# Footnote 1:

This is a proof of the Theorem advertised here. I am assuming standard countably additive probabilities. It is easy to see that continuity is equivalent to saying that there is a random variable X on Ω with uniform distribution (e.g., let X(ω)=max(0,∈f{a:xUa})). Moreover, any real-valued random variable is going to be probabilistically equivalent to some variable of the form \phiX. By Axiom 5, then, we can assume without loss of generality that Ω=[0,1] with Lebesgue measure. Now, suppose a=E(A)<E(B)=b. Let C be a simple function such that AC pointwise and E(C)−E(A)<(ba)/2. Let D be a simple function such that DB pointwise and E(B)−E(D)<(ba)/2. Then E(C)<E(D). If we can show that C<D, then by Axioms 1 and 2 it will follow that A<B, as desired.

Let c=E(C) and d=E(D). Suppose u is a constant large enough that F=u+(dc)/2+C and G=u+D are both non-negative. Note that E(F)<E(G). If we can show that F<G, then −u−(dc)/2+F<−u+G by 3 and 4, and so C<D as desired.

Let f=E(F) and g=E(G). Let d=gf. Without loss of generality (Axiom 5 and rearrangement), F and G are non-negative monotone non-increasing functions on [0,1], bounded above by some constant M. Choose a natural n such that the lower Riemann integral of G for the partition of [0,1] with equal width 1/n is less than d/8 from the Lebesgue integral of G, and the upper Riemann integral of F for the same partition is less than d/8 from the Lebesgue integral of F. There will also be non-negative functions H and K that are constant on each element of the partition, with FH and GK everywhere pointwise, whose values are always integral multiples of d/8, and whose integrals over [0,1] are within d/8 of the upper Riemann integral of F and the lower Riemann integral of G. Then, E(H)+d/2<E(K) and all we need to show is that H<K and we will be done by Axioms 1 and 2.

We can write H and K as the respective sums of the functions hi, 1≤i≤α, and ki, 1≤i≤β, such that each of these functions takes the constant value d/16 on some one of the intervals of the partition and takes the constant value 0 everywhere else. Let h'i, 1≤i≤α, and k'i, 1≤i≤β be functions of the same sort, except that they are as evenly distributed over the intervals of the partition as possible. Thus there will be natural numbers γ and δ such that each interval is the support of γ or γ+1 of the functions h'i and is the support of δ or δ+1 of the functions k'i. Note that the hi, ki, h'i and k'i are all probabilistically equivalent. Let v>0 be small enough that αv<d/16 and βv<d/16. Then by Axioms 3 and 5 we have hi<v+h'i and −v+k'i<ki and by Axioms 2 and 4 we have H<d/16+H' and −d/16+K'<K. Moreover E(H')=E(H) and E(K')=E(K). There will be a constant function H'' dominating H' with E(H'') within d/16 of H' everywhere and a constant function K'' dominated by K' with E(K'') within d/16 of K' everywhere. It is easy to see that E(d/16+H'')<E(−d/16+K'') by the choices of all the inequalities above. Then by Axiom 3 we have d/16+H''<−d/16+K'', and it follows from the other things we've shown that H<K.

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