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I am (self) studying probability theory and measure using the book from Ash, R. et al. [1]. I am trying to solve one of the basic problems (Section 1.2, problem 3) but with no avail...here is the link to the problem:

Let $\Omega$ be a countable infinite set, and let $\mathcal{F}$ be the field consisting of all finite subsets of $\Omega$ and their complements. If $A$ is finite, set $\mu(A) = 0$, and if $A^{c}$ is finite, set $\mu(A) = 1$.

(a) show that $\mu$ is finitely additive but not countable additive.

(b) show that $\Omega$ is the limit of an increasing sequence of sets $A_{n} \in \mathcal{F}$, with $\mu(A_{n}) = 0$ for all $n$, but $\mu(\Omega)=1$

I would like to get some help as where to begin with...

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  • $\begingroup$ Please write the problem formulation directly into the question. $\endgroup$
    – Cettt
    Commented Feb 21, 2018 at 8:26
  • $\begingroup$ Here $\mu$ is not a measure as $\mathcal{F}$ is not a $\sigma$-algebra. The $\sigma$-algebra generated by the finite subsets would be the whole power set. $\endgroup$
    – toe-pose
    Commented Oct 31, 2022 at 9:39

1 Answer 1

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For finite additivity $\mu (A \cup B) =\mu (A) +\mu (B)$ just consider the cases where both sets are finite sets, bot have finite complements and one is finite , the other having finite complement. Now let $x_1,x_2,...$ be the distinct points of the space and note that $\mu (\{x_1\} \cup \{x_2\} \cup ....)=1$ whereas $\sum \mu (\{x_i\})=0$. ( It is understood that '$A$ has finite complement' includes the case when the complement is empty).

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  • $\begingroup$ I have understood that $\mu(A \cup B) = \mu(A) + \mu(B)$ iff $A$ and $B$ are disjoints. Yet, in the problem there is no explicit mention of it....why have you assumed this? $\endgroup$
    – mgbacher
    Commented Feb 21, 2018 at 9:46
  • $\begingroup$ @mgbacher What's the definition of "finitely additive"? $\endgroup$ Commented Feb 21, 2018 at 13:34
  • $\begingroup$ @DavidC.Ullrich, according to the book I am using it says in Theorem 1.2.5: (a) $\mu(\emptyset)=0$, (b) $\mu(A\cup B)+\mu(A\cap B) = \mu(A) + \mu(B)$, for all $A,B \in \mathcal{F}$, (c) if $A,B \in \mathcal{F}$, and $B \subset A$, then, $\mu(A) = \mu(B) + \mu(A-B)$, (d) if $\mu$ is a measure, $\mu(\cup_{n=1}^{\infty}A_{n}) \leq \sum_{n=1}^{\infty}{\mu(A_{n})}$.... $\endgroup$
    – mgbacher
    Commented Feb 21, 2018 at 14:35
  • $\begingroup$ @mgbacher That has nothing to do with my question! What is the definition of the phrase "finitely additive"? $\endgroup$ Commented Feb 21, 2018 at 14:48
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    $\begingroup$ If $A$ and $B$ are infinite and $A,B\in\mathcal F$ then $A$ and $B$ cannot be disjoint. So in proving that $\mu$ is finitely additive the case where $A$ and $B$ are both infinite simply doesn't come up. $\endgroup$ Commented Feb 21, 2018 at 15:10

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