# How do you infer from the axiom of finite additivity that $P$ is a measure?

A probability space is made up of three parts $$(\Omega,$$F$$,$$P$$)$$.

$$\Omega$$, the sample space, is a set. $$F$$ is an algebra or a $$\sigma$$-algebra and $$P$$ is a probabilitiy function. That's to say, $$P: F \mapsto \mathbb{R}$$ verifying:

1) $$P(\Omega) = 1$$

2) $$P(A) \geq 0$$ for every $$A \in F$$

3) If $$A,B \in F$$ and $$A \cap B= \emptyset$$, then $$P(A \cup B)= P(A)+P(B)$$

There's a remark saying that $$P$$ is a measure because of 3).

How do you infer from 3), a sum of two probabilities, that the function $$P$$, (a probability), is a mesure?

The remark I am talking about says literally: $$P$$ is a measure (because of 3), positive (because of 2) and normalized (because of 1).

• Look up the definition of a measure... – Nap D. Lover May 4 at 2:10
• In mathematical analysis, a measure on a set is a systematic way to assign a number to each suitable subset of that set, intuitively interpreted as its size. en.wikipedia.org/wiki/Measure_(mathematics). Do countable subsets have a measure? The subsets are so small that the measure must be zero. – roy212 May 4 at 2:18
• yes, that is all well, but merely descriptive. I meant look at the defining properties of a measure: $1)$ is $\mu(A)\geq 0$, $2)$ that the empty set has zero measure and $3)$ that it is countably additive $\mu(\cup A_i)=\sum_i \mu(A_i)$ (all of which are on the page you linked under the “Definition” section by the way...) – Nap D. Lover May 4 at 2:21
• @LoveTooNap29 Thanks. By Kolmogorov's second axiom $A_i \geq 0$. What happens when there is no $\sigma$-additivity? Is $P$ still a measure? I mean when there is only additivity , for instance only two events $A,B$ and $P(A \cup B)= P(A)+P(B)$ – roy212 May 4 at 2:43
• A measure must be countably additive, and this does not follow from the given condition 3. – littleO May 5 at 6:35

To be clear: A set function $$\mu: \Sigma \to \mathbb{R}$$ where $$\Sigma$$ is a collection of subsets of some set $$X$$ is defined to be measure whenever the following hold:

1. $$\mu$$ is non-negative, i.e. for all $$A\in \Sigma$$, $$\mu(A)\geq 0$$
2. The empty set has zero measure: $$\mu(\emptyset)=0$$
3. $$\mu$$ is $$\sigma$$-additive, i.e. for any disjoint countably infinite collection $$A_i\in \Sigma$$ such that $$\cup_i A_i \in \Sigma$$ one has $$\mu(\cup_i A_i)=\sum_i \mu(A_i)$$

A probability measure is a measure such that $$\mu(X)=1$$. In measure theory, countable almost always means countably infinite, and so $$\sigma$$-additivity implies finite-additivity. [Note that a finite union can be written as a countably infinite union by padding it with empty sets: $$A_1 \cup \dotso \cup A_n \cup \emptyset \cup \emptyset \cup \dotso$$]

Any set function that does not satisfy at least one of the three properties above is not a measure.

• Thanks for your explanation. Had I wrote in 3) If $A,B \in F$ and $A \cap B= \emptyset$, then $P(A \cup B \cup \emptyset \cup \emptyset...$($\bigcup \emptyset$ infinite many times))= $P(A)+P(B)+ \sum_{2}^{\infty}0$. Would then $P$ be a measure? Or how should I write 3) in order to say that $P$ is a measure in a finite sample space? – roy212 May 5 at 13:24
• @roy212 To be honest finite sample spaces are often not the subject of measure theoretical probability theory simply because it is not required—combinatorics suffices. Kolmogorov himself referred to such as “elementary probability measures” and treated them first before quickly moving on and establishing the probabilistic terminology of measure theory for infinite sample spaces in his Foundations of the Theory of Probability. But maybe some others will chime in to elaborate on the confusion as I can’t offer a direct reconciliation at the moment. – Nap D. Lover May 5 at 13:36
• @roy212 in other words: elementary probability measures require finite additivity and are only defined on the subsets of finite sample spaces. Probability measures proper require countable additivity and are defined on general sample spaces (with the general case reducing to the elementary case when the sample space is finite due to the padding argument above). – Nap D. Lover May 5 at 13:42