Finite Borel measure on a compact Metric Space 
Suppose that $\mu$ is a finite Borel measure on a compact metric space $X$ and that $\mu(\{x\}) = 0$ for all $x\in X$. Show that for every $\epsilon > 0$ there is a $\delta > 0$ such that for all $x\in X$, $\mu(B(x,\delta))<\epsilon$ where $B(x,\delta)$ denotes the ball of radius $\delta$ centered at $x$.

Attempted proof - Let $\epsilon > 0$ and suppose there is $\delta > 0$ such that there is an ball $B(x,\delta)$ of radius $\delta$ centered at $x$. Further, suppose that $\mu$ is a finite Borel measure on a compact metric space $X$ and $\mu(\{x\}) = 0$. Then from continuity from above $$\lim_{\delta\rightarrow 0}B(x,\delta) = 0$$ Thus there exists a $\delta_x$ such that $B(x,\delta_x)<\epsilon$. Now, since $X$ is a compact metric space, we can choose points $x_1,\ldots, x_n$ such that $X\subset \bigcup_{1}^{n}B(x_j,\delta_{x_j/2})$.
Let $\delta = \min\{\delta_{x_1/2},\ldots,\delta_{x_n/2}\}$ and suppose $x\in X$. Then for some $j$, $x\in B(x_j,\delta_{x_j/2})$ thus $B(x,\delta)\subset B(x_j,\delta_{x_j})$. Applying monotonicity we have $$\mu(B(x,\delta))\leq \mu(B(x_j,\delta_{x_j}) < \epsilon$$
Not sure if this is correct any suggestions is greatly appreciated.
 A: @Wolfy, your proof is fine. I have only two comments:


*

*As @DavidC.Ullrich pointed out, you don't need "suppose there is $\delta > 0$ such that there is an ball $B(x,\delta)$ of radius $\delta$ centered at $x$". In any metric space $X$, for all $x \in X$ and all $\delta >0$ there is a ball $B(x,\delta)$.

*You used (correctly) the fact that $\mu$ is finite. I suggest you highlight this point. 
Below is you proof with those two minor adjustments (and the correction of some typos)

Suppose that $\mu$ is a finite Borel measure on a compact metric space $X$ and that $\mu(\{x\}) = 0$ for all $x\in X$. Show that for every $\epsilon > 0$ there is a $\delta > 0$ such that for all $x\in X$, $\mu(B(x,\delta))<\epsilon$ where $B(x,\delta)$ denotes the ball of radius $\delta$ centered at $x$.

Proof: Let $\epsilon > 0$ and suppose that $\mu$ is a finite Borel measure on a compact metric space $X$ and $\mu(\{x\}) = 0$. Then, since $\mu$ is finite, from continuity from above $$\lim_{\delta\rightarrow 0}\mu(B(x,\delta))=\mu(\{x\}) = 0$$ Thus there exists a $\delta_x$ such that $\mu(B(x,\delta_x))<\epsilon$. Now, since $X$ is a compact metric space, we can choose points $x_1,\ldots, x_n$ such that $X\subset \bigcup_{1}^{n}B(x_j,\delta_{x_j}/2)$.
Let $\delta = \min\{\delta_{x_1}/2,\ldots,\delta_{x_n}/2\}$ and suppose $x\in X$. Then for some $j$, $x\in B(x_j,\delta_{x_j}/2)$ thus $B(x,\delta)\subset B(x_j,\delta_{x_j})$. Applying monotonicity we have $$\mu(B(x,\delta))\leq \mu(B(x_j,\delta_{x_j}) < \epsilon$$
A: This should be a comment but may be too long to fit. (Your proof is OK  with the adjustments already noted by others.) Another way, although almost the same, is by contradiction : If not, then there exists $e>0$ such that  $$\forall n\in N\; \exists  x_n\in X \; (\mu (B(x_n,1/n)\geq e).$$ Choosing such an $x_n$ for each $n,$ there exists a  subsequence $(x_{j_n})_n$ converging to a point $x\in X.$ But for any $m\in N,$ take $n$ large enough that $d(x,x_{j_n})<1/ 2 m.$ Then $B(x,1/m)\supset B(x_{j_n},1/j_n),$ so $\mu (B(x,1/m))\geq e$ for all $m\in N,$ contradicting continuity from above.
Compactness of $X$ cannot be omitted.For example, let $X$ be the real interval $[0,1)$ with the usual metric; let $f(p)=\tan (p\pi /2)$ for $p\in X;$  let $L$ be Lebesgue measure..... Now for $A\subset X$ and for $A\in$ dom $(l)$ let $\mu (A)=L(f(A)).$
