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I am looking for an example of a probability measure $\mu$ on $[0,1]$ such that

  1. $\mu(A)$ is defined for ALL subsets $A\subset[0,1]$
  2. $\mu$ is finitely additive but not $\sigma$-additive
  3. No atom: If $\mu(A)>0$ then for all $r\in(0,1)$, there is $B\subset A$ such that $\mu(B)=r\mu(A)$

Where can I find an example of such measure? with an explicit formula for $\mu(A)$ for any $A\subset[0,1]$?

Yi-Hsuan Lin
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    See this question – lulu Apr 16 '24 at 12:20
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    If $\mu$ extends the interval length function then I think 3. is automatically true (as the function $t \mapsto \mu([0, t) \cap A)$ will be continuous). It's known that there are finitely additive extensions of the Lebesgue measure - this follows from Hahn-Banach, for example. You can even ask for translation-invariance! See here. It's unlikely that there is a nice explicit formula - I think such a function requires some use of the axiom of choice to construct. – Izaak van Dongen Apr 16 '24 at 13:54
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    See my answer here: https://math.stackexchange.com/a/4892500/465145 . The measure constructed there, when restricted to $[0, 1]$, satisfies the conditions you asked for. Note that the use of axiom of choice is required - I used the existence of ultrafilters there. As @IzaakvanDongen already pointed out, this is unavoidable. In particular, no explicit formula can be given for $\mu$. – David Gao Apr 16 '24 at 17:15

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