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This is related to my measure theory class, but it's not homework. The motivation behind this post is to understand the exact relationship between the three objects mentioned in the title.

Carathéodory's theorem states that given an outer measure $\varphi$ on a space $X$, the set of $\varphi$-measurable sets form a $\sigma$-algebra and $\varphi$ restricted to the measurable sets is a countably additive complete measure. I am wondering if this admits a converse:

Q1. Given a complete measure space $(X, \Sigma, \mu)$, does there always exist an outer measure $\varphi$ satisfying the following conditions?

  1. A set $E \subseteq X$ is $\varphi$-measurable iff $E \in \Sigma$,
  2. $\varphi$ restricted to $\Sigma$ is $\mu$.

A related question is:

Q2. Given a $\sigma$-algebra $\Sigma$ on $X$, can I always define a complete measure $\mu$ on the measurable space $(X, \Sigma)$?

To pinpoint the source of my confusion, I feel that the word measure is used in several related, but seemingly different, ways; e.g., "measure space", "measurable space" and "outer measure". If the answers to both the questions are in the affirmative, then it would become clearer to me that these different usages are very related after all.

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+1: Very nice question. –  Jonas Teuwen Sep 27 '11 at 21:26
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For Q2, one can show this is true trivially by letting $\mu$ be counting measure restricted to $\Sigma$, or, even more trivially, by setting $\mu(A) = 0$ for all $A \in \Sigma$. –  user15464 Sep 28 '11 at 2:05
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Zero measure is only complete if $\Sigma = \mathcal{P}(X)$. $\mu(A) = \infty$ for $A \neq \emptyset$ should work. –  Niels Diepeveen Sep 28 '11 at 8:24
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2 Answers 2

up vote 14 down vote accepted
+200

I'm only going to address question 1 in some generality, since the infinite measure $\mu(E) = \infty$ for $\emptyset \neq E \in \Sigma$ gives a complete measure on $\Sigma$ and thus an obvious answer to question 2 (as was noted by Niels Diepeveen).


Most of the things I'm saying below can be found or extracted from Fremlin's Measure theory, volumes 1 and 2, parts even verbatim, up to minor modifications of notation.


Let me say right away that the situation is somewhat subtle. The answer I'm going to give boils down to:

Every measure $\mu$ is induced by an outer measure $\mu^{\ast}$ canonically obtained from $\mu$, but in general the $\sigma$-algebra of $\mu^{\ast}$-measurable sets can be strictly larger than the one on which $\mu$ is defined. However, the $\mu$-measurable sets are the same as the $\mu^{\ast}$-measurable sets provided that $\mu$ is

  1. complete (obviously necessary) and
  2. $\sigma$-finite (not strictly necessary but sufficient and good enough for many purposes)

and thus $\mu$ can be obtained back from $\mu^{\ast}$ by Carathéodory's construction (or simply restriction, if you prefer).

The $\sigma$-finiteness condition can be weakened to localizability/local determination conditions, but I prefer to formulate the result for $\sigma$-finite measures, as this is a more widely known condition.


Start with a measure space $(X,\Sigma,\mu)$, complete or not.

It seems to me that in this generality, with no more information at hand, the only thing we can do in the first place is to try and associate an outer measure $\mu^{\ast}$ with that situation. The only one that comes to mind is:

For $A \subset X$ put $\mu^{\ast} (A) = \inf{\left\{\mu(E)\,:\,A \subset E, \,E\in \Sigma\right\}}.$

The following points are all very easy but we'll need them later on, so for the record:

  1. This $\mu^{\ast}$ is an outer measure:

    • $\mu^{\ast}(\emptyset) = 0$ and monotonicity $\mu^{\ast}(A) \leq \mu^{\ast}(B)$ for $A \subset B$ are clear from the definition
    • $\sigma$-subadditivity follows from the usual $2^{-n}$-trick:

