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If $\mathcal{U}$ is an ultrafilter on a set $X$, it can be defined a function $\mu_{\mathcal{U}}\colon\mathcal{P}(X)\to\{0,1\}$ such that, for all $A\subseteq X$ it holds $\mu_{\mathcal{U}}(A)=1$ iff $A\in\mathcal{U}$.

Are $\mu_{\mathcal{U}}(\emptyset)=0$ and finite additivity enough to prove that $\mu_{\mathcal{U}}$ is well-defined? Of course, given a subset $E$ of $X$, we can split $E$ into disjoint sets $E_1,\dots,E_n$. If $\mu_{\mathcal{U}}(E)=1$, then exactly one $E_i$ belongs to $\mathcal{U}$, so $\mu_{\mathcal{U}}(E_1)+\dots+\mu_{\mathcal{U}}(E_n)=1$. If $\mu_{\mathcal{U}}(E)=0$, then none of the $E_i$'s belongs to $\mathcal{U}$, so $\mu_{\mathcal{U}}(E_1)+\dots+\mu_{\mathcal{U}}(E_n)=0$. Did I prove that $\mu_{\mathcal{U}}$ is actually a function, hence a finitely additive measure?

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  • $\begingroup$ Closely related question (where you can find answers to what you ask in the title): How can an ultrafilter be considered as a finitely additive measure? $\endgroup$ Commented Jul 6, 2016 at 10:15
  • $\begingroup$ You ask also whether $\mu$ is well-defined. (Although it is less clear to me how your second paragraph relates to this problem.) If you know that for any ultrafilter $\mathcal U$ on $X$ and any $A\subseteq X$ either $A\in\mathcal U$ or $X\setminus A\in\mathcal U$, then you can rewrite the deffinition of this finitely additive measure as $$\mu(A)= \begin{cases} 1 & A\in\mathcal U, \\ 0 & A\notin\mathcal U. \end{cases}$$. From this it is clear that $\mu$ is well-defined. $\endgroup$ Commented Jul 6, 2016 at 10:18

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You just define $\mu(A) = 0$ if $A \notin \mathcal{U}$ and $\mu(A) = 1$ if $A \in \mathcal{U}$. This is well-defined (a set cannot be both in $\mathcal{U}$ and not in $\mathcal{U}$, and so any set has measure $0$ or $1$).

As an ultrafilter does not contain $\emptyset$, $\mu(\emptyset) = 0$, as required.

And if we have finitely many disjoint subsets $E_1,\ldots,E_n$ of $X$, we need to show that for $E = \cup_{i=1}^n E_i$ we have $\mu(E) = \sum_i \mu(E_i)$.

If all $E_i$ are not in $\mathcal{U}$, then $X \setminus E_i \in \mathcal{U}$ for all $i$, which means that $X \setminus E = \cap_{i=1}^n (X \setminus E_i) \in \mathcal{U}$, as filters are closed under finite intersections, and hence $E \notin \mathcal{U}$ as well. So we have equality ($0$'s on both sides) in our equality we have to prove.

On the other hand, if some $E_{i_0} \in \mathcal{U}$, then the other $E_i$ are not in $\mathcal{U}$ by disjointness, and so have measure $0$, and $E$ contains $E_{i_0}$ so also is in $\mathcal{U}$ and has measure $1$. So we have equality as well: $\mu(E) = 1$ and the sum is $1$ as well: one $1$ for $E_{i_0}$ and the others are $0$.

This shows that $\mu$ is indeed a finitely additive measure.

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