8

Other exercise which I found in Dudley's Analysis book:

Show that there is a measure on a infinite set $X$, defined on $2^X$ s.t. is finitely additive, $m(A)=0$ for any finite set and $m(X)=1$.

The solution is very simple using the Frechet filter $\mathcal{F}:=\{A: X\setminus A \text{ finite}\}$ and defining the measure on the ultrafilter $\mathfrak{U}$ containing $\mathcal{F}$, as follows

$$m(A)=\begin{cases} 1& A\in \mathfrak{U}\\ 0&A\notin \mathfrak{U} \end{cases}$$

For the following

Lemma: Let $\mathfrak{U}$ be an ultrafilter of subsets of $X$ and let $m$ defined as above. Then $m$ is finitely additive on $2^X$.

PF: It's clear that $\varnothing\notin \mathfrak{U}$, so $m(\varnothing)=0$. Let $\{A_n\}_{n=1}^N\subset 2^{X}$ and disjoint, and let $A$ be their union. We consider two cases: If all are not in $ \mathfrak{U}$, i.e, $X\setminus A_n\in \mathfrak{U}$ for $n\le N$. Thus $X\setminus A=\bigcap_{n\le N}X\setminus A_n\in \mathfrak{U}$, so $A\notin \mathfrak{U}$. Hence $m(A)=\sum_{n\le N}m(A_n)=0$

Now suppose that at least one is in $\mathfrak{U}$. Let $A_1\in \mathfrak{U}$, so all the other elements are not in $\mathfrak{U}$ since otherwise $\varnothing=A\cap A_i\in \mathfrak{U}$ for $i\not=1$. Since $A_1\subset A$ and $A_1\in \mathfrak{U}$, $A\in\mathfrak{U}$. Thus $m(A)=\sum_{n\le N}m(A_n)=1$. $\Box$

Does someone know if there is hope of a constructive approach? I believe the answer is negative...

Martin Sleziak
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Jose Antonio
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    Different exercises, different questions. Would be better. – leo Oct 14 '14 at 15:46
  • What would be better @leo? Put them in different parts. – Jose Antonio Oct 14 '14 at 15:50
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    Take one exercise out of this question and put it in another question. – Mark Fantini Oct 17 '14 at 01:47
  • But both are not incredible long, I think @MarkFantini – Jose Antonio Oct 17 '14 at 01:47
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    It's not about being long, it's about searchability. Two separate exercises in two separate questions is much easier to find than two exercises in one single question. – Mark Fantini Oct 17 '14 at 01:48
  • The Frechet Filter is not an ultrafilter. It is just a filter. If $\mathcal{U}$ is an ultrafilter over $\mathcal{P}(X)$, then for every $Y \in \mathcal{P}(X)$, $Y \in \mathcal{U}$ or $Y \not \in \mathcal{U}$ (where or is exclusive). The Frechet Filter does not contain any countable or co-countable sets and thus is not an ultrafilter.

    However, your measures is defined correctly and indeed works for any filter and therefore works for any ultrafilter.

     – Kyle Gannon Jul 06 '16 at 13:00

1 Answers1

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It is known that existence of a finitely additive measure defined on the whole powerset $2^{\mathbb N}$ which vanishes on finite sets cannot be proved in ZF. So we cannot hope in a constructive proof of existence of such measures.

Some posts where you can find further references:

Some references where similar questions are studied:

  • Howard, Rubin: Consequences of the Axiom of Choice, AMS, 1998. (See AMS website or Google Books.) This book is accompanied by a website where you can search for implications between various consequences of AC and search for models in which a particular consequence of AC holds (or does not hold). Existence of a nonprincipal measure on $2^{\mathbb N}$ is Form 222 in this book. Existence of non-principal measure is Form 221.
  • E. Schechter: Handbook of Analysis and Its Foundations; It is shown in 29.37 that existence of such measures is equivalent (in ZF) to $\ell_1\subsetneq\ell_\infty^*$ and in 28.38 that it implies existence of a set without Baire property.
  • David Pincus and Robert M. Solovay: Definability of measures and ultrafilters. J. Symbolic Logic 42 (1977), no. 2, 179–190.
  • Eric K. van Douwen. Finitely additive measures on $\mathbb N$. Topology Appl., 47 (3), (1992), 223–268.
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Martin Sleziak
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