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Nowadays, the usual way to extend a measure on an algebra of sets to a measure on a $\sigma$-algebra, the Caratheodory approach, is by using the outer measure $m^* $ and then taking the family of all sets $A$ satisfying $m^* (S)=m^* (S\cap A)+m^* (S\cap A^c)$ for every set $S$ to be the family of measurable sets. It can then be shown that this family forms a $\sigma$-algebra and $m^*$ restricted to this family is a complete measure. The approach is elegant, short, uses only elementary methods and is quite powerful. It is also, almost universally, seen as completely unintuitive (just google "Caratheodory unintuitve" ).

Given that the problem of extending measures is fundamental to all of measure theory, I would like to know if anyone can provide a perspective that renders the Caratheodory approach natural and intuitive.

I'm familiar with the fact that there is a topological approach to the extension problem (see here or link text) for the $\sigma$-finite case due to M.H. Stone (Maharam has actually shown how to extend it to the general case), but it doesn't give much of an insight into why the Caratheodoy approach works and that is what I`m interested in here.

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    $\begingroup$ That particular definition is the whole reason why Lebesgue is given so much credit. If I remember correctly, it was in his Ph D thesis that he gave that definition. The work of Caratheory, I think, is in showing that it can be done in a more general setting. $\endgroup$ – Anweshi Jul 31 '10 at 13:12
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    $\begingroup$ The Caratheodory condition is something about "all sets" decomposing properly. Caratheodory explains in his notes: a known characterization of Lebesgue measurability involved saying "all intervals" decompose properly. So he generalized to settings other than the real line by using "all sets". These notes are found in Caratheodory's collected works. $\endgroup$ – Gerald Edgar Jul 31 '10 at 19:23
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    $\begingroup$ Fun fact: when I googled "Caratheodory unintuitve" now in 2014, the first result was this post. $\endgroup$ – Minethlos Jan 8 '15 at 10:44
  • $\begingroup$ A spin-off of this very question: jmanton.wordpress.com/2017/08/24/… $\endgroup$ – D.R. Oct 7 at 18:10
  • $\begingroup$ maybe? math.stackexchange.com/questions/3385011/… $\endgroup$ – D.R. Oct 21 at 1:37
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Here is an argument that may give some intuition:

Assume that $m^{*}$ is an outer measure on $X$, and let us assume furthermore that this outer measure is finite:

$m^* (X) < \infty$

Define an "inner measure" $m_*$ on $X$ by

$m_* (E) = m^* (X) - m^* (E^c) $

If $m^*$ was, say, induced from a countably additive measure defined on some algebra of sets in $X$ (like Lebesgue measure is built using the algebra of finite disjoint unions of intervals of the form $(a,b]$), then a subset of $X$ will be measurable in the sense of Caratheodory if and only if its outer measure and inner measure agree.

From this viewpoint, the construction of the measure (as well as the $\sigma$-algebra of measurable sets) is just a generalization of the natural construction of the Riemann integral on $\mathbb{R}^n$ - you try to approximate the area of a bounded set $E$ from the outside by using finitely many rectangles, and similarly from the inside, and the set is "measurable in the sense of Riemann" (or "Jordan measurable") if the best outer approximation of its area agrees with the best inner approximation of its area.

The point here (which often isn't emphasized when Riemann integration is taught for the first time) is that the concept of "inner area" is redundant and can be defined in terms of the outer area just as I did above (you take some rectangle containing the set and consider the outer measure of the complement of the set with respect to this rectangle).

Of course, Caratheodory's construction doesn't require $m^*$ to be finite, but I still think that this gives some decent intuition for the general case (unless you think that the construction of the Riemann integral itself is not intuitive :) ).

