# Additive but not $\sigma$-additive function

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?

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

• 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
• @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

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.

• Very neat: you can't get a simpler, clearer example than that. – Tom Collinge Aug 16 '17 at 9:42
• @saz: Why in your example you considered $F=\{A\subset \mathbb N\mid A \text{ or }A^c\text{ finite}\}$ instead of $\tilde F=\{A\subset \mathbb N\mid A \text{ or }A^c\text{ countable}\}$ ? Because what you gave in your post is not a counter-example since $F$ is not a $\sigma -$algebra. So maybe there is something I don't see in your answer. Thank you. – NewMath Dec 13 '18 at 17:47
• 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 at 10:22
• @Menachem $\mu(E) = \mu(O) = \infty$, not $1$ – mm-aops May 18 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 at 13:32

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 $\mu(\bigcup_{n=1}^{\infty} A_n)=\infty \ne 0=\sum_{n=1}^{\infty}\mu(A_n)$.