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Give an example of a measure space $(\Omega, \mathit{F})$ and a function $\mu$ on $\mathit{F}$ that is additive but not $\sigma$-additive, i.e. $\mu(\cup A_i)= \sum\mu(A_i)$ for a finite collection of disjoint $A_i$ but not for some infinite collections.

I know a measure function defined on $\sigma$-algebra is $\sigma$-additive, but I struggle finding a function that would not be additive for infinite collections. Can someone give me an example and show me why?

saz
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Ron Estrin
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3 Answers3

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Hint Consider $\Omega = \mathbb{N}$, the power set $F=\mathcal{P}(\mathbb{N})$ and the mapping $\mu: F\to [0,\infty]$, $$\mu(A) := \begin{cases} 0, & \text{$A$ is a finite set} \\ \infty, & \text{otherwise}. \end{cases}$$

saz
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  • Could you show why it this not $\sigma$ additive? – Boby Sep 23 '14 at 19:55
  • @Alex Just consider $A_n = {n,-n}$ for $n \geq 1$. – saz Sep 23 '14 at 19:58
  • Is it because for any $A \in F$. We have that either $A$ is finite or $A^c$ is finite. Than suppose that we are in case 2 where $A^c$ is finite. Hence $\mu(A)=1$ but on the other hand $\sum_{i=1}^\infty \mu(A_n)=0$ where $A_n$ are singletons. – Boby Sep 23 '14 at 20:03
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    @Alex No, that's not correct. There are $A$ such that $A$ as well as $A^c$ is infinite (e.g. $A = {2n; n \in \mathbb{N}}$). And sorry, there is a typo in my last comment; it should read $A_n := {n}$. Note that $\mu(A_n)=0$, but $$\mu\left( \bigcup_{n \geq 0} A_n \right) = \mu(\mathbb{N})=1 \neq 0 = \sum_{n \geq 0} \mu(A_n).$$ – saz Sep 23 '14 at 20:09
  • Ok. I see. Thanks. – Boby Sep 23 '14 at 20:12
  • But ... this $F$ is not a sigma-algebra, right? E.g., the even-number singletons are in $F$, but their union is not. – Alan Sep 23 '15 at 05:09
  • @Alan Yes, that's correct. $F$ becomes a $\sigma$-algebra if we replace "finite" by "countable" – saz Sep 23 '15 at 05:26
  • @saz: So why in your example you considered $F={A\subset \mathbb N\mid A \text{ or }A^c\text{ finite}}$ instead of $F={A\subset \mathbb N\mid A \text{ or }A^c\text{ countable}}$ ? Because what you gave in your post is not a counter-example. So maybe there is something I don't see in your answer. Thank you. – user621345 Dec 13 '18 at 17:46
  • @NewMath You are right, I fixed it. – saz Dec 13 '18 at 18:22
  • But if you take $\mathcal F={A\subset \mathbb N \mid A\text{ or }A^c\text{ finite}}$ and $\mathbb P(A)=0$ if $A$ finite and $\mathbb P(A)=1$ if $A^c$ finite, is not a counter example additive but not $\sigma -$additive measure ? Because $\mu(A)=\infty $ is not very elegant... (acctually, it would be interesting to have a space that is not a probability space because the only fail is the $\sigma -$ additivity) – user621345 Dec 13 '18 at 18:50
  • @NewMath In the definition of $\mathcal{F}$ you probably mean "countable" and not "finite", right? (You might want to notice that any subset of $\mathbb{N}$ is countable, and therefore $F={A \subseteq \mathbb{N}; A , , \text{or} , , A^c$ coutable$} = \mathcal{P}(\mathbb{N})$) Perhaps it's less elegant with $\infty$ but it's easier to see that it is finitely additive... however, it should work fine with $1$ instead of $\infty$: – saz Dec 13 '18 at 19:08
  • @saz : Yes I mean countable. Thank you very much for your answer (to an old post). Just for your information : on google, if you look for an example of a space $(\Omega ,\mathcal F,\mathbb P)$ s.t. $\mathcal F-$ is a $\sigma -$algebra on $\Omega $, $\mathbb P(\Omega )=1$ and $\mathbb P$ is additive but not $\sigma -$additive, this page is the first one :) (so I'm sure that if you can also make a remark in this sense it would be worth for every one :)) – user621345 Dec 13 '18 at 19:14
  • @NewMath You are welcome. – saz Dec 13 '18 at 19:16
  • Something looks strange... what would be $\mu(2\mathbb N)\in \mathcal P(\mathbb N)$ ? – John Jul 24 '19 at 13:29
  • @saz your latest edit means that $\mu$ is not even finitely additive since $\mu(2 \mathbb{N}) = \mu(2 \mathbb{N} + 1) = \mu(\mathbb{N}) = 1$. – Rhys Steele Jul 24 '19 at 14:50
  • @RhysSteele Yeah, I was too rush about that. – saz Jul 24 '19 at 15:01
  • @John Sorry about that... fixed it in the most obvious way since I've currently no time to think about it. – saz Jul 24 '19 at 15:02
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Take $\Omega$ to be the set of all natural numbers, $F$ to be the family of all subsets of $\Omega$ and let $\mu(A) = 0$ if $A$ is a finite set and $\mu(A) = \infty$ if $A$ is infinite, I leave it to you to check that it's additive but not $\sigma$-additive.

Davide Giraudo
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mm-aops
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  • Very neat: you can't get a simpler, clearer example than that. – Tom Collinge Aug 16 '17 at 09:42
  • I think I'm misunderstanding something. If $E$ is the set of even numbers and $O$ is the set of odd numbers, then $\mu(E)=1$ and $\mu(0)=1$, since both sets are infinite. But if $\mu$ is additive, do we not require that $\mu(E \cup O) = \mu(E)+\mu(O)=2$? – Menachem May 15 '19 at 10:22
  • @Menachem $\mu(E) = \mu(O) = \infty$, not $1$ – mm-aops May 18 '19 at 20:03
  • @NewMath: $F$ you wrote is note a $\sigma -$algebra (take $A_n={2n}\in \mathcal F$, then $\bigcup_{n\in\mathbb N}A_n\notin F$). And $\tilde F$ is $\mathcal P(\mathbb N)$. – John Jul 24 '19 at 13:32
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We set $\Omega=N$, $F$=$2^{N}$, $\mu (A)=0$, if $|A|<\infty$; $\mu (A)=\infty$, if $|A|=\infty$.

We can see that $\mu$ is finite additive. Given a finite collection of disjoint $A_k$,1$\leq$ k $\leq$ n. If every $|A_k|<\infty$, then $|\bigcup_{k=1}^n A_k|<\infty$, so\ $\mu(\bigcup_{k=1}^n A_k)=0=\sum_{k=1}^n \mu(A_k)$; if there is at least one ${A_k}$ has infinite elements, $\mu (A_k)=\infty$, then $\bigcup_{k=1}^n A_k$ also has infinite elements, $\mu(\bigcup_{k=1}^n A_k)=\infty=\sum_{k=1}^n \mu(A_k)=\infty$.

On the other hand, we would like to say that it is not $\sigma$-additive. Let $A_n=\{n\}$ for every integer $n\ge1$, we have $\mu(A_n)=0$ and $\mu(\bigcup_{n=1}^{\infty} A_n)=\infty \ne 0=\sum_{n=1}^{\infty}\mu(A_n)$.

Mike J
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