Show set function is a measure

**TheProphet** · November 12th 2011, 09:48 AM

Hi,

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

Then $\mathcal{A}$ is a $\sigma$ -algebra. Define a measure on this measurable space by $\gamma : \mathcal{A} \to \{0,1\}$ by
\ $\gamma(A) := \begin{cases} 0 \mbox{ if } A \mbox{ is countable}\\ 1 \mbox{ if }A^{c} \mbox{ is countable}\end{cases}$

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

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

Assume all $A_j$ are countable, then the countable union is countable, so $\gamma\left( \bigcup_{j \geq 1} A_j \right) = 0$ .
Also, $\sum_{j \geq 1} \gamma(A_j) = 0 + 0 + \ldots = 0$ , so we have equality in this case.
If we assume that two sets $A_j_{0}, A_j_{1}$ are uncountable, then $A_{j_{0}}^{c}, A_{j_{1}}^{c}$ are countable and since $\bigcap_{j \geq 1} A_{j}^{c} \subset A_{j_{0}}^{c} \cap A_{j_{1}}^{c}$ we have $\left(\bigcup_{j \geq 1}A_{j} \right)^{c} = \bigcap_{j \geq 1}A_{j}^{c}$ is countable.
Thus, $\gamma\left( \bigcup_{j \geq 1} A_j \right) = 1$ , but
$\sum_{j \geq 1} \gamma(A_j) = 2$ ???

I know I am wrong, please point out where.

Thanks

**girdav** · November 12th 2011, 11:26 AM

You have to assume that $X$ is uncountable (if it's not the case $\gamma$ is not well defined). In this case, you cannot have two disjoint set whose complement is countable: if $A$ and $B$ are such sets, then $A^c\cup B^c$ is countable but it's the complement in $X$ of the empty set since $A$ and $B$ are disjoint.

**TheProphet** · November 12th 2011, 01:41 PM

So if there cant exist two disjoint sets whose complements are countable, then the sum of measures can be at most 1 also, and equal to one when just one $A_j$ is uncountable ?

**girdav** · November 12th 2011, 01:45 PM

Yes.

**TheProphet** · November 12th 2011, 01:53 PM

Ok thank you

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