite or divergent, because all functions are infinite on the empty set.
  – Terry Tao
  Aug 22 at 15:03
  
  

  
  
  
  

add a comment | 

up vote
15
down vote



accepted

I have a (possibly idiosyncratic) view that the natural form of measure theory is for finite measure spaces and bounded functions. Other cases are obviously very important, but we have to work harder to get them. You can see this is many of the proofs, where the finite case is easier, and we have to work a bit more to generalize it. For example, the usual proof of the Radon-Nikodym theorem works that way.

In the finite measure space case, with bounded functions, everything can be made symmetric. The symmetry is broken in the general case, because allowing infinity breaks it. In integration, this asymmetry shows up in the way we have to have separate theorems for the non-negative measurable functions and the integrable functions. For bounded functions on finite measure spaces you don't need to impose any extra conditions.

share|cite|improve this answer

answered Aug 22 at 9:26

arsmath

4,19412340

• 
  
  
  

  
  5
  

  
  
  
  
  
  

  
  One very explicit way in which the symmetry is broken is by declaring $0 cdot infty = infty cdot 0$ to equal $0$ rather than $infty$. On the non-negative extended reals $[0,+infty]$, this makes multiplication continuous from below, but not from above: if $a_n,b_n in [0,+infty]$ increase to $a,b$ respectively, then $a_n b_n$ increases to $ab$, but the same is not true for decreasing sequences. I think this already explains much of the asymmetry in Lebesgue measure theory.
  – Terry Tao
  Aug 22 at 14:56
  
  

  
  
  
  


• 
  
  
  

  
  5
  

  
  
  
  
  
  

  
  If one adopted the opposite convention $0 cdot infty = infty cdot 0 = infty$, and reversed the roles of upper and lower integrals, I think one also gets a self-consistent theory, but one that is completely degenerate as soon as one allows $infty$ as a possible value of a function or as a possible value of a measure: all integrals would be infinite or divergent, because all functions are infinite on the empty set.
  – Terry Tao
  Aug 22 at 15:03
  
  

  
  
  
  

add a comment | 

up vote
15
down vote



accepted

up vote
15
down vote



accepted

I have a (possibly idiosyncratic) view that the natural form of measure theory is for finite measure spaces and bounded functions. Other cases are obviously very important, but we have to work harder to get them. You can see this is many of the proofs, where the finite case is easier, and we have to work a bit more to generalize it. For example, the usual proof of the Radon-Nikodym theorem works that way.

In the finite measure space case, with bounded functions, everything can be made symmetric. The symmetry is broken in the general case, because allowing infinity breaks it. In integration, this asymmetry shows up in the way we have to have separate theorems for the non-negative measurable functions and the integrable functions. For bounded functions on finite measure spaces you don't need to impose any extra conditions.

share|cite|improve this answer

answered Aug 22 at 9:26

arsmath

4,19412340

I have a (possibly idiosyncratic) view that the natural form of measure theory is for finite measure spaces and bounded functions. Other cases are obviously very important, but we have to work harder to get them. You can see this is many of the proofs, where the finite case is easier, and we have to work a bit more to generalize it. For example, the usual proof of the Radon-Nikodym theorem works that way.

In the finite measure space case, with bounded functions, everything can be made symmetric. The symmetry is broken in the general case, because allowing infinity breaks it. In integration, this asymmetry shows up in the way we have to have separate theorems for the non-negative measurable functions and the integrable functions. For bounded functions on finite measure spaces you don't need to impose any extra conditions.

share|cite|improve this answer

answered Aug 22 at 9:26

arsmath

4,19412340

share|cite|improve this answer

share|cite|improve this answer

answered Aug 22 at 9:26

arsmath

4,19412340

answered Aug 22 at 9:26

arsmath

4,19412340

answered Aug 22 at 9:26

arsmath

4,19412340

4,19412340

• 
  
  
  

  
  5
  

  
  
  
  
  
  

  
  One very explicit way in which the symmetry is broken is by declaring $0 cdot infty = infty cdot 0$ to equal $0$ rather than $infty$. On the non-negative extended reals $[0,+infty]$, this makes multiplication continuous from below, but not from above: if $a_n,b_n in [0,+infty]$ increase to $a,b$ respectively, then $a_n b_n$ increases to $ab$, but the same is not true for decreasing sequences. I think this already explains much of the asymmetry in Lebesgue measure theory.
  – Terry Tao
  Aug 22 at 14:56
  
