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From Wikipedia

an ultrafilter $U$ on a set $X$ is a collection of subsets of $X$ that is a filter, that cannot be enlarged (as a filter). An ultrafilter may be considered as a finitely additive measure. Then every subset of $X$ is either considered "almost everything" (has measure 1) or "almost nothing" (has measure 0). ... define a function $m$ on the power set of $X$ by setting $m(A) = 1$ if $A$ is an element of $U$ and $m(A) = 0$ otherwise. Then $m$ is a finitely additive measure on $X$, and every property of elements of $X$ is either true almost everywhere or false almost everywhere.

It says "an ultrafilter may be considered as a finitely additive measure." However, I only see how an ultrafilter induces a finitely additive measure, and don't see how a finitely additive measure induces an ultrafilter. In order for a finitely additive measure to induce an ultrafilter, I think it is not enough that for any subset $A$ of $X$, either $m(A) = 1$ and $m(X-A) = 0$ or $m(X-A) = 1$ and $m(A) = 0$, isn't it?

Thanks and regards!

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"Any Y can be considered an X" does not mean the same thing as "any X can be consider a Y." An ultrafilter corresponds with a measure which only takes the values $0$ and $1$. –  Thomas Andrews Jan 4 '13 at 19:29
(Also, the measure must be defined on the entirety of $\mathcal P(X)$, not some proper sub-$\sigma$-algebra.) –  Thomas Andrews Jan 4 '13 at 19:40

2 Answers 2

up vote 2 down vote accepted

It does not say that every finitely additive measure induces an ultrafilter. However, it is true that every non-trivial $\{0,1\}$-valued finitely additive measure on $\wp(X)$ induces an ultrafilter on $X$.

Suppose that $m$ is a $\{0,1\}$-valued finitely additive measure defined on $\wp(X)$ such that $m(X)=1$. Let $\mathscr{U}=\{U\subseteq X:m(U)=1\}$. For each $A\subseteq X$ we have $m(A)+m(X\setminus A)=1$, so exactly one of $A$ and $X\setminus A$ belongs to $\mathscr{U}$. Clearly $V\in\mathscr{U}$ whenever $X\supseteq V\supseteq U\in\mathscr{U}$, so $\mathscr{U}$ is closed under taking supersets. If $U,V\in\mathscr{U}$ and $U\cap V\notin\mathscr{U}$, then

$$1=m(U)=m(U\setminus V)+m(U\cap V)=m(U\setminus V)$$

and similarly $m(V\setminus U=1$, so

$$m(U\cup V)=m(U\setminus V)+m(U\cap V)+m(V\setminus U)=2\;,$$

which is absurd. Thus, $\mathscr{U}$ is closed under taking finite intersections. Finally, for each $A\subseteq X$ we have $m(A)+m(X\setminus A)=1$, so exactly one of $A$ and $X\setminus A$ belongs to $\mathscr{U}$. Thus, $\mathscr{U}$ is an ultrafilter on $X$.

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Yes, it is enough (if we insist that the whole set has measure $1$). We need to check that any superset of a set of measure $1$ has measure $1$ (easy), and that the intersection of two sets of measure $1$ has measure $1$.

So let $A$ and $B$ have measure $1$. Then $A\cup B$ has measure $1$, and is the disjoint union of $A\cap B$, $A\setminus B$, and $B\setminus A$.

If $A\cap B$ does not have measure $1$, then each of $A\setminus B$ and $B\setminus A$ do, contradicting finite additivity.

Remark: Finitely additive $\{0,1\}$-valued measures defined on all subsets of a set $I$, and ultrafilters on $I$ are such close relatives that there is no point in distinguishing between the two.

When we are working with an ultrapower $A^{I}/D$, it is often more natural to say that the functions $f$, $g$ from $I$ to $A$ are equal "almost everywhere" than to say $\{i:f(i)=g(i)\}\in D$.

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+1. Thanks! Both replies are very good and it is hard to pick one to accept. I accept Brian's, because his is a bit clearer, although yours is also clear. But I like yours, because I didn't know about "ultrapower" and thank you for mentioning it. (BTW, I hate myself when I have to judge.) –  Tim Jan 4 '13 at 21:13

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