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    $\begingroup$ Is it still sufficient to require that inner and outer measure if the countably additive measure wasn't defined on an algebra? $\endgroup$ – C-Star-W-Star Sep 1 '14 at 20:00
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    $\begingroup$ For the case $m^*(X) =\infty$, it is useful to note, that for every $A\subseteq X$ the statement "for every set $S$ we have $m^*(S)= m^*(A\cap S)+m^*(A^C\cap S)$" is equivalent to "for every set $S$ with $m^*(S)<\infty$ we have $m^*(A\cap S) = m^*(S) - m^*(A^C\cap S)$." The last expression is basically a "local" form of an inner measure. Requiring a finite outer measure for $S$ is no restriction since for sets of infinite outer measure the (first) equality always holds due to the sub-additivity of $m^*$. $\endgroup$ – Zardo Sep 13 '15 at 23:47
  • $\begingroup$ How do you prove that claim, that is, that if $m^*$ was indeed from a countable additive measure then a subset of $X$ will be measurable in the sense of Caratheodory iff its outer and inner measures agree? $\endgroup$ – Akiva Weinberger Jan 14 '18 at 22:41
  • $\begingroup$ @FreeziiS A counterexample can be found otherwise on $(X,m^*)$ where $X$ has more than two elements and $m^*$ is $0$ on the empty set, $2$ on $X$, and $1$ on any other set. The inner and outer measures agree everywhere there, but no set other than $\emptyset$ and $X$ is measurable in the sense of Carathéodory (and $m^*$ fails to be a measure). $\endgroup$ – Akiva Weinberger Jan 14 '18 at 22:43
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When this definition came up in the (one and only) measure theory course I took as a student, the instructor (Peter Constantin) had this to say about it:

"It says that a set is measurable if you can make change with it."

This explanation has stuck in my mind for the last 15 years, but it is possible that I have remembered it in part because I was never really sure I understood what he meant. Anyway, it sounds good, and if I ever teach a measure theory course (I shudder to imagine the apocalyptic scenario that would necessitate my being called upon to do this: will there be any other mathematicians at all? what color will the sky be?) I might pass it along to my students.

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    $\begingroup$ This answer is hysterical. $\endgroup$ – Steven Gubkin Mar 28 '17 at 12:28
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There is an interesting exposition about the extension measure problem provided by Jun Tanaka and Peter F. McLoughlin in A Realization of Measurable Sets as Limit Points .
Abstract: Starting with a sigma finite measure on an algebra, we define a pseudometric and show how measurable sets from the Caratheodory Extension Theorem can be thought of as limit points of Cauchy sequences in the algebra.

The paper can be downloaded from arxiv at http://arxiv.org/abs/0712.2270

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Although I agree, let's recall that "being intuitive" is quite a relative matter. In the present case, I find that the Carathéodory's settlement is optimal: maximal effect, minimal effort; maximal generality, minimal structure. For whom approaches the subject, and finds it not enough intuitive, I would just say (in the spirit of the celebrated motto by D'Alembert "go ahead and the faith will follow"): It's a good opportunity to train your intuition and, also, to learn some elementary techniques. Measure theory, in its elementary part, is mostly a matter of "$ \epsilon\, 2^{-n} $" (if you know what I mean). Finally, to go even more into the details of the construction, I would recommend to prove that the definition of measurability à la Carathéodory actually comes out characterizing the larger $\sigma$-algebra where an outer measure restricts to a measure. This makes it less out-of-the-hat, if not immediately intuitive.

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  • $\begingroup$ You wrote: "the definition of measurability à la Carathéodory actually comes out characterizing the larger σ-algebra where an outer measure restricts to a measure." By "larger" do you mean "maximal" or "largest"? And do you mean among $\sigma$-algebras that contain the algebra which is used to define the outer measure? I think I can prove your assertion if outer measure is sigma-finite and if the answers to my two questions above are "maximal" and "yes, $\sigma$-algebras that contain the original algebra" But I'm not sure how to do it otherwise. Could you please explain? $\endgroup$ – MichaelGaudreau Dec 29 '18 at 3:59
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I found this link on Terence Tao's blong on the same.

Maharam, Dorothy From finite to countable additivity. Port. Math. 44, 265-282 (1987).

http://purl.pt/3098

The original blog:

http://terrytao.wordpress.com/2009/01/03/254a-notes-0a-an-alternate-approach-to-the-caratheodory-extension-theorem/

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I did this many time ago (as a student), I did it alone (teacher dont care a lot of about), I hope all work well:

0) DEFINITION.\ Let $X\in Set$ and $\mathcal{P}(X)$ the set of parts of $X$. Let $o_fMis(X)$ the class of finite-outer-measure on $X$ i.e. the maps $\mu: \mathcal{P}(X)\to [0, \infty] $ with: $\mu(\emptyset)=0,\ \mu(A) \leq \mu(B) \ for \ A \subset B$ , $\mu(\cup_I A_i) \leq \sum_{i\in I } \mu(A_i)\ for\ I\ finite $, if in the last property $I$ is countable then $\mu$ is said a outer-measure, and these make the subclass $oMis(X)\subset o_fMis(X)$.

