Proof that measure space is complete This exercise is from Achim Klenke - "Probability Theory".
Let $\Omega$ be an uncountably infinite set and let $\omega_0\in\Omega$ be an arbitrary element. Let $\mathcal{A}=\sigma (\{w\}\colon \omega \in\Omega \setminus \{\omega_0\})$. Show that $(\Omega, \mathcal{A}, \delta_{\omega_0})$ is a complete measure space.
First a characterization of $\mathcal{A}$ has to be given analogue to the following $\sigma$-algebra: $\mathcal{A}' = \sigma (\{w\}\colon \omega \in\Omega)=\{A\subset \Omega\colon A\text{ is countable or }A^c\text{ is countable}\}$
I tried the obvious 
$\mathcal{A} = \sigma (\{w\}\colon \omega \in\Omega\setminus \{\omega_0\})=\{A\subset \Omega\setminus \{\omega_0\}\colon A\text{ is countable or }A^c\text{ is countable}\}$
and could finish proving the equation if I could somehow prove that $\omega_0$ can only be in a non-finite set of $\sigma (\{w\}\colon \omega \in\Omega\setminus \{\omega_0\})$.
I am not sure if this is correct, am I on the right track?
 A: Let us prove that $$\sigma(\{w\}/ w\neq w_0) = \{A\subset \Omega,   (w_0\in A \implies A^c \text{ is countable}) \text{ and } (w_0\notin A \implies A \text{ is countable})\}$$
Let $\mathcal A = \{A\subset \Omega,   (w_0\in A \implies A^c \text{ is countable}) \text{ and } (w_0\notin A \implies A \text{ is countable})\}$ and $\mathcal C=\sigma(\{w\}/ w\neq w_0)$.
Consider $A\in \mathcal A$.
If $w_0\in A$, then  $A^c=\bigcup_{w\in A^c} \{w\}$ is a countable union of elements of $\mathcal C$ (because $w_0\notin A^c$). Since $\mathcal C$ is a $\sigma$-algebra, this implies $A^c\in \mathcal C$, hence $A\in \mathcal C$.
If $w_0\notin A$, then $A=\bigcup_{w\in A} \{w\}$ is a countable union of elements of $\mathcal C$ (because $w_0\notin A$). Hence $A\in \mathcal C$.
This implies $\mathcal A \subset \mathcal C$.
To prove $\mathcal C \subset \mathcal A$, it suffices to show that $\mathcal A$ is a $\sigma$-algebra containing all the $\{w\}$ for $w\neq w_0$.
$\bullet$ If $w\neq w_0$, $\{w\}$ is countable, hence $\{w\}\in \mathcal A$.
$\bullet$ $\mathcal A$ is stable under complement. Consider $A\in \mathcal A$. If $w_0\in A$, then $A^c$ is countable and $w_0\notin A^c$, hence $A^c\in \mathcal A$. If $w_0\notin A$, then $w_0\in A^c$ and $(A^c)^c=A$ is countable, hence $A^c\in \mathcal A$.
$\bullet$ $\mathcal A$ is stable under countable union. Consider $(A_i)\in \mathcal A^{\mathbb N}$.
If no $A_i$ contains $w_0$, $w_0\notin \cup_i A_i$ and all the $A_i$ are countable, so $\cup_i A_i$ is countable. Hence  $\cup_i A_i\in \mathcal A$.
If WLOG $w_0\in A_1$, $w_0\in \cup_i A_i$. Note that $A_1^c$ is countable and $(\cup_i A_i)^c = \cap_i A_i^c\subset A_1^c$ is countable. Hence  $\cup_i A_i\in \mathcal A$.

The rest of the exercise if simple. Let $A\in \mathcal A$ be a $\delta_{w_0}$-null set. Then $w_0\notin A$. Given the previous characterization, $A$ is countable.
If $B\subset A$, then $w_0\notin B$ and $B$ is countable, hence $B\in \mathcal A$.
So all subsets of $\delta_{w_0}$-null sets are in $\mathcal A$, so the measure space is complete.
A: The characterisation
$\mathcal{A} =\{A\subset \Omega\setminus \{\omega_0\}\colon A\text{ is countable or }A^c\text{ is countable}\}$
isn't quite correct. Hint: Keep in mind that $ \sigma- $algebras are closed under complement.
Here is why the characterisation fails: 

 Claim: $\mathcal{A}$ contains sets which contain $ \omega_0 $ while $\{A\subset \Omega\setminus \{\omega_0\}\colon A\text{ is countable or }A^c\text{ is countable}\}$ does not.
Proof: Since $ \mathcal{A} $ is a $ \sigma-$algebra, it is closed under complement (i.e. $ A \in \mathcal{A} $ implies $ A^C \in \mathcal{A} $). Now take any $ \omega \neq \omega_0 $. Then $ \{\omega\} \in \mathcal{A} $ (by definition of $ \mathcal{A} $) thus also $ \Omega\setminus\{\omega\} \in \mathcal{A}$. But $ \omega_0 \in \Omega\setminus \{\omega\} $.

A better description of $ \mathcal{A} $:

 $\mathcal{A} = \{ A \subset \Omega : (A \text{ is countable and }\omega_0 \notin A )\text{ or }( A^C \text{ is countable and } \omega_0 \in A)\}$.