  

  
  
  
  


• 
  
  
  

  
  5
  

  
  
  
  
  
  

  
  If one adopted the opposite convention $0 cdot infty = infty cdot 0 = infty$, and reversed the roles of upper and lower integrals, I think one also gets a self-consistent theory, but one that is completely degenerate as soon as one allows $infty$ as a possible value of a function or as a possible value of a measure: all integrals would be infinite or divergent, because all functions are infinite on the empty set.
  – Terry Tao
  Aug 22 at 15:03
  
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  5
  

  
  
  
  
  
  

  
  One very explicit way in which the symmetry is broken is by declaring $0 cdot infty = infty cdot 0$ to equal $0$ rather than $infty$. On the non-negative extended reals $[0,+infty]$, this makes multiplication continuous from below, but not from above: if $a_n,b_n in [0,+infty]$ increase to $a,b$ respectively, then $a_n b_n$ increases to $ab$, but the same is not true for decreasing sequences. I think this already explains much of the asymmetry in Lebesgue measure theory.
  – Terry Tao
  Aug 22 at 14:56
  
  

  
  
  
  


• 
  
  
  

  
  5
  

  
  
  
  
  
  

  
  If one adopted the opposite convention $0 cdot infty = infty cdot 0 = infty$, and reversed the roles of upper and lower integrals, I think one also gets a self-consistent theory, but one that is completely degenerate as soon as one allows $infty$ as a possible value of a function or as a possible value of a measure: all integrals would be infinite or divergent, because all functions are infinite on the empty set.
  – Terry Tao
  Aug 22 at 15:03
  
  

  
  
  
  

5

5

One very explicit way in which the symmetry is broken is by declaring $0 cdot infty = infty cdot 0$ to equal $0$ rather than $infty$. On the non-negative extended reals $[0,+infty]$, this makes multiplication continuous from below, but not from above: if $a_n,b_n in [0,+infty]$ increase to $a,b$ respectively, then $a_n b_n$ increases to $ab$, but the same is not true for decreasing sequences. I think this already explains much of the asymmetry in Lebesgue measure theory.
– Terry Tao
Aug 22 at 14:56

One very explicit way in which the symmetry is broken is by declaring $0 cdot infty = infty cdot 0$ to equal $0$ rather than $infty$. On the non-negative extended reals $[0,+infty]$, this makes multiplication continuous from below, but not from above: if $a_n,b_n in [0,+infty]$ increase to $a,b$ respectively, then $a_n b_n$ increases to $ab$, but the same is not true for decreasing sequences. I think this already explains much of the asymmetry in Lebesgue measure theory.
– Terry Tao
Aug 22 at 14:56

5

5

If one adopted the opposite convention $0 cdot infty = infty cdot 0 = infty$, and reversed the roles of upper and lower integrals, I think one also gets a self-consistent theory, but one that is completely degenerate as soon as one allows $infty$ as a possible value of a function or as a possible value of a measure: all integrals would be infinite or divergent, because all functions are infinite on the empty set.
– Terry Tao
Aug 22 at 15:03

If one adopted the opposite convention $0 cdot infty = infty cdot 0 = infty$, and reversed the roles of upper and lower integrals, I think one also gets a self-consistent theory, but one that is completely degenerate as soon as one allows $infty$ as a possible value of a function or as a possible value of a measure: all integrals would be infinite or divergent, because all functions are infinite on the empty set.
– Terry Tao
Aug 22 at 15:03

add a comment | 

up vote
10
down vote

I think your statement about Jordan is actually wrong. If $m_*(E) = infty$ and $m^*(E) = infty$, then $E$ need not be Jordan measurable. If you talk only about bounded sets $E$, then your characterization is correct. But it is also correct for Lebesgue measure (using Lebesgue inner and outer measure).

The reason for Caratheodory's criterion is to define measurability when even bounded sets could have infinite measure, so that restricting to bounded sets no longer helps. One of Caratheorory's examples was an "arc length" measure for sets in $mathbb R^n$. In that case, there is no obvious way to define inner measure. But we still can define outer measure. And then we need a criterion for measurability that uses only outer measure.

More recent mathematicians have developed a way to start only with an "inner measure" and go from there.

share|cite|improve this answer

answered Aug 22 at 12:57

Gerald Edgar

27k269153

add a comment | 

up vote
10
down vote

I think your statement about Jordan is actually wrong. If $m_*(E) = infty$ and $m^*(E) = infty$, then $E$ need not be Jordan measurable. If you talk only about bounded sets $E$, then your characterization is correct. But it is also correct for Lebesgue measure (using Lebesgue inner and outer measure).

