Is an uncountable sum of measures of finite non-zero sets infinite? Properly showing that uncountable sum of measures finite nonzero sets is infinite, given they are subsets of $X$ of finite measure.
In a question where $(X,\sum,\mu)$ is a finite measure space, I am asked to show that the set $\{x|\mu(\{x\}>0)\}$ is countable at most.
While intuitively this has to be true, because as long as it is countable any series of nonzero measures of sets must converge and this can happen for geometric sequences of measures, but for an uncountable sum, this "can't" be the case, but how to show that- I can't seem to understand.
Firstly, is this the way to solve it? Trying to reach a contradiction, or maybe uncountable sum of elements is generally meaningless as an expression? I'm not sure, what
 A: Hint: Define $A_n := \{x \mid \mu(\{x\}) > 1/n\}$ and consider $B_n := A_n \setminus A_{n - 1}$.
A: Define $A_n = \{x: \ \mu(\{x\}) \ge 1/n\}$. If $A_n$ contains at least $m$ points, then $\mu(A_n) \ge m/n$. Now $\mu(A)\le\mu(X)<\infty$, so $m\le n\mu(X)$. Hence, $A_n$ contains at most $n \mu(X)$ elements. So $A_n$ is finite. The set in question is contained in the union of these $A_n$, so it is countable.
A: I think it is worth noting that the methods suggested by the other posts also work in case of arbitrary sets which are not necessarily singletons.
Namely, if $\mathcal A\subseteq \Sigma$ is an uncountable family of pairwise disjoint, measurable sets of strictly positive measure, then there is a countable subfamily $\mathcal A_0\subseteq \mathcal A$ such that $\bigcup \mathcal A_0$ is infinite, and so $\bigcup \mathcal A$ also has infinite measure (provided it is measurable - otherwise, it has infinite inner measure).
The reason for that is that for some $n$, there are uncountably many (and therefore, infinitely many) sets in $\mathcal A$ of measure at least $\frac{1}{n}$. The conclusion easily follows.
As alluded to in some other posts, this is basically equivalent to the observation that if $(a_i)_i$ is an uncountable sequence of strictly positive reals, then necessarily $\sum_i a_i=\infty$ (for any sensible interpretation of the left hand side - there are at least a couple of those, but all that I can think of coincide in this case).
