Im trying to solve the following exercise: Let $X$ be a compact metric space and let $\mu$ be a finite Borel measure on $X$ such that $\mu(\{x\}) = 0$ for all $x \in X$ (Here $B(x,\delta)$ denotes the open ball with center at $x$ and radius $\delta$.) Show that the measure $\mu$ has the following property: $$\forall x\in X : \forall\epsilon > 0 : \exists\delta > 0 \text{ such that } \mu (B(x,\delta))\leq\epsilon$$
My attempt is: Given any $x\in X$ and $\epsilon > 0$, we want to find a $\delta > 0$ such that $\mu(B(x,\delta)) \leq \epsilon$. Since $\mu$ is a finite measure, for any $\delta > 0$, we have $$\mu (B(x,\delta ))\leq\mu (X)<\infty$$
So we can choose $\delta$ small enough such that $\mu(B(x,\delta)) < \frac{\epsilon}{\mu(X)}$. Then we have $$\mu (B(x,\delta ))\leq \frac{\epsilon}{\mu (X)}\mu (X)=\epsilon$$
Thus, we have found a $\delta$ such that $\mu(B(x,\delta)) \leq \epsilon$, as desired.
But I dont think this is right but cannot see why or what, can anyone see if it is rigth or if I need to redo it all ...
I think your statement "So we can choose $\delta$ small enough such that $\mu(B(x,\delta))<\epsilon \mu(X)$" is begging the question, i.e. it's not obvious that one can do that without assuming the original statement you are trying to prove.
My approach would be the following: Use the so-called continuity from above of finite measures $\mu$: Let $\mu$ be a finite measure on $X$ and let $(A_n)_{n\in\mathbb N}$ be a sequence of measurable sets in $X$ such that $A_{n+1}\subseteq A_n$, i.e. the sequence is decreasing. Then we have $$\mu\bigg(\bigcap_{n\in\mathbb N} A_n\bigg)=\lim_{n\to\infty} \mu(A_n).$$ Now simply note the sequence $A_n:= B(x,\frac{1}{n})$ is such a decrasing sequence and apply this result.