This is a mathematical appendix to my post "Why you must maximize expected utility", giving precise statements and proofs of some results about von Neumann-Morgenstern utility theory without the Axiom of Continuity. I wish I had the time to make this post more easily readable, giving more intuition; the ideas are rather straight-forward and I hope they won't get lost in the line noise!
The work here is my own (though closely based on the standard proof of the VNM theorem), but I don't expect the results to be new.
*
I represent preference relations as total preorders 
on a simplex 
; define 
, 
, 
and 
in the obvious ways (e.g., 
iff both 
and 
, and 
iff but not ). Write 
for the 
'th unit vector in 
.
In the following, I will always assume that satisfies the independence axiom: that is, for all 
and 
, we have if and only if z \prec py + (1-p)z)
. Note that the analogous statement with weak preferences follows from this: holds iff 
, which by independence is equivalent to z \not\prec px + (1-p)z)
, which is just z \preccurlyeq py + (1-p)z)
.
Lemma 1 (more of a good thing is always better). If and 
, then x + py\prec (1-q)x + qy)
.
Proof. Let 
. Then, x + py = \big((1-q)x + py\big) + rx)
and x + qy = \big((1-q)x + py\big) + ry)
. Thus, the result follows from independence applied to 
, 
, x + py\big))
, and 
.
Lemma 2. If 
and 
, then there is a unique 
such that x + qz \prec y)
for )
and x + qz)
for 
.
Proof. Let 
be the supremum of all 
such that x + rz\preccurlyeq y)
(note that by assumption, this condition holds for 
). Suppose that 
. Then there is an 
such that x%20+%20rz%5Cpreccurlyeq%20y)
. By Lemma 1, we have x + qz \prec (1-r)x + rz)
, and the first assertion follows.
Suppose now that 
. Then by definition of , we do not have x%20+%20qz%5Cpreccurlyeq%20y)
, which means that we have x + qz\succ y)
, which was the second assertion.
Finally, uniqueness is obvious, because if both and 
satisfied the condition, we would have x + \frac{p+p'}2z \prec y)
.
Definition 3. is much better than , notation 
or 
, if there are neighbourhoods 
of and 
of (in the relative topology of ) such that we have 
for all 
and 
. (In other words, the graph of 
is the interior of the graph of .) Write 
or 
when 
( is not much better than ), and 
( is about as good as ) when both and 
.
Theorem 4 (existence of a utility function). There is a 
such that for all 
,
Unless for all and , there are 
such that 
.
Proof. Let be a worst and 
a best outcome, i.e. let be such that 
for all 
. If 
, then 
for all 
, and by repeated applications of independence we get 
for all , and therefore again for all , and we can simply choose 
.
Thus, suppose that 
. In this case, let 
be such that for every , 
equals the unique provided by Lemma 2 applied to and . Because of Lemma 1, 
. Let  := (1-r)e^i + re^j)
.
We first show that 
implies . For every , we either have 
, in which case by Lemma 2 we have 
for arbitrarily small 
, or we have 
, in which case we set 
and find )
. Set 
. Now, by independence applied 
times, we have  = f(p+\epsilon))
; analogously, we obtain )
for arbitrarily small 
. Thus, using 
and Lemma 1, \prec f(q-\delta)\preccurlyeq y)
and therefore as claimed. Now note that if 
, then this continues to hold for 
and 
in a sufficiently small neighbourhood of and , and therefore we have 
.
Now suppose that 
. Since we have 
and 
, we can find points and arbitrarily close to and such that the inequality becomes strict (either the left-hand side is smaller than one and we can increase it, or the right-hand side is greater than zero and we can decrease it, or else the inequality is already strict). Then, 
by the preceding paragraph. But this implies that 
, which completes the proof.
Corollary 5. 
is a preference relation (i.e., a total preorder) that satisfies independence and the von Neumann-Morgenstern continuity axiom.
Proof. It is well-known (and straightforward to check) that this follows from the assertion of the theorem.
Corollary 6. is unique up to affine transformations.
Proof. Since is a VNM utility function for , this follows from the analogous result for that case.
Corollary 7. Unless for all , for all 
the set 
has lower dimension than (i.e., it is the intersection of with a lower-dimensional subspace of ).
Proof. First, note that the assumption implies that 
. Let 
be given by 
, 
, and note that is the intersection of the hyperplane 
with the closed positive orthant 
. By the theorem, is not parallel to 
, so the hyperplane 
is not parallel to 
. It follows that 
has dimension 
, and therefore 
can have at most this dimension. (It can have smaller dimension or be the empty set if only touches or lies entirely outside the positive orthant.)