Nice! This, of course, seems like something we should salvage, by e.g. adding an axiom that if A is strictly preferred to B, there should be a lottery strictly between them.
That proposed axiom to add does not work. Consider the function on lotteries over {x,y,z} that gives utility 1 if z is supported, and otherwise gives utility equality to the probability of x. This function is concave but not continuous, satisfies A1-A5 and the extra axiom I just proposed, and cannot be made continuous.
Oh, no, I made a mistake, this counterexample violates A3. However, the proposed fix still doesn’t work, because you just need a function that is decreasing in probability of x, but does not hit 0, and then jumps to 0 when probability of x is 1.
Nice! This, of course, seems like something we should salvage, by e.g. adding an axiom that if A is strictly preferred to B, there should be a lottery strictly between them.
That proposed axiom to add does not work. Consider the function on lotteries over {x,y,z} that gives utility 1 if z is supported, and otherwise gives utility equality to the probability of x. This function is concave but not continuous, satisfies A1-A5 and the extra axiom I just proposed, and cannot be made continuous.
Oh, no, I made a mistake, this counterexample violates A3. However, the proposed fix still doesn’t work, because you just need a function that is decreasing in probability of x, but does not hit 0, and then jumps to 0 when probability of x is 1.
Oh, nvm, that is fine, maybe it works.
I think the way I would rule out my counterexample is by strengthening A3 to if A≻B and B≻C then there is p∈(0,1)…
That does not rule out your counterexample. The condition is never met in your counterexample.
Oh, derp. You’re right.