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

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

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

Prove that  \mu is a measure.

Proof.

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

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

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

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

Then we have  |F|=|F_1|+|F_2|+...

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

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

Therefore  \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!