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$X:= \Bbb Q$ and $\mathcal S:=\mathcal I^1 | \Bbb Q := \{]a,b] \cap \Bbb Q | a,b \in \Bbb R , a\le b\} $ and define the content $$\mu : \mathcal S \to [0,\infty[, \:\:\: \mu(]a,b] \cap \Bbb Q ) := b-a \:\:\:\:\:\:\:(a \le b)$$

I've already shown that $\mathcal S$ is a semiring and that $\mu$ is a finite content, but now I have to show that $\mu$ is NOT a premeasure, i.e. i have to find a sequence of disjoint sets $A_i \in \mathcal S, \: \: \:\bigcup_{i=1}^\infty A_i \in \mathcal S$ for which the following is true: $$\mu \Big(\bigcup_{i=1}^\infty A_i \Big ) \not = \sum_{i=1}^\infty \mu(A_i)$$

Normally contetns aren't premeasures because of some set that would deliver infinity somewhere, but I can't really see where I can show this for $\mu$.

Any ideas or tipps? Thanks in advance!

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Let $I = (a,b] \cap \Bbb Q$ for some arbitrary $a,b \in \Bbb R$. Since $I$ is made only of (infinitely many) rational numbers, and we know that these are countably many, it follows that we may write $I$ as $\{q_i \mid i \in \Bbb N\}$. Take then $A_i = \{q_i\}$ and notice that, assuming that $\mu$ is countably-aditive, we have

$$b-a = \mu (I) = \mu \left( \bigcup _{i \in \Bbb N} A_i \right) = \sum _{i \in \Bbb N} \mu (A_i) = \sum _{i \in \Bbb N} 0 = 0$$

which is blatantly impossible. Therefore $\mu$ is not a pre-measure.

To show that the measure of a single rational point $\{q\}$ is $0$ just notice that

$$\{q\} = \bigcap _{n \ge 1} I_n$$

with

$$I_n = (q-\frac 1 n, q] \cap \Bbb Q ,$$

so that

$$\mu (\{q\}) = \mu \left( \bigcap _{n \ge 1} I_n \right) = \inf _{n \ge 1} \mu (I_n) = \inf _{n \ge 1} \frac 1 n = 0 .$$

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  • $\begingroup$ Ah i see, thanks a lot :D $\endgroup$
    – DeltaChief
    Oct 30, 2016 at 14:02
  • $\begingroup$ Why is the intersection of enumerable sets in S? Wouldn't you need to show that S is a $\sigma$-ring first? $\endgroup$ Feb 2, 2018 at 13:56

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