Let $a\_0>a\_1>\cdots>0$ have the property that, for each positive $a<\sum\_{n\in\Bbb N}a\_n$ (admitting $\infty$ for the sum), there is $A\subset\Bbb N$ such that $a=\sum\_{n\in A}a\_n$ . Are there known necessary and sufficient conditions on the $a\_n$ (not involving arbitrary partial sums) for this property? To illustrate, $a\_n=1/(n+1)$ and $a\_n=1/2^n$ possess the required property, but $a\_n=1/(2+\varepsilon)^n$ does not for any $\varepsilon>0$.


(This question is adapted from one asked on [Mathematics Stack Exchange](https://math.stackexchange.com/questions/3596636/surjective-sum-measure) two months ago which received no answer.)



Is there a known condition for partial sums of a decreasing positive sequence to take all values up to the total sum?
It is necessary and sufficient that



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> $\lim\_{n \rightarrow \infty}a\_n = 0$, and
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> $a\_n \leq \sum\_{m > n}a\_m$ for all $n$.
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In other words: the terms go to zero, and no term is bigger than the sum of all the following terms.


*Necessity*: First, it is necessary that $\lim\_{n \rightarrow \infty}a\_n = 0$ because you cannot form any sum smaller than $\lim\_{n \rightarrow \infty}a\_n$.


Suppose $a\_n > \sum\_{m > n}a\_m$ for some $n$. Let $\varepsilon$ be some number with $0 < \varepsilon < a\_n - \sum\_{m > n}a\_m$. Then I claim there is no $A$ such that $\sum\_{m \in A}a\_m = a\_1+a\_2+\dots+a\_{n-1}+a\_n - \varepsilon$. To see this, just consider two cases: (1) if $\{a\_1,\dots,a\_n\} \subseteq A$, then $\sum\_{m \in A}a\_m$ is too big, because $\varepsilon > 0$, and (2) if $a\_i \notin A$ for some $i \leq n$, then $\sum\_{m \in A}a\_m$ is too small, because
$\sum\_{m \in A}a\_m \,\leq\, (a\_1+a\_2+\dots+a\_n) - a\_i +\sum\_{m > n}a\_m \,\leq\, (a\_1+a\_2+\dots+a\_n) -a\_n + \sum\_{m > n}a\_m < (a\_1+a\_2+\dots+a\_n) - \varepsilon.$


*Sufficiency:* Suppose $a\_n \leq \sum\_{m > n}a\_m$ for all $n$, and let $c$ be any number with $0 \leq c \leq \sum\_{m \in \mathbb N}a\_m$. Then we can construct the desired $A \subseteq \mathbb N$ recursively, as follows. If it has already been decided for all $m < n$ whether $m \in A$ or not, then put $n \in A$ if and only if $a\_n + \sum\_{m \in A \cap \{1,2,\dots,n-1\}} a\_m \leq c$. (In other words, put $n \in A$ if and only if putting $n \in A$ does not make the sum too big.) Once we have built $A$ according to this rule, it is clear that $\sum\_{m \in A}a\_m \leq c$, because none of the finite partial sums exceeds $c$.


Now suppose, aiming for a contradiction, that $\sum\_{m \in A}a\_m < c$, and let $\varepsilon = c - \sum\_{m \in A}a\_m$. There is some $N$ such that $a\_n < \varepsilon$ for all $n \geq N$. For each such $n$, we have $\sum\_{m \in A \cap \{1,2,\dots,n-1\}} a\_m \leq \sum\_{m \in A} a\_m = c - \varepsilon$, and hence $a\_n + \sum\_{m \in A \cap \{1,2,\dots,n-1\}} a\_m < c$. By our rule for constructing $A$, this means $n \in A$ for all $n \geq N$. In other words, $A$ is a co-finite subset of $\mathbb N$.


Let $n$ denote the largest member of $\mathbb N \setminus A$. (Note that $\mathbb N \setminus A \neq \emptyset$, because $\sum\_{m \in A}a\_m < c \leq \sum\_{m \in \mathbb N}a\_m$.) Then $\left( \sum\_{m \in A \cap \{1,2,\dots,n-1\}} a\_m \right)+a\_n \leq \left( \sum\_{m \in A \cap \{1,2,\dots,n-1\}} a\_m \right)+ \sum\_{m > n}a\_m = \sum\_{m \in A}a\_m < c$. This is the contradiction we were after, because this tells us that we should have had $n \in A$, although $n$ was supposed to be the largest number not in $A$.


