Consider a set X, define $\displaystyle f:X \rightarrow [0, \infty ] $ by $\displaystyle f(x)=1 \ \ \ \ \forall x \in X $.

Let $\displaystyle E \subset \mathbb {P}(X) $, the set of all subsets in X.

Define $\displaystyle \mu (E)= \sum _{x \in E } f(x) = sup \{ \sum _{x \in F } f(x) : F \subset E, F \ finite \ \} $

Prove that $\displaystyle \mu $ is a measure.

Proof.

Let $\displaystyle E_1,E_2,... \subset X $, with all $\displaystyle E_n$ disjoint.

I claim that $\displaystyle \mu ( \bigcup ^{ \infty }_{n=1}E_n)= \sum ^{ \infty }_{n=1} \mu (E_n) $

Now suppose that $\displaystyle F \subset \bigcup ^{ \infty } _{n=1} E_n $ is the largest finite subset.

Then $\displaystyle F = F_1 \bigcup F_2 \bigcup F_3 \bigcup . . . = \bigcup _{n=1}^{ \infty } $ for $\displaystyle F_n \subset E_n$ the biggest subset. Note that $\displaystyle F_n$ are disjoint since each $\displaystyle E_n$ is disjoint.

Then we have $\displaystyle |F|=|F_1|+|F_2|+... $

It follows that $\displaystyle \mu ( \bigcup ^ \infty _{n=1} E_n ) = sup \{ \sum _{x \in F } f(x) : F \subset \bigcup _{n=1}^ \infty En, F \ finite \ \} $$\displaystyle = \sum _{x \in F } f(x) $

$\displaystyle =|F|=|F_1|+|F_2|+...= \sum _{x \in F_1 } f(x) + \sum _{x \in F_2 } f(x) + . . . = $$\displaystyle \mu (E_1)+ \mu (E_2) + . . . = \sum ^ \infty _{n=1} \mu (E_n) $

Therefore $\displaystyle \mu $ is a measure.

Is this okay? I know that idea should be right, but it is just the part about choosing the largest finite subset could be a bit verge.

Thank you!