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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

Asaf Karagila
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טל אברבנאל
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  • Do you have any additional hypotheses about $(X, \Sigma, \mu)$? I don't see why one would believe that ${ x\in X: \mu(x) > 0}$ is a countable set. – Matt E. Nov 10 '21 at 14:04
  • I am trying to reach a contradiction, so that if the set of elements whose singletons are of positive measure is uncountable that would somehow contradict the fact the X is of finite measure. – טל אברבנאל Nov 10 '21 at 14:06
  • In this answer, it is shown that an uncountable sum is finite only if all but a countable number of the terms is $0$. – robjohn Nov 11 '21 at 20:25

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Hint: Define $A_n := \{x \mid \mu(\{x\}) > 1/n\}$ and consider $B_n := A_n \setminus A_{n - 1}$.

user7427029
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  • Looks great I am going to give it a try, thank you! – טל אברבנאל Nov 10 '21 at 14:09
  • You are welcome. I learnt the proof in a different setting (series in $\mathbb{R}$) in Functional Analysis and it was a "Woah!" moment of pure joy. – user7427029 Nov 10 '21 at 14:11
  • Just a question: $B_i$ are disjoint, and so the measure of their union (not including $A_i$) is bigger than $\sum {1\over n}$ which is divergent, is it enough to say that the each $\mu(B_i)$ is bigger than $\mu({x}$ for $x\in B_i$? – טל אברבנאל Nov 10 '21 at 14:28
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    The sets $A_n$ contain finitely many points only. Why consider those $B_n$'s? – daw Nov 10 '21 at 14:34
  • It contains finitely many points because otherwise there was a countable collection of (obviously disjoint) singletons whose measures are $1 \over n$, which is a contradiction? – טל אברבנאל Nov 10 '21 at 14:42
  • @טלאברבנאל Yes, if for some $n$ there were $n+1$ singletons with measure $>\mu(X)/n$ then the measure of the union of these singletons would be greater than $\mu(X).$ Agree with daw that disjointifying the $A_n$ seems unnecessary. The key thing is that if ${x:\mu({x})>0}$ is uncountable, then one of the $A_n$ needs to be infinite, which gives the aforementioned contradiction. – spaceisdarkgreen Nov 10 '21 at 15:01
  • @Question author (left-right / right-left writing does not work well in referencing) That the sequence $\mu(B_i)$ is greater than $\mu({x, x \in B_i})$ does not suffice, since you get something like $\sum \mu({x, x \in B_i}) \leq \sum \mu(B_i) = \infty$, i. e. LHS is not guaranteed to be $\infty$. Or did I understand you wrongly? – user7427029 Nov 11 '21 at 16:00
  • You simply do not have $\forall x \in \mathbb{R} \exists i \in \mathbb{N}: \mu({x} \geq \mu(B_i)$. You only have $\leq$. – user7427029 Nov 11 '21 at 16:02
  • @daw: You are considering an union of sets. The $A_i$ are not disjoint, therefore we are not allowed to write $\mu(\Bigcup A_i) = \sum \mu(A_i) = \infty$, only $\leq$. – user7427029 Nov 11 '21 at 16:05
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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.

daw
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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).

tomasz
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