Dominance and countably infinite fair lotteries

Suppose we have a finitely-additive probability assignment p (perhaps real, perhaps hyperreal) for a countably infinite lottery with tickets 1, 2, ... in such a way that each ticket has infinitesimal probability (where zero counts as an infinitesimal). Now suppose we want to calculate the expected value or previsio E_pU of any bounded wager U on the outcome of the lottery, where we think of the wager as assigning a value to each ticket, and the wager is bounded if there is a finite M such that |U(n)| < M for all n.

Here are plausible conditions on the expected value:

1. Dominance: If U₁ < U₂ everywhere, then E_pU₁ < E_pU₂.

2. Binary Wagers: If U is 0 outside A and c on A, then E_pU = cP(A).

3. Disjoint Additivity: If U₁ and U₂ are wagers supported on disjoint events (i.e., there is no n such
   that U₁(n) and U₂(n) are both non-zero), then E_p(U₁+U₂) = E_pU₁ + E_pU₂.

But we can’t. For suppose we have it. Let U(n) = 1/(2n). Fix a positive integer m. Let U₁(n) be 2 for n ≤ m + 1 and 0 otherwise. Let U₂(n) be 1/m for n > m + 1 and 0 for n ≤ m + 1. Then by Binary Wagers and by the fact that each ticket has infinitesimal probability, E_pU₁ is an infinitesimal α (since the probability of any finite set will be infinitesimal). By Binary Wagers and Dominance, E_pU₂ ≤ 1/(m+1). Thus by Disjoint Additivity, E_p(U₁+U₂) ≤ α + 1/(m+1) < 1/m. But U < U₁ + U₂ everywhere, so by Dominance we have E_pU < 1/m. Since 0 < U everywhere, by Dominance and Binary Wagers we have 0 < E_pU.

Thus, E_pU is a non-zero infinitesimal β. But then β < U(n) for all n, and so by Binary Wagers and Dominance, β < E_pU, a contradiction.

I think we should reject Dominance.