Hi,

Let $\displaystyle X $ be a set.

Define $\displaystyle \mathcal{A} := \{ A \subset X : |A| \leq |\mathbb{N}| \mbox{ or } |A^{c}| \leq |\mathbb{N}| \}$, where $\displaystyle | \cdot | $ means cardinality.

Then $\displaystyle \mathcal{A} $ is a $\displaystyle \sigma$-algebra. Define a measure on this measurable space by $\displaystyle \gamma : \mathcal{A} \to \{0,1\} $ by

\$\displaystyle \gamma(A) := \begin{cases} 0 \mbox{ if } A \mbox{ is countable}\\ 1 \mbox{ if }A^{c} \mbox{ is countable}\end{cases}$

Show that $\displaystyle \gamma $ is a measure, i.e. show that $\displaystyle \gamma(\emptyset) = 0 $ and that $\displaystyle \gamma $ is $\displaystyle \sigma $-additive.

That $\displaystyle \gamma(\emptyset) = 0$ I guess comes easily since $\displaystyle \emptyset $ is countable. If $\displaystyle (A_j)_{j \geq 1} \subset \mathcal{A}$, pairwise disjoint:

Assume all $\displaystyle A_j$ are countable, then the countable union is countable, so $\displaystyle \gamma\left( \bigcup_{j \geq 1} A_j \right) = 0$.

Also, $\displaystyle \sum_{j \geq 1} \gamma(A_j) = 0 + 0 + \ldots = 0 $, so we have equality in this case.

If we assume that two sets $\displaystyle A_j_{0}, A_j_{1} $ are uncountable, then $\displaystyle A_{j_{0}}^{c}, A_{j_{1}}^{c}$ are countable and since $\displaystyle \bigcap_{j \geq 1} A_{j}^{c} \subset A_{j_{0}}^{c} \cap A_{j_{1}}^{c} $ we have $\displaystyle \left(\bigcup_{j \geq 1}A_{j} \right)^{c} = \bigcap_{j \geq 1}A_{j}^{c} $ is countable.

Thus, $\displaystyle \gamma\left( \bigcup_{j \geq 1} A_j \right) = 1 $, but

$\displaystyle \sum_{j \geq 1} \gamma(A_j) = 2 $ ???

I know I am wrong, please point out where.

Thanks