## Measure of sums

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!