Demystifying the Caratheodory Approach to Measurability - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-19T21:21:10Z http://mathoverflow.net/feeds/question/34007 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/34007/demystifying-the-caratheodory-approach-to-measurability Demystifying the Caratheodory Approach to Measurability Michael Greinecker 2010-07-31T11:58:13Z 2011-08-19T10:27:08Z <p>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" ). </p> <p>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.</p> <p>I'm familiar with the fact that there is a topological approach to the extension problem (see <a href="http://terrytao.wordpress.com/2009/01/03/254a-notes-0a-an-alternate-approach-to-the-caratheodory-extension-theorem/" rel="nofollow">here</a> or <a href="http://arxiv.org/abs/0712.2270" rel="nofollow">link text</a>) for the $\sigma$-finite case due to M.H. Stone (Maharam has actually shown how to extend it to the <a href="http://purl.pt/3098/1/" rel="nofollow">general case</a>), but it doesn't give much of an insight into why the Caratheodoy approach works and that is what I`m interested in here.</p> http://mathoverflow.net/questions/34007/demystifying-the-caratheodory-approach-to-measurability/34011#34011 Answer by Pietro Majer for Demystifying the Caratheodory Approach to Measurability Pietro Majer 2010-07-31T12:54:33Z 2010-07-31T13:00:10Z <p>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, can be summarized just as: $$\epsilon\, 2^{-n}.$$ You know what I mean: it's as simple as that... 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. </p> http://mathoverflow.net/questions/34007/demystifying-the-caratheodory-approach-to-measurability/34029#34029 Answer by Mark Schwarzmann for Demystifying the Caratheodory Approach to Measurability Mark Schwarzmann 2010-07-31T15:02:07Z 2010-07-31T15:02:07Z <p>Here is an argument that may give some intuition:</p> <p>Assume that $m^{*}$ is an outer measure on $X$, and let us assume furthermore that this outer measure is finite:</p> <p>$m^* (X) &lt; \infty$</p> <p>Define an "inner measure" $m_*$ on $X$ by</p> <p>$m_* (E) = m^* (X) - m^* (E^c)$</p> <p>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.</p> <p>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.</p> <p>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).</p> <p>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 :) ).</p> http://mathoverflow.net/questions/34007/demystifying-the-caratheodory-approach-to-measurability/34040#34040 Answer by Leandro for Demystifying the Caratheodory Approach to Measurability Leandro 2010-07-31T17:45:06Z 2010-07-31T17:45:06Z <p>There is an interesting exposition about the extension measure problem provided by Jun Tanaka and Peter F. McLoughlin in <i> A Realization of Measurable Sets as Limit Points </i>.<br> <b>Abstract:</b> 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. </p> <p>The paper can be downloaded from arxiv at <a href="http://arxiv.org/abs/0712.2270" rel="nofollow">http://arxiv.org/abs/0712.2270</a></p> http://mathoverflow.net/questions/34007/demystifying-the-caratheodory-approach-to-measurability/38512#38512 Answer by alephomega for Demystifying the Caratheodory Approach to Measurability alephomega 2010-09-12T21:00:31Z 2010-09-12T21:00:31Z <p>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.</p> http://mathoverflow.net/questions/34007/demystifying-the-caratheodory-approach-to-measurability/62709#62709 Answer by Pete L. Clark for Demystifying the Caratheodory Approach to Measurability Pete L. Clark 2011-04-23T06:15:22Z 2011-04-24T04:40:42Z <p>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:</p> <p>"It says that a set is measurable if you can make change with it."</p> <p>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.</p> http://mathoverflow.net/questions/34007/demystifying-the-caratheodory-approach-to-measurability/62735#62735 Answer by Buschi Sergio for Demystifying the Caratheodory Approach to Measurability Buschi Sergio 2011-04-23T12:11:53Z 2011-04-24T14:25:34Z <p>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:</p> <p>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 $oMis(X)\subset o_fMis(X)$. </p> <p>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</p> <p>$(A_1\setminus A_2) \Delta (B_1\setminus B_2)= [(A_1\setminus B_1)\cap(B_2\setminus A_2)]\cup[(B_1\setminus A_1)\cap(A_2\setminus B_2)]\subset$</p> <p>$\subset [(A_1\setminus B_1)\cup(B_1\setminus A_1)]\cap[(A_2\setminus B_2)\cup(B_2\setminus A_2)]=(A_1\Delta B_1)\setminus (A_2\Delta B_2)$ </p> <p>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 $[\mu&lt; \infty]:={A\subset X| \mu(A)&lt;\infty}$ we have the pseudo-metric $d_µ$ equivalent to the restriction of $\rho_\mu$.