I'm sure there is a better place to post a wall of math than this forum.
It's published as an appendix, in the discussion section. People aren't expected to read the appendices unless they are interested in why something is assumed in the main paper or the details happen to be important to their work somehow. I approve of splitting of the mathematical details into an appendix like this---the only downside is the cost to the author of producing the work.
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.)