For $\mu \in o_fMis(X)$ define $d_\mu: \mathcal{P}(X)\times \mathcal{P}(X) \to [0, \infty]$ as $d_\mu(A, B):=\mu(A\Delta B)$ (where $A\Delta B:= (A\setminus B)\cup(B \setminus A) $) and $\rho_\mu:\mathcal{P}(X)\times \mathcal{P} (X)\to [0,1]$ as $\rho_\mu(A,B):= d_\mu(A,B)/(1+ d_\mu(A,B))$ (let $\infty/\infty:=1$) this is a is a pseudo-metric and from $d(A\Delta S,A\Delta T)=d(S, T)$ this pseudo-metric is additive. Further, from

$(A_1\setminus A_2) \Delta (B_1\setminus B_2)= [(A_1\setminus B_1)\cap(B_2\setminus A_2)]\cup[(B_1\setminus A_1)\cap(A_2\setminus B_2)]\subset$

$\subset [(A_1\setminus B_1)\cup(B_1\setminus A_1)]\cap[(A_2\setminus B_2)\cup(B_2\setminus A_2)]=(A_1\Delta B_1)\setminus (A_2\Delta B_2)$

follow that the map $(A,B) \mapsto A\setminus B$ is uniformly continuous, then there are also the maps: $(A,B) \mapsto A\cap B=A\setminus (A\setminus B)$, $(A,B) \mapsto A\cup B=X\setminus (X\setminus A \cap X\setminus B)$, $(A,B) \mapsto A\Delta B=A\cup B\setminus (A\cap B)$ ; then the (boolean) ring $(\mathcal{P}(X), \Delta,\cup, 0,1)$ is a uniformly ring. In the the subspace $[\mu< \infty]:=\{A\subset X| \mu(A)<\infty\}$ we have the pseudo-metric $d_µ$ equivalent to the restriction of $\rho_\mu$.

1) Let $\mu\in oMis(X)$ and fixed the pseudo-metric $\rho_\mu$.

We prove that a $Cauchy$-sequence $ (C_n)_n$ converging to the inferior limit: $inf.lim_nC_n:=\cup_n(\cap_{h\geq n} C_h)$ and to the superior limit $ sup.lim_n\ C_n:= \cap_n(\cup_{h\geq n} C_h)$: \ For $\epsilon >0$ let $ N(\epsilon )>0$ such that $ d(C_n,C_m)<\epsilon /2\ for\ n,m\geq N(\epsilon )$ ; let $F_n:= C_{ N(1/2^n)}$ and put $E:= \cup_n(\cap_{h\geq n} F_h)$. We have that:

$(\cap_{k\geq m} F_k) \Delta F_m= F_m\setminus(\cap_{k\geq m} F_k)= (\cup_{ k>m } F_m\setminus F_k)\subset$

$\subset (\cup_{ k>m } (F_m\setminus F_{ m-1} \cup\ldots\cup F_{k+1} \setminus F_k)$.

And

$E\Delta(\cap_{k\geq m} F_k)= E\setminus(\cap_{k\geq m}F_k) =$

$\bigcup_n[(\cap_{h\geq n} F_h)\setminus( \cap_{ k\geq m }F_k)]=$

$= \bigcup_ n[(\cap_{h\geq n} F_h)\setminus (\cap_{n>k\geq m} F_k) \bigcap (\cap_{h\geq n} F_h)]=$

$=\bigcup_{ n>m } [(\cap_{h>n} F_h)\setminus( \cap_{n>k\geq m} F_k)] =$

$= \bigcup_{ n>m } [(\cap_{h>n} F_h)\setminus F_m \cup (\cap_{h>n} F_h)\setminus F_{ m+1})\ldots \cup (\cap_{h>n} F_h)\setminus F_n)]\subset$

$\subset \bigcup_{ n>m } (F_{ m+1} \setminus F_m) \cup (F_{ m+2} \setminus F_{ m+1}) \ldots \cup (F_{ n+1} \setminus F_n)$.