The reason for Caratheodory's criterion is to define measurability when even bounded sets could have infinite measure, so that restricting to bounded sets no longer helps. One of Caratheorory's examples was an "arc length" measure for sets in $mathbb R^n$. In that case, there is no obvious way to define inner measure. But we still can define outer measure. And then we need a criterion for measurability that uses only outer measure.

More recent mathematicians have developed a way to start only with an "inner measure" and go from there.

share|cite|improve this answer

answered Aug 22 at 12:57

Gerald Edgar

27k269153

add a comment | 

up vote
10
down vote

up vote
10
down vote

I think your statement about Jordan is actually wrong. If $m_*(E) = infty$ and $m^*(E) = infty$, then $E$ need not be Jordan measurable. If you talk only about bounded sets $E$, then your characterization is correct. But it is also correct for Lebesgue measure (using Lebesgue inner and outer measure).

The reason for Caratheodory's criterion is to define measurability when even bounded sets could have infinite measure, so that restricting to bounded sets no longer helps. One of Caratheorory's examples was an "arc length" measure for sets in $mathbb R^n$. In that case, there is no obvious way to define inner measure. But we still can define outer measure. And then we need a criterion for measurability that uses only outer measure.

More recent mathematicians have developed a way to start only with an "inner measure" and go from there.

share|cite|improve this answer

answered Aug 22 at 12:57

Gerald Edgar

27k269153

I think your statement about Jordan is actually wrong. If $m_*(E) = infty$ and $m^*(E) = infty$, then $E$ need not be Jordan measurable. If you talk only about bounded sets $E$, then your characterization is correct. But it is also correct for Lebesgue measure (using Lebesgue inner and outer measure).

The reason for Caratheodory's criterion is to define measurability when even bounded sets could have infinite measure, so that restricting to bounded sets no longer helps. One of Caratheorory's examples was an "arc length" measure for sets in $mathbb R^n$. In that case, there is no obvious way to define inner measure. But we still can define outer measure. And then we need a criterion for measurability that uses only outer measure.

More recent mathematicians have developed a way to start only with an "inner measure" and go from there.

share|cite|improve this answer

answered Aug 22 at 12:57

Gerald Edgar

27k269153

share|cite|improve this answer

share|cite|improve this answer

answered Aug 22 at 12:57

Gerald Edgar

27k269153

answered Aug 22 at 12:57

Gerald Edgar

27k269153

answered Aug 22 at 12:57

Gerald Edgar

27k269153

27k269153

add a comment | 

add a comment | 

up vote
6
down vote

I think, the reason is that if the ground space has infinite measure, you can not define the measurable sets as those for which inner measure equals the outer measure: it may happen that both are infinite, while the set is still not measurable.

Note also (this may be related) that outer and inner regarity behave differently in general. For example, the sigma-finite Borel measure on the Polish space is inner regular, but not always outer regular (example: counting measure of rational numbers as a measure on $mathbbR$.)

share|cite|improve this answer

answered Aug 22 at 7:17

Fedor Petrov

44.6k5107211

add a comment | 

up vote
6
down vote

I think, the reason is that if the ground space has infinite measure, you can not define the measurable sets as those for which inner measure equals the outer measure: it may happen that both are infinite, while the set is still not measurable.

Note also (this may be related) that outer and inner regarity behave differently in general. For example, the sigma-finite Borel measure on the Polish space is inner regular, but not always outer regular (example: counting measure of rational numbers as a measure on $mathbbR$.)

share|cite|improve this answer

answered Aug 22 at 7:17

Fedor Petrov

44.6k5107211

add a comment | 

up vote
6
down vote

up vote
6
down vote

I think, the reason is that if the ground space has infinite measure, you can not define the measurable sets as those for which inner measure equals the outer measure: it may happen that both are infinite, while the set is still not measurable.

Note also (this may be related) that outer and inner regarity behave differently in general. For example, the sigma-finite Borel measure on the Polish space is inner regular, but not always outer regular (example: counting measure of rational numbers as a measure on $mathbbR$.)

share|cite|improve this answer

answered Aug 22 at 7:17

Fedor Petrov

44.6k5107211

I think, the reason is that if the ground space has infinite measure, you can not define the measurable sets as those for which inner measure equals the outer measure: it may happen that both are infinite, while the set is still not measurable.

