Let $(X,v,S_0)$ a measure space and define $$\mu^*(E)=\inf\{\sum_nv(A_n):A_n \in S_0,E \subseteq \bigcup_nA_n\}$$
It is easy to see that $\mu^*$ is an outer measure and is called $\textbf{the outer measure induced by}$ $v$.
Also again it is not difficult to prove that $\mu^*(E)=\min\{v(A):A \supseteq E\}$
We denote $\Sigma_{\mu^*}$ the Caratheodory sigma algebra of $\mu^*-$measurable sets.
Now with some work it is proved that if $\mu$ (the restriction of $\mu^*$ to $\Sigma_{\mu^*}$) is sigma finite then $(X,\Sigma_{\mu^*},\mu)$ is the completion of $(X,v,S_0)$
Now assume that $X$ is an uncountable set and $S_0=\{A \subseteq X: \text{$A$ is countable or $A^c$ is countable}\}$ and $v=$counting measure $|A|$
$X$ is not sigma finite with respect to this measure since it is uncountable.
I want to prove that $\Sigma_{\mu^*}=P(X)$ where $m^*$ is the outer measure induced by the counting measure.
Here is my work:
Clearly $(X,|\cdot|,S_0)$ is a measure space.
Now let $A \subseteq X$ and $E \subseteq X$
We always have the inequality $\mu^*(E)\leq \mu^*(E \cap A)+\mu^*(E \cap A^c)$
So if $E$ is infinite exists $B \in S_0$ such that $B \supset E$ and $+\infty=|B|=\mu^*(E) \leq\mu^*(E)\leq \mu^*(E \cap A)+\mu^*(E \cap A^c)$
Thus the equality is satisfied with like this: $+\infty=+\infty$
Now assume that $E$ is finite.
Then $$\mu^*(E)=|E|=|E\cap A|+|E \cap A^c| = \mu(A \cap E)+\mu(E \cap A^c)=\mu^*(A \cap E)+\mu^*(E \cap A^c)$$
since $E,A \cap E,A^c \cap E$ are finite sets so belong to $S_0$.
In every case the desired equality is satisfied,thus $\Sigma_{\mu^*}=P(X)$
Is this correct,or am i missing something?
Thank you in advance.
Best Answer
If $X$ is uncountable then the counting measure $\nu$ on the collection of countable and cocountable sets is not $\sigma$-finite so that identification with completion is not okay here.
Anyway, a counting measure is always complete (already). This because its only null-set is the empty set (which belongs to every $\sigma$-algebra).
edit:
Any outer measure $\mu^*$ on a set $X$ is a function on $\wp(X)$ with special properties and induces a complete measure on a $\sigma$-algebra $\Sigma_{\mu^*}$.
This complete measure is then $\mu^*$ itself but restricted to this $\sigma$-algebra and $A$ is an element of $\Sigma_{\mu^*}$ iff:$$\forall E\in\wp(X)\;[\mu^*(E)=\mu^*(E\cap A)+\mu^*(E\cap A^{\complement})]\tag1$$
If we observe that $\mu^*$ is already a measure itself on $\sigma$-algebra $\wp(X)$ then from that we can draw the immediate conclusion that $(1)$ is satisfied for all $A\in\wp(X)$ so that $\Sigma_{\mu^*}=\wp(X)$ and the complete measure induced by $\mu^*$ is just $\mu^*$ itself.
You are in that situation here because it is easy to verify that the map that sends finite sets to their cardinality and sends infinite sets to $+\infty$ is indeed a measure on $\wp(X)$.
So your proof (which is okay) is a bit redundant.