      Let $A = \bigcup_{n=1}^{\infty} A_n$ and $\varepsilon \gt 0$. We may assume that $\mu^{\ast}(A_n)$ is finite for all $n$. For each $A_n$ choose $E_n \in \Sigma$ with $A_n \subset E_n$ and $\mu(E_n) \leq \mu^{\ast}(A_n) + \varepsilon \cdot 2^{-n}$ then $A \subset E = \bigcup_{n=1}^{\infty} E_n$ and $\mu(E) \leq \sum \mu(E_n) \leq \sum \mu^{\ast}(A_n) + \varepsilon$. Since $\varepsilon \gt 0$ was arbitrary, we conclude. $\mu^{\ast}(A) \leq \sum \mu^{\ast}(A_n)$.
  2. For all $E \in \Sigma$ we have $\mu(E) = \mu^{\ast}(E)$.

  3. Notice that the infimum in the definition of $\mu^{\ast}$ is actually a minimum: choose $E_n \in \Sigma$, $A \subset E_n$ such that $\mu(E_n) \leq \mu^\ast(A) + 1/n$. Then $E = \bigcap_{n=1}^{\infty} E_n \in \Sigma$, we have $A \subset E$ and $\mu^{\ast}(A) = \mu(E)$.

  4. The usual Carathéodory extension theorem applied to $\Sigma$ and $\mu$ shows that all the sets $E \in \Sigma$ are measured by $\mu^{\ast}$, so $$\Sigma \subset \Sigma_{C} = \{E \subset X\,:\,\mu^{\ast}(A) = \mu^{\ast}(A \cap E) + \mu^{\ast}(A \smallsetminus E)\text{ for all } A \subset X\}.$$ Appealing to Carathéodory's extension theorem here is clearly overkill: Let $F \in \Sigma$. By subadditivity we have for all $A \subset X$ that $\mu^{\ast}(A) \leq \mu^{\ast}(A \cap F) + \mu^{\ast}(A \smallsetminus F)$ and if $A$ has infinite outer measure then $\mu^{\ast}(A) \geq \mu^{\ast}(A \cap F) + \mu^{\ast}(A \smallsetminus F)$ holds vacuously, so assume that $\mu^{\ast}(A) \lt \infty$. According to the previous point we can choose $E \in \Sigma$, with $A \subset E$ and $\mu^{\ast}(A) = \mu(E)$. Moreover, $A \cap F \subset E \cap F \in \Sigma$ and $A \smallsetminus F \subset E \smallsetminus F \in \Sigma$ so that $\mu^{\ast}(A \cap F) + \mu^{\ast}(A \smallsetminus F) \leq \mu^{\ast}(E \cap F) + \mu^{\ast}(E\smallsetminus F) = \mu(E \cap F) + \mu(E\smallsetminus F) = \mu(E) = \mu^{\ast}(A)$.

So here's what we have so far: given a measure space $(X,\Sigma,\mu)$ we find an outer measure $\mu^{\ast}$ on $X$, and the associated space $(X,\Sigma_{C},\mu_C)$, where $\mu_{C}$ is the complete measure obtained from $\mu^{\ast}$ by Carathéodory's construction. We argued that $\Sigma \subset \Sigma_C$ and for $E \in \Sigma$ we have $\mu(E) = \mu_C(E)$, so $\mu_C$ is an extension of $\mu$.

This gives a first partial affirmative answer to Q1: there always exists an outer measure $\mu^{\ast}$ such that the restriction of $\mu^{\ast}$ to $\Sigma$ is $\mu$ itself. That is, $\mu_{C}$ satisfies requirement 2. of your Q1.


Recall that every measure space $(X,\Sigma,\mu)$ has a completion: there is a smallest $\sigma$-algebra $\check{\Sigma}$ and a complete measure $\check{\mu}$ such that $\Sigma \subset \check{\Sigma}$ and $\mu(E) = \check{\mu}(E)$ for $E \in \Sigma$.