Note also (this may be related) that outer and inner regarity behave differently in general. For example, the sigma-finite Borel measure on the Polish space is inner regular, but not always outer regular (example: counting measure of rational numbers as a measure on $mathbbR$.)

share|cite|improve this answer

answered Aug 22 at 7:17

Fedor Petrov

44.6k5107211

share|cite|improve this answer

share|cite|improve this answer

answered Aug 22 at 7:17

Fedor Petrov

44.6k5107211

answered Aug 22 at 7:17

Fedor Petrov

44.6k5107211

answered Aug 22 at 7:17

Fedor Petrov

44.6k5107211

44.6k5107211

add a comment | 

add a comment | 

up vote
1
down vote

Concerning Tao comment that the symmetry is broken by declaring 0⋅∞ = ∞⋅0 = 0, I would like to add that this is the reason why Lebesgue integral does not satisfy Newton-Leibniz formula. Namely, for Cantor-Lebesgue function f, f(1)–f(0) = 1 but ∫01f’ = 0 because f’ = ∞ on Cantor set C which has measure 0 (and f’ = 0 on its complement). But if we realize that the measure of C is 0 = (1, 2/3, 4/9, ... ) and f’ = ∞ = (1, 3/2, 9/4, ...) then we see that this particular 0⋅∞ is not 0 but exactly 1, as it should be by Newton-Leibniz formula. We have the similar problem with countable additivity of limiting frequencies, which is usually contradicted by an infinite lottery with tokens 1,2,3,4,…. This contradiction also depends on ∞⋅0 = 0 and disappears if we really calculate the relevant ∞⋅0 ( https://www.fsb.unizg.hr/matematika/download/ZS/clanci/ZS-a_note_on_probability_frequency.pdf )

share|cite|improve this answer

answered Aug 28 at 18:05

Zvonimir Sikic

111

add a comment | 

up vote
1
down vote

Concerning Tao comment that the symmetry is broken by declaring 0⋅∞ = ∞⋅0 = 0, I would like to add that this is the reason why Lebesgue integral does not satisfy Newton-Leibniz formula. Namely, for Cantor-Lebesgue function f, f(1)–f(0) = 1 but ∫01f’ = 0 because f’ = ∞ on Cantor set C which has measure 0 (and f’ = 0 on its complement). But if we realize that the measure of C is 0 = (1, 2/3, 4/9, ... ) and f’ = ∞ = (1, 3/2, 9/4, ...) then we see that this particular 0⋅∞ is not 0 but exactly 1, as it should be by Newton-Leibniz formula. We have the similar problem with countable additivity of limiting frequencies, which is usually contradicted by an infinite lottery with tokens 1,2,3,4,…. This contradiction also depends on ∞⋅0 = 0 and disappears if we really calculate the relevant ∞⋅0 ( https://www.fsb.unizg.hr/matematika/download/ZS/clanci/ZS-a_note_on_probability_frequency.pdf )

share|cite|improve this answer

answered Aug 28 at 18:05

Zvonimir Sikic

111

add a comment | 

up vote
1
down vote

up vote
1
down vote

Concerning Tao comment that the symmetry is broken by declaring 0⋅∞ = ∞⋅0 = 0, I would like to add that this is the reason why Lebesgue integral does not satisfy Newton-Leibniz formula. Namely, for Cantor-Lebesgue function f, f(1)–f(0) = 1 but ∫01f’ = 0 because f’ = ∞ on Cantor set C which has measure 0 (and f’ = 0 on its complement). But if we realize that the measure of C is 0 = (1, 2/3, 4/9, ... ) and f’ = ∞ = (1, 3/2, 9/4, ...) then we see that this particular 0⋅∞ is not 0 but exactly 1, as it should be by Newton-Leibniz formula. We have the similar problem with countable additivity of limiting frequencies, which is usually contradicted by an infinite lottery with tokens 1,2,3,4,…. This contradiction also depends on ∞⋅0 = 0 and disappears if we really calculate the relevant ∞⋅0 ( https://www.fsb.unizg.hr/matematika/download/ZS/clanci/ZS-a_note_on_probability_frequency.pdf )

