$\left\{x:\mu(\left\{x\right\})>0\right\}$ is countable.

Question

For all Baire measure with $\mu(X)<\infty$ with $X$ compact Hausdorff, the set $A:=\left\{x:\mu(\left\{x\right\})>0\right\}$ is countable.

I have this.

$A=\bigcup_n A_n$ with $A_n=\left\{x:\mu(\left\{x\right\})\geq \frac{1}{n}\right\}$

Now, $A_n=\bigcup_{x\in A_n} \left\{x\right\}$ and $A_n\subset X$ then $|A_n|\frac{1}{n}=\sum_{x\in A_n} \frac{1}{n}\leq \sum_{x\in A_n}\mu(\left\{x\right\})=\mu(\bigcup_{x\in A_n}\left\{x\right\})=\mu(A_n)\leq \mu(X)$ Then $|A_n|\leq \mu(X)n$ then $A_n$ is countable.

But according to me there is an error in what I have done. The union $\bigcup_{x\in A_n}\left\{x\right\}$ is not countable ...then $\mu(\bigcup_{x\in A_n})=\sum_{x\in A_n} \mu(\left\{x\right\})$ is incorrect...

How can I prove it?

Answer 1

We do have $|A_n|\leq n\mu(X)$. For if $|A_n|>n\mu(X)<\infty$ then $A_n$ has a finite subset $B$ with $|B|>n\mu(X).$ But then $$\mu(X)\geq \mu(B)= \sum_{b\in B}\mu (\{b\})\geq$$ $$\geq \sum_{b\in B}(1/n)=|B|(1/n)>\mu(X),$$ which is absurd. So each $A_n$ is finite. So $A$ is a countable union of finite sets.

Answer 2

Each $A_n$ is finite, otherwise its measure is infinite. So you go through all of $A_1$, then all of $A_2$, and so on.