# Properly showing that uncountable sum of measures finite nonzero sets is infinite, given they are su

Pattab 2022-07-15 Answered
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 $\left(X,\sum ,\mu \right)$ is a finite measure space, I am asked to show that the set $\left\{x|\mu \left(\left\{x\right\}>0\right)\right\}$ 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
You can still ask an expert for help

Expert Community at Your Service

• Live experts 24/7
• Questions are typically answered in as fast as 30 minutes
• Personalized clear answers

Solve your problem for the price of one coffee

• Available 24/7
• Math expert for every subject
• Pay only if we can solve it

## Answers (2)

Ordettyreomqu
Answered 2022-07-16 Author has 22 answers
Define ${A}_{n}:=\left\{x\mid \mu \left(\left\{x\right\}\right)>1/n\right\}$ and consider ${B}_{n}:={A}_{n}\setminus {A}_{n-1}$.
###### Did you like this example?
Sylvia Byrd
Answered 2022-07-17 Author has 6 answers
Define . If ${A}_{n}$ contains at least m points, then $\mu \left({A}_{n}\right)\ge m/n$. Now $\mu \left(A\right)\le \mu \left(X\right)<\mathrm{\infty }$, so $m\le n\mu \left(X\right)$. Hence, ${A}_{n}$ contains at most $n\mu \left(X\right)$ elements. So ${A}_{n}$ is finite. The set in question is contained in the union of these ${A}_{n}$, so it is countable.
###### Did you like this example?

Expert Community at Your Service

• Live experts 24/7
• Questions are typically answered in as fast as 30 minutes
• Personalized clear answers

Solve your problem for the price of one coffee

• Available 24/7
• Math expert for every subject
• Pay only if we can solve it