share|cite|improve this answer

answered Aug 28 at 18:05

Zvonimir Sikic

111

Concerning Tao comment that the symmetry is broken by declaring 0⋅∞ = ∞⋅0 = 0, I would like to add that this is the reason why Lebesgue integral does not satisfy Newton-Leibniz formula. Namely, for Cantor-Lebesgue function f, f(1)–f(0) = 1 but ∫01f’ = 0 because f’ = ∞ on Cantor set C which has measure 0 (and f’ = 0 on its complement). But if we realize that the measure of C is 0 = (1, 2/3, 4/9, ... ) and f’ = ∞ = (1, 3/2, 9/4, ...) then we see that this particular 0⋅∞ is not 0 but exactly 1, as it should be by Newton-Leibniz formula. We have the similar problem with countable additivity of limiting frequencies, which is usually contradicted by an infinite lottery with tokens 1,2,3,4,…. This contradiction also depends on ∞⋅0 = 0 and disappears if we really calculate the relevant ∞⋅0 ( https://www.fsb.unizg.hr/matematika/download/ZS/clanci/ZS-a_note_on_probability_frequency.pdf )

share|cite|improve this answer

answered Aug 28 at 18:05

Zvonimir Sikic

111

share|cite|improve this answer

share|cite|improve this answer

answered Aug 28 at 18:05

Zvonimir Sikic

111

answered Aug 28 at 18:05

Zvonimir Sikic

111

answered Aug 28 at 18:05

Zvonimir Sikic

111

111

add a comment | 

add a comment | 

up vote
0
down vote

My two cents. Outer measures are sub-additive on countable coverings: $$Asubset cup _jinmathbbN A_jquad Rightarrow quad mu( A)le sum_jinmathbbNmu(A_j)$$
which is somehow a nicer and more practical property than the analogous dual property of super-additivity for inner measures (even in the case of finite measures). It gives a bound on the set $A$ in terms of the supposedly simpler sets $A_j$. Also, we like sub-additivity more than super-additivity, because it recalls norms.

share|cite|improve this answer

answered Aug 28 at 19:50

Pietro Majer

37.9k277180

add a comment | 

up vote
0
down vote

My two cents. Outer measures are sub-additive on countable coverings: $$Asubset cup _jinmathbbN A_jquad Rightarrow quad mu( A)le sum_jinmathbbNmu(A_j)$$
which is somehow a nicer and more practical property than the analogous dual property of super-additivity for inner measures (even in the case of finite measures). It gives a bound on the set $A$ in terms of the supposedly simpler sets $A_j$. Also, we like sub-additivity more than super-additivity, because it recalls norms.

share|cite|improve this answer

answered Aug 28 at 19:50

Pietro Majer

37.9k277180

add a comment | 

up vote
0
down vote

up vote
0
down vote

My two cents. Outer measures are sub-additive on countable coverings: $$Asubset cup _jinmathbbN A_jquad Rightarrow quad mu( A)le sum_jinmathbbNmu(A_j)$$
which is somehow a nicer and more practical property than the analogous dual property of super-additivity for inner measures (even in the case of finite measures). It gives a bound on the set $A$ in terms of the supposedly simpler sets $A_j$. Also, we like sub-additivity more than super-additivity, because it recalls norms.

share|cite|improve this answer

answered Aug 28 at 19:50

Pietro Majer

37.9k277180

My two cents. Outer measures are sub-additive on countable coverings: $$Asubset cup _jinmathbbN A_jquad Rightarrow quad mu( A)le sum_jinmathbbNmu(A_j)$$
which is somehow a nicer and more practical property than the analogous dual property of super-additivity for inner measures (even in the case of finite measures). It gives a bound on the set $A$ in terms of the supposedly simpler sets $A_j$. Also, we like sub-additivity more than super-additivity, because it recalls norms.

share|cite|improve this answer

answered Aug 28 at 19:50

Pietro Majer

37.9k277180

share|cite|improve this answer

share|cite|improve this answer

answered Aug 28 at 19:50

Pietro Majer

37.9k277180

answered Aug 28 at 19:50

Pietro Majer

37.9k277180

answered Aug 28 at 19:50

Pietro Majer

37.9k277180

37.9k277180

add a comment | 

add a comment | 

 

draft saved

draft discarded

 

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

StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmathoverflow.net%2fquestions%2f308856%2fwhy-is-lebesgue-measure-theory-asymmetric%23new-answer', 'question_page');

);

Post as a guest

Name Email

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

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

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

Name Email

Name Email

Name

Email

Name Email

Name

Email