Let X be non-empty and consider $a : X → [0,+\infty]$. We define $\mu : 2^X → [0,\infty] $ by $$\mu(E)=\sum_{x\in E} a(x)$$ for every $E\subset X$. The measure (proposition easy to prove) on $(X,2^X)$ defined above is called the point-mass distribution on X induced by the function a.
I am trying to solve the following exercise
Let $X\neq\emptyset$. Prove that every measure $\mu$ on $(X,2^X)$ is a point-mass distribution.
So we want to define a function $a$ on X such that $$\mu(E)=\sum_{x\in E} a(x)$$ for every $E\subset X$.
If the set X is at most countable this obvious. We just have to take $a(x)=\mu(\{x\})$, for every $x\in E$ and use the countable additivity of the measure to gain the result, since $E=\cup_{x\in E}\{x\}$. But we can't use the same argument in the general case.
Which is my question.
Also, in order to solve the exercise, I tried to find the point-mass distribution for a specific measure on a general non-empty set X. So I considered the measure $\mu(E)=0$, if $E$ is countable or else is $+\infty$. Which confused me more.
The question appears as exercise #15 in page 31 of Notes on Measure Theory The discussion preceding the exercise addresses point-mass-distributions.
On page 23, the measure of a set E is defined as the supremum of the sum of measures of finite subsets of E, not very different from what was suggested by drhab.
This definition is not standard in measure theory and if adopted it implies that a measure on a power set must be a point mass measure. In this sense the exercise is true.
However, without this assumption the claim is wrong. Take, for example, a non-countable set and define a measure on its subsets so that countable sets have measure 0 and non-countable sets have an infinite measure.