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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?

Thanks in advance!

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  • $\begingroup$ 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. $\endgroup$ Nov 1, 2018 at 18:25

2 Answers 2

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$\infty + \infty$ 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

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

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  • $\begingroup$ $a_n$ are not necessarily natural numbers. $\endgroup$ Oct 28, 2018 at 15:13
  • $\begingroup$ Thanks, you are right, I correct it $\endgroup$
    – ictibones
    Oct 28, 2018 at 15:20
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$\infty+\infty$ 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.

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  • $\begingroup$ 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? $\endgroup$ Oct 28, 2018 at 15:21
  • $\begingroup$ 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. $\endgroup$ Oct 28, 2018 at 15:28
  • $\begingroup$ My point is that the sequence $\{a_n\} $does not belong necessarily to $\mathbb{N}$ once the measure is defined $\tau:\mathscr{P}(N)\to\mathbb{R}^+$. $\endgroup$ Oct 28, 2018 at 15:31
  • $\begingroup$ $\tau$ is defined on $P(\mathbb{N})$ but $A$ is not included in $\mathbb{N}$ so you can't take $\tau(A)$ $\endgroup$
    – ictibones
    Oct 28, 2018 at 15:34
  • $\begingroup$ You are assuming that $a_n\subset\mathbb{N}$. $\endgroup$ Oct 28, 2018 at 15:34

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