Then

$E\Delta F_m= (E\Delta (\cap_{k\geq m} F_k)) \Delta ((\cap_{k\geq m} F_k) \Delta F_m)\subset$

$\subset (E\Delta (\cap_{k\geq m} F_k)) \bigcup (( \cap_{k\geq m} F_k) \Delta F_m)\subset$

$\subset (F_m\Delta F_{ m+1}) \bigcup ( F_{ m+1} \Delta F_{ m+2})\cup\ldots$

By countable subadditivity follow that the sequence $(F_n)_n$ (and then the sequence $(C_n)_n)$) converging to $E=inf.lim_n C_n$. Applying this to the sequence $(\widetilde{C_n})_n)$ (where put $\widetilde{C}:=X\setminus A$) from $A\Delta B= \widetilde{A}\Delta \widetilde{B})$ follow that this sequence converging to $\cup_n(\cap_{h\geq n} \widetilde{C_h})$ then applying the (uniform) map $C \mapsto \widetilde{C}$ follow that the sequence $(C_n)_n)$ converging to the superior limit $sup.lim_n\ C_n:= \cap_n(\cup_{h\geq n} C_h)$ too, and this sequence is equivalent (i.e. has the some limit) the increasing sequence $(\cap_{n\leq k} C_k)_n$, and if $(C_n)_n)$ is a increasing $Chauchy$ sequence then it converging to its union $C=\cup_n C_n$ .

2) If $\mu \in o_fMis(X)$ for subadditivity we have $|\mu(A)-\mu(B)|\leq \mu(A\Delta B)$ for $ A,B\in [\mu < \infty]$ then $\mu: [\mu < \infty]\to [0, \infty[$ is uniformly continuous. If $\mu \in oMis(X)$ then from $(1)$ follow that $\mu: \mathcal{P}(X)\to [0,\infty]$ is continuous: it's enough show that for $\mu(C)=\infty$ and $C_n$ increasing sequence with union $C$ then $sup_n (C_n)=\infty$: we have that $lim_{n\to\infty}\mu(C\setminus C_n)=0$ then follow from $\mu(C) \leq \mu(C\setminus C_n)+\mu(C_n)$.

3) Let $\mu\in oMis(X)$ and $\mathcal{ R }$ a ring of subset of $X$ with $\mu: \mathcal{ R } \to[0,\infty]$ additive. Let $\mathcal{R}(\mu)=${$ A\subset X | \forall \epsilon \geq 0 \exists R\in \mathcal{ R } : \mu (A\Delta R)\leq \epsilon $} the topological closure of $\mathcal{R}$, it is also the Cauchy completion, then is still a (boolean topological) ring ; and from last observation on $(1)$ the ring $\mathcal{R}(\mu)$ is a $\sigma $-Ring and the continuous extension $\mu:\mathcal{R}(\mu)\to[0, \infty]$ is still additive, and in particular continuous for the increasing sequences, then it is $\sigma $-additive.

EDIT:

Let $ \mathcal{R}\subset \mathcal{P}(X)$ a ring and $\mu: \mathcal{R} \to [0, \infty[$ a measure (then $\sigma$-addittive ), and suppose that $X$ is a countable unions of elements $\mathcal{R}$. Define the Lebesgue extension $µ_L\in oMis(X)$: $ \mu_L (A):=inf_{\ A\subset \cup_n R_n} \sum_n \mu(R_n)$ where the 'Inf' is on the countable families $(R_n)_n$ such that $A\subset \cup_n R_n$. Let $Mis(\mu_L)$ the class of the Caratheodory $\mu_L$-measurable sets, this is a $\sigma$-ring that containing $\mathcal{R}$ and $\mu_L$ is strong-regular: for any $A\subset X$ there exist a measurable set $E_A\in Mis(\mu_L)$ such that $A\subset E_A$ and $\mu_L(A)=\mu_L(E_A)$.

d) Let $\mu^\star\in ofMis(X)$, let $\mu^\star$ additive and finite on a subsets algebra $\mathcal{R}\subset \mathcal{P}(X)$. For $A\subset X$ considering the following property:

i) $\mu^\star (X)= \mu^\star (A) + \mu^\star (X\setminus A)$

ii) $A\in Mis(\mu^\star)$

iii) $A\in \mathcal{R}(\mu^\star)$.

