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My textbook's definition of measure goes like this

A set function $\mu$ defined on an algebra $\mathscr A$ is called measure if:
a) $\mu ( \emptyset ) = 0 $
b) $\mu$ is countably additive: i.e., if $\{A_n\}^\infty_{n=1} $ is a countable collection of sets in $\mathscr A$ such that (i) $A_n \bigcap A_m = \emptyset$ for $ n \neq m$, and (ii) $ A = \bigcup^\infty_{n=1} A_n \in \mathscr A $, then $\mu(A) = \sum^\infty_{n=1} \mu(A_n$).

However, in the definition of algebra, the book defined it to be closed under finite unions. To cross-check, this maths stack exchange post confirms that an algebra and sigma algebra's difference is the fact that algebra may not be closed over infinite unions.
Therefore, $A$ in the definition of measure may not belong to the algebra which doesn't make sense as I think the set function $\mu$ is $\mathscr A \rightarrow [0, \infty] $ (closed braces on infinity as given in my book). Can someone please solve my confusion? I think it may be that the set function is defined from smallest sigma field containing $\mathscr A$ then it would make sense but it is not apparent.

Henno Brandsma
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Black Jack 21
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2 Answers2

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The book explicitly specifies that this only applies if the infinite union happens to be in the algebra. If it isn't, then it doesn't assert that the measure of that set exists.

Matt Samuel
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  • Does this apply to the definition of outer measure as well? $ \mu^{*}(E) = inf{\sum_{n=1}^\infty \mu(A_n): E \subseteq \bigcap_{n=1}^\infty A_n, A_n \in \mathscr A \forall n } $ The union of $A_n$ be outside the algebra, would that mean that the $A_n$ choosen is not valid or would I still consider this combination while taking inf? – Black Jack 21 Sep 22 '19 at 10:38
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    @Red It doesn't seem to be specifying that outer measure isn't defined if it isn't in the algebra, and I seem to recall from measure theory that you can usually take the outer measure even if it isn't measurable. – Matt Samuel Sep 22 '19 at 13:13
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The condition b) says that if we happen to have a disjoint family of members of $\mathcal{A}$ such that the union is also in $\mathcal{A}$ (this is indeed not guaranteed to be the case (algebras are only required to be closed under finite unions), but it could be the case and the condition only applies when it does), in that case we know that $\mu(A)$ is the sum of the $\mu(A_n)$. If the union is not in $\mathcal{A}$, too bad, and there is nothing to check in that case.

A simple example: let $X=\mathbb{N}$ and $\mathcal{A}$ is the algebra of finite and co-finite (complement is finite) subsets of $\mathbb{N}$ and $\mu(A)$ is $+\infty$ for $A$ cofinite, and the number of elements of $A$ for $A$ finite. We only need to check the condition for situations like e.g. $A_n = \{2n,2n+1\}$ for $n=0,1,2,\ldots$ where $\bigcup_n A_n = \Bbb N \in \mathcal{A}$ but not for situations like when $A_n = \{2n\}$ where the union is neither finite not co-finite (though it does hold then, but that's irrelevant for the definition).

Henno Brandsma
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