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

Question

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!

Answer 1

$\infty + \infty$ 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.

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

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

Answer 2

$\infty+\infty$ 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.