Examples of measures that assign positive measure to countable sets of $\mathbb{R}^n$? 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?
 A: 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.
A: The questions is: 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 $A\subset\mathbb{R}^n$?
Answer:
Using the Borel (or the Lebesgue) $\sigma$-algebra on $\Bbb R^n$, let $\nu_1$ be any measure (not signed measure) such that $\nu_1$ is not identically zero and, for all $x\in \Bbb R^n$, $\nu_1(\{x\})=0$.
Let $C$ be any non-empty countable subset of $\Bbb R^n$ and let $f: C \rightarrow (0,+\infty]$ be a function. For any measurable set $A$, define $\nu$ by
$$ \nu(A)= \nu_1(A) + \sum_{x\in A\cap C} f(x) $$
Then $\nu$ is a measure. For all measurable set $A$, $\nu(A)\geqslant 0$ and, for all $E \subseteq C$, if $E \neq \emptyset $ then $E$ is countable and $\nu(E) >0$.  However $\nu$ is not a Dirac measure.
Remark 1:  The construction of the measure $\nu$ is a general construction. Any measure constructed in this way will a measure positive for some countable subset of $\Bbb R^n$, but not a Dirac measure.
Remark 2: We can generalize the answer even further:
Assume your $\sigma$-algebra is such that all one-point sets are measurable.
Let $\nu_1$ be any measure (not signed measure) such that, for all $x\in \Bbb R^n$, $\nu_1(\{x\})=0$, $C$ be any non-empty countable subset of $\Bbb R^n$ and $f: C \rightarrow (0,+\infty]$ be a function. Then $\nu$ defined by, for all $A$ measurable,
$$ \nu(A)= \nu_1(A) + \sum_{x\in A\cap C} f(x) $$
is a measure that is positive on a countable set. The Dirac measures are just special cases where $\nu_1$ is identically zero and $C$ is an one-point set.
Moreover, in the "reverse direction", given  any any measure $\nu$, such that $\nu$ is positive on a countable set, and  for all $x\in \Bbb R^n$, $\nu(\{x\})<+\infty$, then $\nu$ can be obtained from the above construction with $f: C \rightarrow (0,+\infty)$.
