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*:*x*∈*U*_{a}})). Moreover, any real-valued random variable is going to be probabilistically equivalent to some variable of the form \*p**h**i*∘*X*. 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 *A*≤*C* pointwise and *E*(*C*)−*E*(*A*)<(*b*−*a*)/2. Let *D* be a simple function such that *D*≤*B* pointwise and *E*(*B*)−*E*(*D*)<(*b*−*a*)/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*+(*d*−*c*)/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*−(*d*−*c*)/2+*F*<−*u*+*G* by 3 and 4, and so *C*<*D* as desired.

Let *f*=*E*(*F*) and *g*=*E*(*G*). Let *d*=*g*−*f*. 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 *F*≤*H* and *G*≤*K* 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 *h*_{i}, 1≤*i*≤α, and *k*_{i}, 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 *h*_{i}, *k*_{i}, *h*'_{i} and *k*'_{i} are all probabilistically equivalent. Let *v*>0 be small enough that α*v*<*d*/16 and β*v*<*d*/16. Then by Axioms 3 and 5 we have *h*_{i}<*v*+*h*'_{i} and −*v*+*k*'_{i}<*k*_{i} 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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