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Let

$$\Omega = \{2, 3, 5, 7, 11, 13, 17, 19, ...\}$$

be the set of all prime numbers and

$$\epsilon := \{\{x\}: x \in \Omega\}$$

be a generating system.

Determine $\sigma(\epsilon)$ and decide whether the measure

$$\lambda : \sigma(\epsilon) \rightarrow [0, \infty], \lambda(A) := \sum_{x \in A} {1 \over x^2}$$

is finite or not.

$\sigma(\epsilon)$ must be the smallest $\sigma$-algebra on $\Omega$ that contains every element of $\epsilon$. Therefore, I would say that

$\sigma(\epsilon) = P(\Bbb N)$.

It is $\sigma$-finite and finite because

1) There exists a countable sequence $(A_n) \in P(\Bbb N)$ such that $\bigcup_{n = 0}^{\infty} A_n = \Omega$ (one just has to unite all the elements of $\epsilon$ here).

2) $\lambda(A_n) = \sum_{x \in A_n} {1 \over x^2} \le \sum_k^{\infty} {1 \over k^2} < \infty$.

Is my argumentation correct?

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    Yes, it is correct . – 2017-02-12
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    Why is the measure not finite? As far as I can see it holds: $\lambda(\Omega) = \sum_{k \in \Omega} \frac{1}{k^2} \le \sum_{k \in \mathbb{N}} \frac{1}{k^2} < \infty$ – 2017-02-12
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    I guess there is a difference between a measure being finite and $\sigma$-finite. But in this case, "finite" probably means "$\sigma$-finite" anyway, so yeah. – 2017-02-12
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    Ah, I looked up the exact definitions again. "finite" does refer to my second argument, "$\sigma$"-finite to my first argument. So it is "finite" because of 2) and $\sigma$-finite because of 1). Thanks for pointing that out! – 2017-02-12
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    why should be e.g. $1\in\sigma(\epsilon)$? – 2017-02-12
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    $\sigma(\epsilon) \neq P(\Bbb N)$ – 2017-02-12
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    My solution is wrong then? – 2017-02-14
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    Ah, maybe it would be wiser to choose the power set of the prime numbers, so $\sigma(\epsilon) = P(\Omega)$? – 2017-02-14

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