Generalized Counting Measure Let $E$ be an uncountable set. Fix a specific countable subset $M$ from $E$, namely
$$M:=\{x_1,x_2,\ldots\}\subset E.$$
Fix a sequence $\{c_i\}_{i=1}^\infty\in\ell_1(\mathbb{N})$. For any subset $S$ of $E$, we define the measure as
$$\mu(S):=\sum\limits_{x_i\in S}|c_i|.$$
So is this meansure well-defined? If it is, for any function $f$ on $E$, the integration with the measure above should be
$$\int_E fd\mu:=\sum\limits_{n=1}^\infty f(x_i)\cdot |c_i|.$$
Is this correct? Thanks.
 A: The answer to both questioins is YES and this is stanadrd in Probailtiy Theory. Counatble addiitvity of $\mu$ follows from the fact that we can always interchange two infinite sums when the terms are non-negative. The last part holds whenver $f$ is non-negative or integrable w.r.t. $\mu$. (The measure $\mu$ is defined on the power set and every function is measurable).
A: Yes, the measure $\mu$ is well-defined in the $\sigma$-algebra $2^E$. An easy way to see this is to note that, for any $S \subseteq E$,
$$\mu(S)=\mu(S \cap M) = \sum\limits_{x_i\in S \cap M}|c_i|$$
It is then immediate to see that $\mu(\emptyset)=0$ and that $\mu$ is countably additive.
Since the $\sigma$-algebra is $2^E$, it follows that any function $f$ defined on $E$ is measurable.
Now, note that $\mu(E\setminus M)=0$ and, for all $x_i \in M$,  $\mu({x_i}) =|c_i|$.  It follows that
$$\int_E fd\mu=\int_{E \setminus M} fd\mu+\int_M fd\mu=\int_M fd\mu=\sum\limits_{n=1}^\infty f(x_i)\cdot \mu({x_i})=\sum\limits_{n=1}^\infty f(x_i)\cdot |c_i|.$$
