Radon measure taking integer values has discrete support Let $X$ be a Polish space and $M(X)$ be the set of Radon measures on $X$. Can you explain why if $L \in M(X)$ is integer valued, then $L$ is supported on a discrete set?
 A: A Radon measure is inner regular. This means that if $A$ is Borel then $L(A)=\sup\{L(K): K\text{ is compact}, K\subset A\}$.
If $L\neq0$ then there is a ball $U$ (use a metric on $X$ which makes it complete) with $L(U)\neq0$. By inner regularity there is a compact $K\subset U$ such that $L(U)-\frac{1}{3}<L(K)\leq L(U)$. Since $L$ takes only integer values then $L(K)=L(U)$.
Now we begin chasing the support points inside these compact. Stand on a point $a\in K$. 
If for all balls $V$ centered at $a$ we have $L(V)\neq0$ then $L(\{a\})=\lim L(V_n)\neq0$, where $V_n$ is a decreasing sequence of balls with $a$ as intersection. The inequality follows since the $L(V)$ are non-zero integers. Moreover, there is some $n_0$ such that for $n>n_0$ we have $L(V_n)=L(\{a\})$. Therefore $a$ is isolated from other density points.
If there is a ball $a\in V\subset U$ such that $L(V)=0$ we consider $K\setminus V\subset K$ as the new compact which satisfies $L(K\setminus V)=L(K)=L(U)$. The intersection of all compacts found in this way is non-empty. So, some point in that intersection falls in the previous case of having all balls around it with non-zero measure.