</p> <p>1) Let $\mu\in oMis(X)$ and fixed the pseudo-metric $\rho_\mu$. </p> <p>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 $d(C_n,C_m)&lt;\epsilon /2\ for\ n,m\geq N(\epsilon )$ ; let $F_n:= C_{ N(1/2^n)}$ and put $E:= \cup_n(\cap_{h\geq n} F_h)$. We have that:</p> <p>$(\cap_{k\geq m} F_k) \Delta F_m= F_m\setminus(\cap_{k\geq m} F_k)= (\cup_{ k>m } F_m\setminus F_k)\subset$</p> <p>$\subset (\cup_{ k>m } (F_m\setminus F_{ m-1} \cup\ldots\cup F_{k+1} \setminus F_k)$.</p> <p>And </p> <p>$E\Delta(\cap_{k\geq m} F_k)= E\setminus(\cap_{k\geq m}F_k) =$</p> <p>$\bigcup_n[(\cap_{h\geq n} F_h)\setminus( \cap_{ k\geq m }F_k)]=$</p> <p>$= \bigcup_ n[(\cap_{h\geq n} F_h)\setminus (\cap_{n>k\geq m} F_k) \bigcap (\cap_{h\geq n} F_h)]=$</p> <p>$=\bigcup_{ n>m } [(\cap_{h>n} F_h)\setminus( \cap_{n>k\geq m} F_k)] =$</p> <p>$= \bigcup_{ n>m } [(\cap_{h>n} F_h)\setminus F_m \cup (\cap_{h>n} F_h)\setminus F_{ m+1})\ldots \cup (\cap_{h>n} F_h)\setminus F_n)]\subset$</p> <p>$\subset \bigcup_{ n>m } (F_{ m+1} \setminus F_m) \cup (F_{ m+2} \setminus F_{ m+1}) \ldots \cup (F_{ n+1} \setminus F_n)$.</p> <p>Then</p> <p>$E\Delta F_m= (E\Delta (\cap_{k\geq m} F_k)) \Delta ((\cap_{k\geq m} F_k) \Delta F_m)\subset$ </p> <p>$\subset (E\Delta (\cap_{k\geq m} F_k)) \bigcup (( \cap_{k\geq m} F_k) \Delta F_m)\subset$</p> <p>$\subset (F_m\Delta F_{ m+1}) \bigcup ( F_{ m+1} \Delta F_{ m+2})\cup\ldots$</p> <p>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 $C=\cup_n C_n$ .</p> <p>2) If $\mu \in o_fMis(X)$ for subadditivity we have $|\mu(A)-\mu(B)|\leq \mu(A\Delta B)$ for $A,B\in [\mu &lt; \infty]$ then $\mu: [\mu &lt; \infty]\to [0, \infty[$ 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 $\mu(C) \leq \mu(C\setminus C_n)+\mu(C_n)$.</p> <p>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 addictive, and in particular continuous for the increasing sequences, then it is $\sigma$-additive.</p> <p>EDIT:</p> <p>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 $\mu_L(A)=\mu_L(E_A)$.</p> <p>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:</p> <p>i) $\mu^\star (X)= \mu^\star (A) + \mu^\star (X\setminus A)$</p> <p>ii) $A\in Mis(\mu^\star)$</p> <p>iii) $A\in \mathcal{R}(\mu^\star)$.</p> <p>Then $(ii)\Rightarrow (i)$ and $(iii)\Rightarrow (i)$ ; and $(i)\Leftrightarrow (ii)$ if $\mu$ is strong-regular and $\mu^\star (A)&lt;\infty$ . 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 $\mu^\star(A\Delta R)=\mu^\star((X\setminus A)\Delta(X\setminus R))&lt;\epsilon$ ; we have that $\mu^\star(A)\leq \mu^\star(R)+\epsilon$ (from $A\subset A\Delta R \cup R$) </p> <p>and $\mu^\star(X\setminus A)\leq \mu^\star(X\setminus R)+\epsilon$ </p> <p>and follow that </p> <p>$\mu^(X) - \mu^\star(X\setminus A) \geq \mu^\star(X) - \mu^\star(X\setminus R)-\epsilon \geq \mu^\star(A) -2\epsilon$ </p> <p>the last follow from $\mu^\star(X)+\epsilon \geq \mu^\star(X)-\mu^\star(R)+ \mu^\star(A)=\mu^\star(X\setminus R)+ \mu^\star(A)$.</p> <p>$(i)\Rightarrow (ii):$ 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 $\mu^\star (E)= \mu^\star (E \cap M)+ \mu^\star (E\setminus M)= \mu^\star (E \cap A)+ \mu^\star (E\setminus A)$.</p> <p>$(ii)\Rightarrow(iii):$ For $\epsilon >0$ let $A\subset \cup_n B_n$ with $B_n\in \mathcal{R}$ with $\sum_n \mu(B_n)&lt; \mu^\star(A)+\epsilon /2$ and let $N>0$ a integer such that $\sum_{n>N} \mu (B_n)&lt;+\epsilon /2$, let $F:=\cup_{1\leq k\leq n } B_n$, then $A\setminus F \subset \cup_{ n>N } B_n$ </p> <p>and $\mu^\star(A\setminus F)&lt;\epsilon /2$, from $F\setminus A \subset \cup_n \ B_n\setminus A$ </p> <p>follow $\mu^\star (F\setminus A)\leq \mu^\star (\cup_n B_n\setminus A) =\ ^{\mu^\star\ is\ \sigma-addittive\ on\ measurables}$=</p> <p>$= \mu^\star (\cup_ n\ B_n)-\mu^\star (A)\leq \sum_{1\leq i\leq n } \ \mu(B_i)\ -\ \mu^\star(A)&lt; \epsilon /2$ then $\mu^\star (A\Delta F) &lt;\epsilon$.</p> http://mathoverflow.net/questions/34007/demystifying-the-caratheodory-approach-to-measurability/73213#73213 Answer by Ramesh Kadambi for Demystifying the Caratheodory Approach to Measurability Ramesh Kadambi 2011-08-19T10:27:08Z 2011-08-19T10:27:08Z <p>I found this link on Terence Tao's blong on the same.</p> <p>Maharam, Dorothy From finite to countable additivity. Port. Math. 44, 265-282 (1987).</p> <p><a href="http://purl.pt/3098" rel="nofollow">http://purl.pt/3098</a></p> <p>The original blog:</p> <p><a href="http://terrytao.wordpress.com/2009/01/03/254a-notes-0a-an-alternate-approach-to-the-caratheodory-extension-theorem/" rel="nofollow">http://terrytao.wordpress.com/2009/01/03/254a-notes-0a-an-alternate-approach-to-the-caratheodory-extension-theorem/</a></p>