Finitely additive measure over $\mathbb{N}$, under AD. Under the axiom of choice, there are lots of non-trivial (that is, a measure vanishes every finite set) finitely-additive measures over $\mathbb{N}$: for example, ultrafilters over $\mathbb{N}$ or measures occur in other answers. However, as far as I know, $\mathsf{AD}$ proves every ultrafilter is $\aleph_1$-complete and hence every ultrafilter over $\mathbb{N}$ is principal.
My question is: does the same result hold for (real-valued) measures? That is, I wonder whether $\mathsf{AD}$ proves every finitely-additive measure is $\sigma$-additive. Thanks for any help.
 A: You can show that every finitely additive measure on $\Bbb N$ corresponds to a continuous functional on $\ell^\infty$ which is not induced by a sequence in $\ell^1$.
Consequently, such a measure shows there is a set without the Baire property. This, of course, contradicts $\sf AD$. 
You can find the proof in Schechter's Handbook of Analysis and its Foundations. 
A: I'm not sure how you define a "nontrivial" finitely additive measure over $\mathbb N.$ I will assume that it's a finitely additive set function $\mu:\mathcal P(\mathbb N)\to[0,1]$ such that $\mu(\mathbb N)=1$ and $\mu(\{x\})=0$ for all $x\in\mathbb N.$ I will show how to construct an undetermined game from such a measure $\mu.$
Consider an infinitely long game where two players, White and Black, take turns choosing finite sets $W_1,B_1,W_2,B_2,\dots$ of previously unchosen elements of $\mathbb N;$ White goes first, but that won't matter. Let $W=\bigcup_{n\in\mathbb N}W_n,$ the set of numbers chosen by White, and let $B=\bigcup_{n\in\mathbb N}B_n,$ the set of numbers chosen by Black. The payoff is that Black wins if $\mu(B)\ge\frac25,$ while White wins if $\mu(B)\lt\frac25.$ As there are countably many finite subsets of $\mathbb N,$ this is equivalent to a game where the players choose natural numbers. I claim that neither player has a winning strategy, contrary to AD.
Suppose White has a winning strategy, i.e., a strategy which guarantees that $\mu(B)\lt\frac25.$ A simple modification (choose one more number at each turn, so as to make sure that all numbers are eventually chosen) gives White a strategy which guarantees that $\mu(W)\gt\frac35.$ However, since finite sets have measure zero, the first move confers no advantage; Black could adopt the same strategy and thereby guarantee that $\mu(B)\gt\frac35.$ Since $\mu(W)+\mu(B)\gt\frac65$ is impossible, White has no winning strategy.
Suppose Black has a winning strategy, which guarantees that $\mu(B)\ge\frac25.$ Now consider a game played by three players, Red, Green, and Yellow, who take turns choosing disjoint finite sets $R_1,G_1,Y_1,R_2,G_2,Y_2,\dots,$ and let $R=\bigcup_{n\in\mathbb N}R_n$ be the set of numbers chosen by Red, $G=\bigcup_{n\in\mathbb N}G_n$ the set of numbers chosen by Green, and $Y=\bigcup_{n\in\mathbb N}Y_n$ the set of numbers chosen by Yellow. If each player imagines that he is playing Black (and using the winning strategy) in a two-person game, with his two adversaries together playing White, then the game will end up with $\mu(R)+\mu(G)+\mu(Y)\ge\frac65.$ This absurdity shows that Black has no winning strategy.
