# Proving a measure is addtive but not $\sigma-\text{addtive}$

Let $$\mathbb{N}$$ be set positive integers, $$\mathscr{P}(\mathbb{N})$$ the class of $$\mathbb{N}$$ (parts of $$\mathbb{N}$$) and $$\sum_\limits{n=1}^{\infty}a_n$$ a series of positive terms which is convergent. Define a set function $$\tau:\mathscr{P}(\mathbb{N})\to\mathbb{R}^+=[0,+\infty]$$ by $$\tau(e)=\begin{cases} \sum_\limits{n\in E}^{}a_n, & \mbox{ if }E\subset\mathbb{N}\:\: \text{is finite}\\ +\infty, & \mbox{ if }\text{otherwise} \end{cases},$$

Show that $$\tau$$ is additive but not $$\sigma-\text{additive}$$.

Consider $$E_j$$ converging to the finite set $$E\subset\mathbb{N}$$.If $$E_j\subset\mathbb{N}\forall j\in J$$ so that $$E_j$$ and $$J$$ are finite.($$J$$ is finite because $$E$$ is finite) $$E_{j+1}\supset E_j\forall j$$

Then $$\lim(\tau(E))=\lim\sum_\limits{n\in E_j}^{}a_n=\sum_\limits{j\in J}\sum_\limits{n\in E_j}^{}a_n=\sum_\limits{n\in E}^{}a_n=\tau(E)$$

Since $$E$$ is arbitrary and finite $$\tau$$ is additive. It is not sigma additive because any infinite set measure is infinite and $$\infty + \infty$$ is not determined.

Questions:

Is my proof right?

What are other alternative proves?

• If we had defined $\tau(E) = \sum_{n \in E} a_n$ for all $E \subseteq \mathbb{N}$, we would have had a finite $\sigma$-additive measure on all subsets of $\mathbb{N}$. This is also simpler. Nov 1, 2018 at 18:25

$$\infty + \infty$$ is not a problem. Consider the existence of the trivial "infinite" measure space: take any measurable space $$(X, \mathcal{S)}$$ and define

$$\begin{equation*} \mu(E) := \begin{cases} 0 & E=\emptyset\\ \infty & \text{otherwise} \end{cases} \end{equation*}$$

$$(X,\mathcal{S}, \mu)$$ is a measure space.

$$\tau$$ is not $$\sigma$$-additive because

$$\infty=\tau\left(\bigcup_{n\in\mathbb{N}}\{n\}\right)\ne\sum_{n\in\mathbb{N}} \tau(\{n\})=\sum_{n\in\mathbb{N}}a_n< \infty$$

• $a_n$ are not necessarily natural numbers. Oct 28, 2018 at 15:13
• Thanks, you are right, I correct it Oct 28, 2018 at 15:20

$$\infty+\infty$$ is not the problem. Consider $$E_n=\left \{ 1,2\cdots,n \right \}$$.

Then, $$E=\bigcup _n E_n=\left \{ 1,2\cdots, \right \}$$ and $$\tau(A)=\infty.$$

But $$\tau(E_n)=\sum^n_{i=1}a_n$$ and this converges by assumption, so

$$\tau(E)\neq \lim \tau(E_n)$$ and so $$\tau$$ is not countably additive.

• In your proof the fact $\tau(A)\neq\lim\tau(A_n)$ violates the Measure continuity theorem hence the measure is not sigma-additive. But the problem is that you are assuming the $a_n\forall n$ to be natural numbers, while there is no condition assuring that. What were you thinking? Oct 28, 2018 at 15:21
• I do not understand your comment, sorry. I think the proof is ok. I only use the fact that $1)\ \tau=\infty$ on sets of infinite cardinality and $2).\$ the sum of the $a_n$ converges. I certainly am NOT assuming that the $a_n$ are integers. Oct 28, 2018 at 15:28
• My point is that the sequence $\{a_n\}$does not belong necessarily to $\mathbb{N}$ once the measure is defined $\tau:\mathscr{P}(N)\to\mathbb{R}^+$. Oct 28, 2018 at 15:31
• $\tau$ is defined on $P(\mathbb{N})$ but $A$ is not included in $\mathbb{N}$ so you can't take $\tau(A)$ Oct 28, 2018 at 15:34
• You are assuming that $a_n\subset\mathbb{N}$. Oct 28, 2018 at 15:34