A naive question.

Let $S$ be a set and let $[0,1]^S$ the set of functions from $S$ to the closed interval $[0,1]$.

Suppose given some function $P \colon [0,1]^S \to [0,1]$ satisfying the following three conditions:

- If $f \geq g$ everywhere on $S$, then $P(f) \geq P(g)$;
- $P(\min(f,g)) \geq \min(P(f),P(g))$;
- $P(1-f) = 1 - P(f)$.

This is supposed to model a situation each point in $S$ has a "degree of belief" in some proposition, which yields a function $f$ in $[0,1]^S$; then $P$ is a process which takes all these degrees of belief and aggregates them into a "consensus" degree of belief $P(f)$.

Of course, this is meant to mimic the definition of an ultrafilter, which I think is given by the above definition with [0,1] replaced by {0,1}.

Certainly you have "principal" $P$, which just evaluate $f$ at some point $s$ of $S$. I suppose you could get other $P$ by sending $f$ to its limit with respect to some non-principal ultrafilter.

Is that it?

**Added:** Actually, the second condition above is perhaps too strong. I don't see an option for "hide question until I've thought about a bit more about what the best version of the question is" so I will just append this remark.

**Added:** Thanks, guys, for all the great answers. I now think the formulation of (2) was misguided (at least if the definition is meant to model consensus about degrees of belief) and I don't know what the "right" formulation is. One might well, for instance, want P to behave well when f and g refer to independent propositions; that would ask that P(fg) = P(f)P(g), which in the case of {0,1}-valued functions again agrees with the ultrafilter definition. This rules out averages but leaves in evaluation at ultrafilters.

finitelyadditive measure. – Nate Eldredge Mar 7 '11 at 13:13