Explicitly, $\check{\Sigma}$ consists of the sets of the form $E \cup N$ where $E \in \Sigma$ and $N$ is a subset of a $\mu$-null set in $\Sigma$. It is not hard to show that $\check{\Sigma}$ is a $\sigma$-algebra and that every complete measure $\bar{\mu}$ extending $\mu$ must extend $\check{\mu}$. More precisely, if $\bar{\mu}$ is a complete measure defined on a $\sigma$-algebra $\bar{\Sigma} \supset \Sigma$ and $\bar{\mu}(E) = \mu(E)$ for all $E \in \Sigma$ then $\bar{\Sigma} \supset \check{\Sigma}$ and $\bar{\mu}(\check{E}) = \check{\mu}(\check{E})$ for all $\check{E} \in \check{\Sigma}$.

Observe: If $(X,\Sigma,\mu)$ is already complete then $\check{\Sigma} = \Sigma$ and $\check{\mu} = \mu$. If $\mu$ isn't complete, then clearly $\check{\Sigma} \supsetneqq \Sigma$ and $\check{\mu} \neq \mu$ as you already observed by assuming $(X,\Sigma,\mu)$ to be complete.

Finally, it is not very hard to check that $(\check{\mu})^{\ast} = \mu^{\ast}$ so that passing to the completion doesn't change the associated outer measure.


Summarizing what we know so far, we have three measure spaces:

  1. the original measure space $(X,\Sigma,\mu)$ we started with,

  2. its completion $(X,\check{\Sigma},\check{\mu})$,

  3. its extension $(X,\Sigma_{C},\mu_{C})$ obtained via Carathéodory's construction from the outer measure $\mu^{\ast}$.

We know that $\Sigma \subset \check{\Sigma} \subset \Sigma_{C}$ and that $\check{\mu}$ is an extension of $\mu$ and that $\mu_{C}$ is an extension of $\check{\mu}$.

Your question becomes, in my interpretation above:

Are $\check{\Sigma} = \Sigma_C$ and $\check{\mu} = \mu_{C}$?

This turns out to be wrong in general.


Here's an artificial, very degenerate, but easily tractable example (I learned this from Fremlin's treatise Measure theory, volume 2, exercise 213Ya, page 31):

Let $X$ be a countable set and consider the outer measure $\varphi(A) = \sqrt{\# A}$.

The $\sigma$-algebra of $\varphi$-measurable sets is $\Sigma = \{\emptyset,X\}$: If $\emptyset \neq E \neq X$ choose $A = \{e, x\}$ with $e \in E$ and $x \in X \smallsetminus E$. Then $2 = \varphi(A \cap E) + \varphi(A \smallsetminus E) \gt \varphi(A) = \sqrt{2}$ so that $E$ isn't $\varphi$-measurable. The thing “going wrong here” is of course that $\varphi$ is strictly subadditive on sets of finite non-zero measure.

Clearly $\mu(\emptyset) = 0$ and $\mu(X) = \infty$ is the measure obtained from $\varphi$ by the Carathéodory construction.

Now $\mu^{\ast}$ is the infinite (outer) measure $\mu^{\ast}(\emptyset) = 0$ and $\mu^{\ast}(A) = \infty$ if $A \neq \emptyset$. The $\sigma$-algebra of $\mu^{\ast}$-measurable sets is $\mathcal{P}(X)$ and $\mu_{C} = \mu^{\ast}$.

In particular $\Sigma \subsetneqq \Sigma_{C}$ and $\mu \neq \mu_{C}$.

The take-away is of course:

If $\mu$ is induced from some outer measure $\varphi$ then it need not be identical to the measure $\mu_{C}$ given by its associated outer measure $\mu^{\ast}$.


The counterexample above is admittedly very artificial and not something we really care about.

