# What exactly is the “Carathéodory Extension theorem”?

I have read four texts introducing a theorem so-called "Carathéodory's Extension Theorem", and they all differ.

Here is the statement of the Carathéodory Extension Theorem in Wikipedia:

Let $\mathfrak{R}$ be a ring of subsets of $X$ Let $\mu:\mathfrak{R} \rightarrow [0,\infty]$ be a premeasure. Then, there exists a measure on the σ-algebra generated by $\mathfrak{R}$ which is a extension of $\mu$.

I like this statement since it is very simple and clear.

However mainstream textbooks don't introduce the Caratheodory Extension Theorem as this.

For example, Royden's Real Analysis (4th edition) defines premeasure as a set function which is finitely additive, countably monotone (is this a widely used term?), which is different from the definition in wikipedia. Then, he proves a theorem so-called Carathéodory-Hahn Extension Theorem. This theorem does not imply the 'Carathéodory Extension Theorem in Wikipedia' but is deeper than that in Wikipedia, in my opinion. Since he defined premeasure differently, I am quite hesitant to memorize this. I don't want to be out of the mainstream. He even doesn't require the domain of a given set function to have an empty set.

Another example, Folland's Real Analysis states Carathéodory's Theorem as follows:

If $\mu^*$ is an outer measure on $X$, the collection $\mathcal{M}$ of $\mu^*$-measureable sets is a σ-algebra, and the restriction of $\mu^*$ to $\mathcal{M}$ is a complete measure.

Even though this is critical in any proof for any kind of Extension Theorem, this is obviously not 'the Carathéodory Extension Theorem'.

What is the Carathéodory Extension Theorem?

It is off the topic, but I have one more question.

Why Royden tries to include set functions whose domain does not contain an empty set?

If the emptyset is not in a domain, can't we just extend a given set function by defining $\mu(\emptyset)=0$? Does this sometimes break structure of a given set? E.g. ring of sets, algebra of sets or whatever.

• From what I remember from Folland, there is an earlier theorem about constructing an outer measure from a premeasure, and a corollary or exercise proving that it is compatible with the premeasure, which makes it essentially the same as the Wikipedia version. – mathematician Jan 5 '14 at 18:28
• I'm kind of being picky here, but there aren't too many functions whose domains don't contain the empty set. In any case, it's hard to say what you're talking about until you tell us the details. – tomasz Jan 5 '14 at 18:30
• On a different note, it's common in mathematics to use the name of one theorem to denote a wide variety of similar, sometimes by far more general theorems. I've had a course where the professor insisted on calling every Haar measure the Lebesgue measure and the Holder's theorem (or something closely related, I don't remember exactly) the Pythagorean theorem. – tomasz Jan 5 '14 at 18:33

The condition for the theorem for Wikipedia and Royden are the same, just phrase in different language. Countably monotone together with finitely additive immediately imply countable additive, so not much different there. Empty set is always in the ring due to intersection of a set with its complement. By finite additivity immediately can conclude that once you extend to a measure empty set is always measure 0.

The conclusions are indeed different. I think the version where extension extend beyond just the $\sigma$-algebra and produce a complete measure is the one more widely use, because all of my text have it. But uniqueness of extension to $\sigma$-algebra seems standard too.

Folland appears to start with outer measure instead of premeasure. Presumably, it is so that the theorem can by applied to any outer measure, not just one that come from premeasure. However, the conclusion is weaker, and will fail to give certain result such as uniqueness, or even the assurance that everything in the ring will be measurable.

• Extending to a complete measure or just a measure does not matter at all. If a function once extended to a measure, then it can always be extended to a complete measure. and I'm wondering how could countably monotone and finitely additive imply countable additivity – Number 9 Jan 6 '14 at 16:03
• @Number9: yes it doesn't matter much, but there is no harm stating the strongest version available for the sake of completeness when all you need is insert in a 2 lines proof. – Gina Jan 6 '14 at 22:28
• @Number9:For your other question. Consider a countable mutual disjoint collection $A_{i}$. Then $\mu(\bigcup\limits_{i=1}^{\infty}A_{i})$ $=\mu(\bigcup\limits_{i=n+1}^{\infty}A_{i})+\sum\limits_{i=1}^{n}\mu(A_{i})$ $\geq\sum\limits_{i=1}^{n}\mu(A_{i})$ for all $n$ so $\mu(\bigcup\limits_{i=1}^{\infty}A_{i})\geq\sup\limits_{n}(\sum\limits_{i=1}^{n}\mu(A_{i}))=\sum\limits_{i=1}^{\infty}\mu(A_{i})$. And countably monotone is the opposite direction:$\mu(\bigcup\limits_{i=1}^{\infty}A_{i})\leq\sum\limits_{i=1}^{\infty}\mu(A_{i})$ – Gina Jan 6 '14 at 22:31
• Since the domain is a semiring, $\mu(\bigcup_{i=1}^\infty A_i)$ is not defined. – Number 9 Jan 7 '14 at 1:16
• @Number9:It does not NECESSARILY defined. If it is not defined, then we don't even need to worry about it. But I can certainly write $]0,1]=\bigcup\limits_{i=1}^{\infty}]2^{-i},2^{-i+1}]$ and if somehow $]0,1]$ was given measure 0.9 while the rest got the usual measure, we obviously got a problem. We just want to make sure that such a case would not happen. All the text agree on the requirement on what the domain is already, it's not the point of contention here; we just need to make sure that the additivity of the premeasure are defined the same despite different language. – Gina Jan 7 '14 at 2:27

There's basically two ways of stating the theorem:

Either you choose to state all the critical steps one after the other step by step or you crunch them altogether as one big theorem in wildest generality.

In the situation here it comes to state either in steps that every premeasure on semiring extends to a nonempty semiring by patching the empty set to it then Hahn-Kolmogorov extension and then Caratheodory's construction. Or one states that every premeasure on a semiring extends to a complete measure.

In my personal opinion the former is clearer and focuses on the critical steps while the latter one hides it under biggest generality which is not always preferable. Also when proving things you don't try to merge always all in one since it gets messy but split it into LEMMATA!! And that's what the wiki version did ;)

The problem is that the most important part of Carateodory theorem (and sometimes of many theorems) is not the exact statement but the main idea(s) in the proof. In this theorem's proof there is a key idea to remember, which is what sets to call measurable for the (complete) measure it defines. These are the sets that are partitioned additively (for the outer measure) by all other sets.

Another idea to remember could be to remember how to build an outer measure. But this is kind of natural, and therefore easy to remember. Learn the proof and you will know all the statements you have seen.