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Choose a sequence $(x_n)_{n\in\mathbb N}$ of nonnegative real numbers with finite sum $x=\sum\limits_{n\in\mathbb N}x_n$ and consider the set $X=\{x_I\mid I\subseteq \mathbb N\}$ where, for every $I\subseteq \mathbb N$, $x_I=\sum\limits_{n\in I}x_n$. Thus, $X\subseteq[0,x]$.

Question: Can the set $X$ fail to be closed?

A motivation from probability theory is explained there. Note that $X=[0,x]$ if $x_n=1/2^n$ and that $X\neq [0,x]$ if $x_n=a^n$ for some $a$ in $(0,1/2)$ (yielding measure zero Cantor sets), but that in all these cases $X$ is closed.

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Since $x_{(\cdot)}:2^\Bbb N\to [0,\infty)$ is a finite positive measure, it is interesting whether a more general result hold. For any finite positive measure $\mu$ on a measurable space $(X,\mathfrak B_X)$ its range $\mu(\mathfrak B_X)$ is a closed subset of $\Bbb R$. –  Ilya Mar 16 '13 at 17:18
The downvote boggles the mind. –  cardinal Mar 16 '13 at 17:38
@Ilya Endow the sigma-algebra $\mathfrak F$ with the metrics $d(A,B)=\mu(A\Delta B)$ (or rather, the quotient of $\mathfrak F$ by the negligible sets). Is $(\mathfrak F,d)$ a compact metric space? –  Did Mar 16 '13 at 17:57
@cardinal: Yup. –  Did Mar 16 '13 at 17:59
@Did: I guess, since $\mu$ is continuous w.r.t. $d$, compactness of $\mathfrak F$ implies the positive answer to the general answer. Sequential compactness, using boundness of $\mu$, perhaps holds indeed - though I don't see it now. Before I left the first comment, I was rather thinking about the following, related problem: if $\lim_n \mu(A_n) = m$, can we assume that $\{A_n\}$ is a monotone sequence? (for example, w.l.o.g. $\{\mu(A_n)\}$ is monotone) I tried to work with $\liminf/\limsup$ but didn't come up with a proof or found counterexamples. –  Ilya Mar 16 '13 at 19:14
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2 Answers

up vote 13 down vote accepted

Consider the compact metric space $(A,d)$ where $A$ is the set of binary sequences with metric $$d(a,b) = \sum_{n=1}^{\infty} \frac{ |a_i - b_i|}{2^n}.$$ Define $f:A\to [0,x]$ by $f(a) = \displaystyle \sum_{n=1}^{\infty} a_n x_n.$ Then $f$ is continuous so $f(A)=X$ is compact.

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@DavideGiraudo Suppose $a^{(m)}\to a$ and let $\epsilon>0.$ There exists an $N$ such that $\sum_{n>N} x_n < \epsilon.$ So $$| f( a^{(m)} ) - f(a) | < \epsilon+ \sum_{n=1}^N | a^{(m)}_n - a_n | x_n .$$ Letting $m\to\infty$ gives $\limsup_{m\to\infty} |f(a^{(m)}) - f(a)| \leq \epsilon.$ –  Ragib Zaman Mar 16 '13 at 11:15
@RagibZaman Thanks for this nice answer. Any idea about the more general question asked in a comment? –  Did Mar 16 '13 at 18:10
@Did I've had no luck showing that space is compact, sorry. –  Ragib Zaman Mar 17 '13 at 4:52
@RagibZaman: perhaps, because it's not compact –  Ilya Mar 17 '13 at 21:54
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The range of an $n$-dimensional vector measure is always closed. See the paper (and references therein)

P. R. Halmos (1948), The range of a vector measure, Bull. Amer. Math. Soc. 54, 416–421.

So the set will always be closed.

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Thanks for the (freely available) reference. –  Did Mar 20 '13 at 14:13
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