Then $(ii)\Rightarrow (i)$ and $(iii)\Rightarrow (i)$ ; and $(i)\Leftrightarrow (ii)$ if $\mu$ is strong-regular and $\mu^\star (A)<\infty$ . These are all equivalent if $\mu^\star= µ_L$ (where $\mu $ is $\sigma $-addittive and $X$ $\sigma $-finite).\ DIM. $(ii)\Rightarrow(i)$: Obvious. $(iii)\Rightarrow(i):$ If $\mu\star(X)=\infty$ follow by subaddittivity, otherwise for $\epsilon >0$ let $ R\in \mathcal{R}$ with $\mu^\star(A\Delta R)=\mu^\star((X\setminus A)\Delta(X\setminus R))<\epsilon $ ; we have that $\mu^\star(A)\leq \mu^\star(R)+\epsilon$ (from $ A\subset A\Delta R \cup R $)

and $ \mu^\star(X\setminus A)\leq \mu^\star(X\setminus R)+\epsilon $

and follow that

$ \mu^(X) - \mu^\star(X\setminus A) \geq \mu^\star(X) - \mu^\star(X\setminus R)-\epsilon \geq \mu^\star(A) -2\epsilon $

the last follow from $\mu^\star(X)+\epsilon \geq \mu^\star(X)-\mu^\star(R)+ \mu^\star(A)=\mu^\star(X\setminus R)+ \mu^\star(A)$.

$(i)\Rightarrow (ii):$ We have $\mu^\star(X)= \mu^\star (A)+ \mu^\star (X\setminus A)$ and let $A \subset M\in Mis(\mu^\star)$ with $\mu^\star (A)= \mu^\star (M)$, from $\mu^\star (A) = \mu^\star (A\cap M) + \mu^\star (M\setminus A)= \mu^\star (A)+ \mu^\star (M\setminus A)$ follow $\mu^\star (M\setminus A)=0$ for $E\subset X$ we have $\mu^\star (E\cap M) \leq \mu^\star (E\cap A)+ \mu^\star (E\cap (M\setminus A)) = \mu^\star (E\cap A)$ then $\mu^\star (E \cap M)= \mu^\star (E \cap A)$ and from $E\setminus M \subset E\setminus A=(E\setminus M) \cup (E \cap (M\setminus A))$ follow $\mu^\star (E\setminus M) = \mu^\star (E\setminus A)$ then $\mu^\star (E)= \mu^\star (E \cap M)+ \mu^\star (E\setminus M)= \mu^\star (E \cap A)+ \mu^\star (E\setminus A)$.

$(ii)\Rightarrow(iii):$ For $\epsilon >0$ let $A\subset \cup_n B_n$ with $B_n\in \mathcal{R}$ with $\sum_n \mu(B_n)< \mu^\star(A)+\epsilon /2$ and let $N>0$ a integer such that $\sum_{n>N} \mu (B_n)<+\epsilon /2$, let $F:=\cup_{1\leq k\leq n } B_n$, then $A\setminus F \subset \cup_{ n>N } B_n$

and $\mu^\star(A\setminus F)<\epsilon /2$, from $F\setminus A \subset \cup_n \ B_n\setminus A$

follow $\mu^\star (F\setminus A)\leq \mu^\star (\cup_n B_n\setminus A) =\ ^{\mu^\star\ is\ \sigma-addittive\ on\ measurables}$=

$= \mu^\star (\cup_ n\ B_n)-\mu^\star (A)\leq \sum_{1\leq i\leq n } \ \mu(B_i)\ -\ \mu^\star(A)< \epsilon /2$ then $\mu^\star (A\Delta F) <\epsilon $.

