# Measure vs outer measure

Basically I feel like I'm having a chicken before egg problem.

An outer measure is a set function defined on some set X with the properties that empty returns 0, of sub additivity and of monotonicity. From this outer measure we get a sigma algebra of measurable sets. On this sigma algebra the outer measure is a measure with the property of additivity for unions of disjoint sets.

But we can also define a measure , a set function, on an algebra of sets with the properties that empty returns zero and of additivity of unions of disjoint sets. Then via caratheodory construction we can make an outer measure from the measure for the whole set.

So what comes first, or does neither come first?

Note: that this all began with a sample final question that I'm still stuck on that asked "if a measure is defined on the whole power set is it also an outer measure?" Which is why I marked this as homework.

Note 2: also is the answer to the question the following: Any measure can be transformed into an outer measure via caratheodory?

Thanks.

Edit: deleted the bit that is talked about below

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"This algebra is clearly a sigma algebra, then via caratheodory construction we can make an outer measure from the measure for the whole set." I don't see what family is clearly a sigma algebra. The usual Caratheodory construction starts with an outer measure or a measure on a generating algebra. – Michael Greinecker Feb 26 '13 at 17:28
First thanks for your patience, I'm a lowly and rather dense undergrad. If the algebra on which a measure is defined isn't a sigma algebra wouldn't that imply that the measure isn't necessarily a measure? Since a disjoint sequence of sets from an algebra won't necessarily be contained in the algebra? – Michael Summerville Feb 26 '13 at 18:22
You are kind of right. You star with a nonnegative set function on an algebra that is countably addiive in the sense that whenever the union of a disjoint sequence of sets in the algebra is also in the algebra, the value of the union is the sum of the individual values. You can construct an outer measure from this functon. If you then apply the Caratheodory construction you get the measure on a $\sigma$-algebra that includes the original algebra and the measure extends the original set function. – Michael Greinecker Feb 26 '13 at 18:29
But if we start with an algebra that is countably additive isn't it already a sigma algebra? – Michael Summerville Feb 26 '13 at 18:42
It is the set function that is countably additive, not the algebra. So if $\mathcal{A}$ is the algebra, $\mu$ the set function, $B_1,B_2,\ldots$ be disjoint sets in $\mathcal{A}$ and $\bigcup_n B_n\in\mathcal{A}$, then we require $\sum_n \mu(B_n)=\mu\big(\bigcup_n B_n\big)$. This is not a restriction if $\bigcup_n B_n\notin\mathcal{A}$, which can happen since $\mathcal{A}$ is only an algebra and not a $\sigma$-algebra. – Michael Greinecker Feb 26 '13 at 18:49

Here is standard way for constructing measures on a measurable space $(S,\mathcal{S})$ that can for example be used to construct Lebesgue measure; various closely related approaches exist. For simplicity, we look at finite measures only.
You start with an algebra $\mathcal{A}$ on $S$ such that $\mathcal{S}$ is the smallest $\sigma$-algebra that contains all elements of $\mathcal{A}$. You then define a finite, nonnegative and finitely additive set-function $\mu_0$ on $\mathcal{A}$ that is countably additive in the sense that if $(A_n)$ is a sequence of disjoint sets in $\mathcal{A}$ and $\bigcup_n A_n\in\mathcal{A}$, then $\sum_n \mu_0(A_n)=\mu_0\big(\bigcup_n A_n\big)$. This assumption is easier to check than for $\sigma$-algebras since it might well be the case that $\bigcup_n A_n\notin\mathcal{A}$.
Then you define an outer measure $\mu^*$ on all of the powerset $2^S$ by letting $$\mu^*(B)=\inf\bigg\{\sum_{A\in\mathcal{C}}\mu_o(A):\mathcal{C}\textrm{ is a countable subset of }\mathcal{A}\textrm{ such that }B\subseteq\bigcup_{A\in\mathcal{C}}A\bigg\}.$$ Let $$\mathcal{S}_\mu=\Big\{B\in 2^S:\mu^*(A)=\mu^*(A\cap B)+\mu^*(A\cap S\backslash A)\textrm{ for all }A\in 2^S\Big\}.$$ It turns out that $\mathcal{S}_\mu$ is a $\sigma$-algebra and the restriction of $\mu^*$ to it, $\mu^*|\mathcal{S}_\mu$, is a measure. Moreover, $\mathcal{S}_\mu\supseteq \mathcal{S}\supseteq\mathcal{A}$, so $\mu=\mu^*|\mathcal{S}$ is a measure on $\mathcal{S}$ and moreover, $\mu^*|\mathcal{A}=\mu_0$.