So, we're looking for conditions under which $\check{\mu} = \mu_C$. Here's a natural one, which gives a affirmative answer to Q1 under the additional hypothesis that $(X,\Sigma,\mu)$ is $\sigma$-finite, that is to say $X = \bigcup_{n=1}^{\infty} E_{n}$ with $\mu(E_n) \lt \infty$.

The following proposition is a minor adaptation of Fremlin's more general proposition 213C on p.24 of volume 2 of his Measure theory:

Proposition. If $(X,\Sigma,\mu)$ is complete and $\sigma$-finite then $\mu = \mu_{C}$.

In other words, assuming completeness and $\sigma$-finiteness, $\mu$ is induced by the outer measure $\mu^{\ast}$ given by $$\mu^{\ast} (A) = \inf{\left\{\mu(E)\,:\,A \subset E, \,E\in \Sigma\right\}}$$ and $\Sigma$ coincides with the $\mu^{\ast}$-measurable sets in the sense of Carathéodory.

It suffices to prove that $\Sigma = \Sigma_C$. Recall that $\Sigma \subset \Sigma_C$ where $$\Sigma_C = \{E \subset X\,:\,\mu^{\ast}(A) = \mu^{\ast}(A \cap E) + \mu^{\ast}(A \smallsetminus E)\text{ for all } A \subset X\}.$$ Let $F \in \Sigma_C$. Since we assume that $X = \bigcup_{n =1}^{\infty} E_n$ with $E_n \in \Sigma$ and $\mu(E_n) \lt \infty$ we can put $F_n = E_n \cap F$ and it suffices to show that $F_n \in \Sigma$ because $F = \bigcup_{n=1}^{\infty} F_n$.

Choose $G_1 \in \Sigma$ with $F_n \subset G_1 \subset E_n$ and $\mu(G_1) = \mu^{\ast}(E_n \cap F_n)$ and $G_2 \in \Sigma$ with $E_n \smallsetminus F_n \subset G_2 \subset E_n$ and $\mu(G_2) = \mu^{\ast}(E_n \smallsetminus F_n)$. Since $F_n \in \Sigma_C$ we have by definition of $\Sigma_C$ that $$\mu(G_1) + \mu(G_2) = \mu^{\ast}(E_n \cap F_n) + \mu^{\ast}(E_n \smallsetminus F_n) = \mu^{\ast}(E_n) = \mu(E_n).$$ Since $\mu(E_n) \lt \infty$ and $E_n = G_1 \cup G_2$ we have that $\mu(G_1 \cap G_2) = \mu(G_1) + \mu(G_2) - \mu(E_n) = 0.$ But notice that $G_1 \smallsetminus F_n \subset G_1 \cap G_2$, so $G_1 \smallsetminus F_n$ is a null-set, hence $\mu$-measurable since $\mu$ is complete, and $F_n = G_1 \smallsetminus (G_1 \smallsetminus F_n)$ shows that $F_n \in \Sigma$, as desired.

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Thanks t.b. for your detailed answer. I have understood the answer pretty much I think, but I'll take a few days to study it carefully. Actually, I did anticipate some portions of your answer, but couldn't express them so coherently. :) And it was great to see the sufficient condition (sigma-finite); it didn't occur to me to ask for one in my question since I was implicitly expecting the answers to both the questions to be "True, of course". –  Srivatsan Sep 28 '11 at 17:21
    
@Srivatsan: You're welcome, of course. Take your time, and maybe I'll find the time to fix the numerous typos by then :) –  t.b. Sep 28 '11 at 19:49
    
A less artificial counterexample is the restriction of the counting measure to countable and cocountable sets on an uncountable set. The Caratheodory extension is the whole counting measure. –  scineram Sep 29 '11 at 22:57
    
@scineram: Thanks, that's right. The problem with that example is that it seems to show a little less: I don't see how to obtain counting measure on the countable-cocountable $\sigma$-algebra from Carathéodory's construction from some outer measure. –  t.b. Sep 30 '11 at 9:40
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I am posting this because I think it's quite relevant to my question, even if it doesn't directly address it. This is inspired by t.b.'s answer and Niels' comment, and it hopefully complements them both.

