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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.

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  • $\begingroup$ It seems everyone is assuming the measure is bounded. Do you know what happens when we drop that assumption? $\endgroup$
    – Lxm
    Commented Dec 15, 2021 at 23:53
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    $\begingroup$ @Lxm You can reproduce bof's proof by taking any infinite subset $X\subseteq \mathbb{N}$ such that $0<\mu(X)<\infty$ and restricting $\mu$ to $\mathcal{P}(X)$. $\endgroup$
    – Hanul Jeon
    Commented Dec 16, 2021 at 4:16
  • $\begingroup$ I meant to say an unbounded finitely additive signed measure...so that we don't have monotonicity $\endgroup$
    – Lxm
    Commented Dec 16, 2021 at 5:13
  • $\begingroup$ @Lxm I am not familiar with unbounded signed measure, and so I do not know in that case. $\endgroup$
    – Hanul Jeon
    Commented Dec 16, 2021 at 5:23
  • $\begingroup$ As Asaf says bounded signed measures correspond to functionals on $\ell^\infty$, so consistently they are just $\ell^1$, and this really doesn't use the full power of AD. But that's for bounded measures, so I wonder what about unbounded ones. $\endgroup$
    – Lxm
    Commented Dec 16, 2021 at 5:34

2 Answers 2

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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.

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  • $\begingroup$ You can also find the proof in the note I wrote about Zorn's lemma in functional analysis, which you can find on my homepage. $\endgroup$
    – Asaf Karagila
    Commented Apr 21, 2017 at 5:03
  • $\begingroup$ Existence of such continuous functional is a motivation of my question! I will read your posts and accept the answer, but time does not allow me to read it now. I will check it at night. $\endgroup$
    – Hanul Jeon
    Commented Apr 21, 2017 at 5:08
  • $\begingroup$ @bof: It is my special nickname for Rene-Louis Baire. Specifically, that's how my phone likes to write his name. $\endgroup$
    – Asaf Karagila
    Commented Apr 21, 2017 at 10:34
  • $\begingroup$ @Hanul: You should accept the other answer. $\endgroup$
    – Asaf Karagila
    Commented Apr 21, 2017 at 17:54
  • $\begingroup$ I accepted your answer because it points out the motivation of my question, but I agree with your comment. bof's proof is interesting and easy to understand, so I follow your suggestion :) $\endgroup$
    – Hanul Jeon
    Commented Apr 21, 2017 at 18:02
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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.

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  • $\begingroup$ Nice trick. White still uses the original strategy, imagining the extra numbers she is playing are actually part of the sets played by Black. $\endgroup$ Commented Apr 21, 2017 at 11:32
  • $\begingroup$ Or White can just ignore the extra numbers, computing the strategic response to the sets actually chosen by Black, and then adding extra numbers to fill in the gaps., $\endgroup$
    – bof
    Commented Dec 16, 2021 at 5:19

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