Let $A\subset\mathbb{R}^n$ be a countable (and finite) set of Real-valued $N$-dimensional vectors. The Lebesgue measure of this set $A$, denoted $\lambda(A)$, is just $\lambda(A)=0$ because the Lebesgue measure assigns measure zero to all countable subsets. The Dirac measure of this set $A$, denoted $\delta(A)$, may or may not be zero, since:
\begin{gather}
\delta_x(A)=
\begin{cases}
1&\text{if }x\in A\\
0&\text{if }x\notin A
\end{cases}
\end{gather}
for some $x\in\mathbb{R}^n$. My first question is this: in case they exist, can you provide examples of other measures that are not the Dirac one, that can be positive for some countable set $A\subset\mathbb{R}^n$?
BONUS QUESTION: consider some (possibly uncountable and infinite) set $B\subset\mathbb{R}^N$ and consider this construction:
\begin{gather}
\delta_B'(A)=
\begin{cases}
1&\text{if }A\cap B\neq\emptyset\\
0&\text{if }A\cap B=\emptyset
\end{cases}
\end{gather}
My bonus question is as follows: is the construction $\delta_B'(A)$ a measure? Why or why not?
EDIT: While $\delta_B'$ satisfies the properties of 'non-negativity' and 'null empty set', I think it fails to satisfy 'countable additivity'. Consider two sets $A,A'\subset\mathbb{R}^N$ such that $A\cap A'=\emptyset$ yet $A\cap B\neq\emptyset$ and $A'\cap B\neq\emptyset$. Then, $\delta_B'(A)=1$, $\delta_B'(A')=1$ and $\delta_B'(A\cup A')=1$. However, $\delta_B'(A)+\delta_B'(A')=2$, thus implying $\delta_B'(A\cup A')\neq\delta_B'(A)+\delta_B'(A')$ and therefore countable additivity fails to be satisfied. Is this reasoning correct?
Best Answer
Your analysis of $\delta_B'$ is accurate.
Seperately, measures of the sort you seek do not really exist. Suppose $C$ is a countable set with $\mu(C)>0$. Suppose moreover that, for each $x\in C$, the set $\{x\}$ is measurable. Then, by countable additivity, we must have $\mu=\sum_{x\in C}{\mu(\{x\})\delta_x}$ (as measures on $C$). So any countably-supported measure on $\mathbb{R}^n$ with the Lebesgue $\sigma$-algebra is expressible in terms of Dirac masses.
But that caveat about $\sigma$-algebras cannot be removed! For example, consider the set $\mathbb{Z}^+\times\{0,1\}$, with the $\sigma$-algebra $\{S\times\{0,1\}:S\subseteq\mathbb{Z}^+\}$. (That is, any measurable set cannot determine the second coordinate.) Dirac masses aren't measures with respect to this $\sigma$-algebra, so (for example) the measure $$\mu(S\times\{0,1\})=\sum_{s\in S}{2^{-s}}$$ cannot be written in terms of them.