# Show set function is a measure

• Nov 12th 2011, 09:48 AM
TheProphet
Show set function is a measure
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
• Nov 12th 2011, 11:26 AM
girdav
Re: Show set function is a measure
You have to assume that $\displaystyle X$ is uncountable (if it's not the case $\displaystyle \gamma$ is not well defined). In this case, you cannot have two disjoint set whose complement is countable: if $\displaystyle A$ and $\displaystyle B$ are such sets, then $\displaystyle A^c\cup B^c$ is countable but it's the complement in $\displaystyle X$ of the empty set since $\displaystyle A$ and $\displaystyle B$ are disjoint.
• Nov 12th 2011, 01:41 PM
TheProphet
Re: Show set function is a measure
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 $\displaystyle A_j$ is uncountable ?
• Nov 12th 2011, 01:45 PM
girdav
Re: Show set function is a measure
Yes.
• Nov 12th 2011, 01:53 PM
TheProphet
Re: Show set function is a measure
Ok thank you