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This question arose when I saw that some people define a discrete probability distribution using the probability function.

Let $p: \mathbb{R} \to [0,1]$ be a probability function. I.e., $S:=\{p \neq 0\}$ is countable and $\sum_{x \in S} p(x) = 1$. Is it true that there exists a probability measure $\mu: \mathcal{R} \to [0,1]$ such that $\mu(\{x\}) = p(x)?$

I think the answer is yes.

Define $\mu: \mathcal{R} \to [0,1]: A \mapsto \sum_{x \in A \cap S} p(x)$. The function is well defined by absolute convergence of the series (and hence commutative convergence)

Then, $\mu(\mathbb{R}) = \sum_{x \in S} p(x) = 1$

and if $(A_n)_{n\geq1}$ is a sequence of disjoint sets, then:

$$\mu\left(\bigcup_{n=1}^\infty A_n\right) = \sum_{x \in S \cap\bigcup A_n} p(x) = \sum_{x \in \bigcup (A_n \cap S)} p(x) = \sum_{n=1}^\infty \sum_{x \in A_n\cap S}p(x) = \sum_{n=1}^\infty \mu(A_n)$$

and hence $\mu$ is a probability measure that satisfies the condition.

Is this correct? As a follow up question, is a probability measure $\mu$ satisfying the condition I wrote down unique?

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  • $\begingroup$ Yes, you are correct. This is true for arbitrary $\sigma$-algebras on the real line. The measure you defined is unique; if there was two of them their defining characteristic would immediately show that they are equal on arbitrary sets. $\endgroup$ – Joker123 May 6 '18 at 10:25
  • $\begingroup$ I don't quite understand the uniqueness. Can you elaborate? $\endgroup$ – user370967 May 6 '18 at 11:04
  • $\begingroup$ Imagine there was two measures $\mu_1$ and $\mu_2$ with all the characteristics you mentioned above. Then, for every $A$ in your $\sigma$-algebra, you have, $\mu_1(A) = \sum_{x \in A \cap S} p(x) = \mu_2(A)$. This equality holds per definition, since both measures are defined via this equality. Since both measures are the same on arbitrary sets, they are equal. $\endgroup$ – Joker123 May 6 '18 at 19:40
  • $\begingroup$ You assume that $\mu(A)$ is given by that sum. But I wonder if there exists a measure, not necessarily given by such a sum, which has the properties I described. $\endgroup$ – user370967 May 6 '18 at 20:28
  • $\begingroup$ Now I understand what you mean. $\endgroup$ – Joker123 May 7 '18 at 20:16
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You are right and uniqueness holds. To see this, let $\mu: \mathcal{R} \to [0,1]$ be a probability measure such that $\mu(\{x\}) = p(x)$. Then, for every $A \in \mathcal{R}$, we have $$ \mu (A) = \mu ((A \cap S) \cup (A \cap S^{c})) = \mu ((A \cap S)) + \mu (A \cap S^{c}) = \mu ((A \cap S)) + 0 = \mu( \bigcup_{x \in A \cap S} \{x\}) = \sum_{x \in A \cap S} \mu(\{x\}) $$

In this chain of equations, we have used the fact that probability measures are $\sigma$- additive (and therefore also finitely additive) two times.

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