Interpreting my second question broadly, we want to understand whether a measurable space admits a measure (with additional restrictions). Of course, the zero measure trivially works for all measurable spaces $(X, \Sigma)$. Slightly more interestingly,

  • For an arbitrary $x_0 \in X$, the measure $$ \mu_{x_0}(A) = \begin{cases} 1, &x_0 \in A, \\ 0, &x_0 \notin A, \end{cases} $$ is a probability measure (consequently also $\sigma$-finite).
  • The measure $$ \mu(A) = \begin{cases} \infty, &A \neq \emptyset, \\ 0, &A = \emptyset, \end{cases} $$ suggested by Niels is a complete measure, but it fails to be $\sigma$-finite (and fails quite badly too). Of course, this technically answers my second question.

However, one could further ask whether every measurable space admits a measure that is both complete and sigma-finite. This combination of properties is attractive because this is precisely the sufficient condition given by t.b.'s answer for the measure to be identical to that given by the associated outer measure. Unfortunately, we show that this is too much to hope for in general.


As a counter-example, take $I$ to be an uncountable index set, and $\{ X_\alpha \}_{\alpha \in I}$ to be a collection of disjoint 2-element sets. Define $X = \bigcup_{\alpha \in I} X_\alpha$, and consider the $\sigma$-algebra $$ \Sigma = \{ A \subseteq X \,:\, \forall \alpha \in I, \text{ either } A \cap X_{\alpha} = \emptyset \text{ or } X_\alpha \subseteq A \}. $$ Suppose $\mu$ is a complete measure on $(X, \Sigma)$. Since none of the proper subsets of $X_\alpha$ are in $\Sigma$, completeness requires that $\mu(X_{\alpha}) > 0$ for all $\alpha \in I$.

Now, any countable partition of $X$ into measurable sets must be of the form $$ \left\{ \bigcup_{\alpha \in I_n} X_{\alpha} \right\}_{n \in \mathbb N} $$ where $\{I_n\}_{n \in \mathbb N}$ forms a partition of $I$. Since $I$ is uncountable, it follows that there exists $n$ such that $J := I_n$ is uncountable. We show that $\bigcup_{\alpha \in J} X_\alpha$ has infinite measure. (This will complete the proof that the measure $\mu$ is not $\sigma$-finite.)

The remaining parts of the proof also follow standard ideas. Notice that, by the Archimedean property of the reals, we can write $J$ as a countable union $$ J = \bigcup_{m \in \mathbb N} \left \{ \alpha \in J \,:\, \mu(X_\alpha) \geq \frac{1}{m} \right\}. $$ Since $J$ is uncountable, it follows that there exists $m \in \mathbb N$ such that $$ J_m := \left \{ \alpha \in J \,:\, \mu(X_\alpha) \geq \frac{1}{m} \right\} $$ is infinite (in fact, uncountable as well). Since each $\alpha \in J_m$ is such that $X_\alpha$ accounts for at least $1/m$ measure, and all the $X_\alpha$'s are disjoint, it follows that the total measure of the set $$ \bigcup_{\alpha \in J_m} X_\alpha $$ is infinite. Finally, since $J \supseteq J_m$, we are done by the monotonicity of $\mu$.

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Just a small comment: your measure space in the counterexample could also be described as the product measure space $I \times \{0,1\}$ where $I$ is an uncountable set equipped with the $\sigma$-algebra $\mathcal{P}(I)$ and $\{0,1\}$ is equipped with the trivial $\sigma$-algebra $\{\emptyset, \{0,1\}\}$. –  t.b. Sep 30 '11 at 10:36
    
@t.b. Yes, that's indeed much simpler way to describe it. Thanks. –  Srivatsan Sep 30 '11 at 17:58
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