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    $\begingroup$ Thanks for the effort but it is impossible to follow the argument mainly because of the latex problems and the notation. Could you try cleaning it a little bit? $\endgroup$ – Nishant Chandgotia Apr 24 '11 at 7:18
  • $\begingroup$ Two comments: first, this looks interesting, but it's close to unreadable. Without devoting half an hour to reading it carefully, I would have no idea what you are proving. Second: without spending the half-hour, I think you are verifying that some method (perhaps the Caratheodory one?) extends the measure to a complete $\sigma$-additive one. I don't think that's what the question was asking at all: validity is not in doubt, only the motivation behind the manipulations. $\endgroup$ – Ryan Reich Apr 24 '11 at 14:48
  • $\begingroup$ Hi, excuse me for my poor latex and formatting and English. I presents an approach (equivalent to a classical Caratheodory) extension of a measure defined on a ring to a $\sigma$-ring, based on the simple completion of a metric space built on subsets. It 's a classic argument, but I always found it only as exercises, and I decided to tackle the various exercises and unify the various topic, then set out here, if you want I can send the original latex or PDF (I think where it is more readable) (sergiobuschi@yahooo.com) $\endgroup$ – Buschi Sergio Apr 24 '11 at 21:46
  • $\begingroup$ Then I showed a (more intuitive, or more natural) approach to the extension of measure, but equivalent to the Carathodory exposure (for finite measures) – Buschi Sergio 0 secs ago $\endgroup$ – Buschi Sergio Apr 24 '11 at 21:53
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The notion of exterior measure seems quite natural to me after having seen the definition of the Lebesgue exterior measure as the infimum of the measures associated to countable covers of the sets by intervals or rectangles.

Let us first assume that we are on a space $X$ of finite measure, say the unit cube. Let us try to single out a class of sets for which $\sigma$-additivity may holds. Let $A$ be a set in such a class. If we want to measure $A$, we also want to measure its complement in $X$ and the two sets are disjoints so that at least the following equality should be satisfied.

$$\mu^*(A)+\mu^*(X\backslash A) = \mu^*(X).$$

This property can be shown to be equivalent to Caratheodory definition of measurability if $X$ is of finite measure and the measure is regular (for all subsets $A$ of $X$, there exists a measurable set $B$ containing $A$ with the same outer measure, a fact that can be guaranted by replacing $\mu^*(A)$ by $inf\{\mu^*(B) \mid A \subset B, \ B \ \mu^*-\hbox{measurable}\}$), and it looks quite natural to me.

If $X$ is not of finite measure, this definition is not very restrictive because $everything + \infty = \infty$. In particular, it is satisfied for all bounded sets in ${\bf R}^d$ wrt to Lebesgue exterior measure. So we need to be slightly more restrictive and asks that the previous property holds in restriction, say, to balls $B_N$ of radius $N$, $N>0$ if they are of finite measure.

$$ \mu^*(B_N) = \mu^*(B_N \cap A) + \mu^*(B_N \cap A^c), \quad \forall N.$$

And here again, this can be shown to be equivalent to Caratheodory's definition.

If we are not on ${\bf R}^d$ but on some general abstract set $X$ with infinite measure, the only choice here is to restrict to all sets $E$ of finite measure.

$$ \mu^*(E) = \mu^*(E \cap A) + \mu^*(E \cap A^c), \quad \forall E \subset X \hbox{ such that } \mu^*(E)<\infty.$$

Note that this definition is obviously satisfied if $E$ is of infinite measure and we get the general definition of Caratheodory.

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  • $\begingroup$ You say that, if $X$ is a space of finite measure, then $\mu^*(A)+\mu^*(X\setminus A)=\mu^*(X)$ iff $A$ satisfies Carathéodory's condition. This is not true: see here. $\endgroup$ – Akiva Weinberger Jan 14 '18 at 22:45
  • $\begingroup$ @Weinberger I am assuming that the measure is regular: for all subset $A$ of $X$, there exists a measurable set $B$ containing $A$ with the same outer measure. Given an exterior measure $m^*$, we can always build a regular measure coinciding with $m^*$ on $m^*$-measurable sets, so it is not a big restriction. Let me edit my post. $\endgroup$ – coudy Jan 15 '18 at 9:02
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I find Caratheodory's definition to be intuitive: for one thing, it satisfies what we would think of as the measure of the intersection of 2 sets. Intuitively we should have, for $A$ a measurable set, $m(A \bigcap B)= m(B)-m(B\backslash A)$ whether the intersection is empty or not: if it empty then it is 0 and if not then it is that little remaining piece.

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    $\begingroup$ But in Caratheodory's criterion, $B$ doesn't have to be measurable. $\endgroup$ – Willie Wong Apr 23 '11 at 12:10
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    $\begingroup$ I agree with Willie Wong what one wants is a restriction of the outer measure to become a additive but this is not required for all sets! $\endgroup$ – C-Star-W-Star Sep 1 '14 at 18:59

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