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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:

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

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.

2 texified

A naive question.

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

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

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

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

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, $P$, which just evaluate f $f$ at some point s $s$ of S. $S$. I suppose you could get other P $P$ by sending f $f$ 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.

1

# "Probabilistic ultrafilters?"

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: [0,1]^S -> [0,1] satisfying the following three conditions:

1. If f >= g everywhere on S, then P(f) >= P(g);
2. P(min(f,g)) >= min(P(f),P(g));
3. 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.