# Fourth axiom in Kolmogorov's probability axiom system?

I'm reading Jaynes's "Probability Theory: The Logic of Science". In appendix A he contrasts the theory he develops in the first few chapters to the more conventional measure theory based system, A.K.A. Kolmogorov's. He lists four axioms used in this system. $F$ is a $\sigma$-algebra (He calls it a "$\sigma$-field")

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

(2) Non-negativity: $P(f_{i}) ≥ 0$ for all $f_{i}$ in $F$

(3) Additivity: if ${f_{1}\ldots f_{n}}$ are disjoint elements of $F$ then $P( f ) = \sum_{i}P(f_{i})$, where $f =\cup_{j}f_{j}$

(4) Continuity at zero: if a sequence $f_{1} \supseteq f_{2} \supseteq f_{3} \supseteq\ldots$ tends to the empty set, then $P( f_{j} ) → 0$.

I remembered use of only the first three axioms, was the fourth once used and abandoned or is there something else going on?

• (3) and (4) are often replaced by (3') in which it is the same statement as (3) but for countably many elements instead of finitely many. You are probably remembering using (1), (2) and (3'). Commented Mar 10, 2018 at 11:33
• @YAlexandrov Thanks. The equivalence is not immediately obvious to me, would you care to elaborate or alternatively reference to a source that does? Commented Mar 10, 2018 at 11:36
• It is an straight forward exercise. Just look at the sets $f_n=\cup_{k=1}^{\infty}A_k\setminus\cup_{k=1}^{n}A_k$. They decrease to the empty set. Apply (1)-(4) to get that $P(\cup_{k=1}^{\infty}A_k)=\sum_nP(a_n)$, which is (3'). Conversely, (3') implies (3) by putting almost all $A_n=\emptyset$. To get (4) put $A_n=f_{n-1}\setminus f_n$. Commented Mar 10, 2018 at 11:39

There is an alternate set of axioms that appear in my book (An Introduction to Mathematical Statistics and Its Applications by Larsen and Marx) which more clearly explain what types of functions can qualify as probability measures. They replace your fourth axiom by the following:

1. Let $$A_1$$, $$A_2$$, $$\cdots$$, be events defined over $$\Omega$$. If $$A_i \cap A_j = \emptyset$$ for each $$i \neq j$$, then $$P\left(\bigcup_{i=1}^{\infty} A_i\right) = \sum_{i=1}^{\infty} P(A_i)$$

In conjunction with the third axiom, this one is trying to say that the probability measure chosen must work for countably infinite disjoint decompositions of $$\Omega$$ as well. The third axiom does not automatically guarentee this unless the chosen probability measure converges to zero as the sequence of $$f_i$$ tends to the empty set. In short, the fourth axiom from your list or from my list is required to allow infinite decompositions of the sample set.

• Thank you for your answer, this is essentially what YAlexandrov explained in his comment. It would be interesting to see an example where there is finite additivity but countable additivity fails. (Since there is no continuity at zero) Commented Jan 22, 2019 at 17:41
• @H.Rappeport edited Commented Jan 24, 2019 at 0:43
• In this case axiom (3) is unnecessary, as the finite additivity is recovered from the countable additivity by letting $A_j := \emptyset$ for $j > n$ and using $P(\emptyset)=0$, where the latter is also recovered from the countable additivity. Commented Nov 29, 2022 at 19:15