# Proving the sum of the point masses of a complex Borel measure is in the dual of $M(X)$.

Let $C_0(X\subset \mathbb{R}^n)$ denote the space of continuous functions on $X$ that vanish at $\infty$ (meaning for all $\epsilon>0, \{x: |f(x)|\ge \epsilon\}$ is compact). For $\mu$ a complex Radon measure on $X$, that is for $\mu\in M(X)$, consider $$\Phi(\mu)=\sum_{x\in X} \mu(\{x\}).$$

I would like to show that this is a well-defined function on the dual $M(X)^*$, and if there is a $\mu\not\equiv 0$, a complex Radon measure, such that $\mu(\{x\})=0$ for all $x\in X$, then $\Phi$ is not in the image of $C_0(X)$ in $M(X)^*\cong C_0(X)^{**}$.

This is a slightly more specific case of part (a) of Folland, 7.20.

Work so far:

To show this function is well-defined I would like to show that the sum $\sum_{x\in X} \mu(\{x\})$ is absolutely convergent. We know that $\mu$ is bounded, since this is true of all complex Borel measures, but given that the sum is possibly uncountable I'm not sure how to proceed.

If the sum makes sense, then linearity is obvious and I think it is clear that $\Phi$ is bounded by $|\mu|(X)$. Is this obvious?

Now suppose we have shown that $\Phi$ is in the dual of $M(X)$, and suppose there is a nonzero $\mu\in M(X)$ such that $\mu(\{x\})=0$ for all $x\in X$, and consider $|\mu|$. We know that there is some $f\in C_0(X)$ such that $\Phi(|\mu|)=\int f d|\mu|>0$, but I'm not sure how to proceed from there.

Any help would be greatly appreciated.

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I assume that $X$ is a locally compact subspace of $\mathbb R^n$. (which is not true for all subspaces of $\mathbb R^n$). In order to show the first part, it suffices to prove that $\{x: \mu(\{x\})\ne 0\}$ is countable. Suppose not, then there is an uncountable subset $A\subseteq X$ so that $|\mu(\{x\})|\geq 1/n$, shrinking $A$ we may assume that $\mu(\{x\})\geq 1/n$ (the case $\mu(\{x\})\leq -1/n$ is similar). There is an $r>0$ so that $\{ x\in A: \|x\|\leq r\}$ is uncountable. Let $\overline{B_r}$ denote the closed ball or radius $r$.
Now note that if $\{x_1,...,x_m\}\subseteq A\cap \overline{B_r}$ then $\mu(\{x_1,...,x_m\})\geq m/n$. It follows that $\mu(\overline{B_r})=\infty$ which is a contradiction.
Ok, I think I've got it. As shown in the other answer, $\Phi$ is well-defined. It is obviously linear and $|\Phi(\mu)|\le ||\mu\||,$ so $\Phi\in M(X)^*$.
By the Riesz Representation Theorem, $\Phi(\mu)=\int f\ d\mu$ for some $f\in C_0(X)$. We can verify that $f$ is identically $1$ by considering $\mu$ to be the point mass at any point. Thus if $\mu$ is nonzero, but zero on points, then $\Phi(\mu)=0=\mu(X)$, which is a contradiction, therefore $\Phi$ is not in the image of $C_0(X)$.