Proving completeness of $\sigma$-algebra Let $\Omega$ be uncountably infinite, and  $\mathcal{A}=\sigma(\{\omega\}: \omega\in \Omega\setminus\{\omega_0\})$. 
I'm supposed to give a characterisation of $\mathcal{A}$ and then prove that $(\Omega,\mathcal{A},\delta_{\omega_0})$ is complete. 
My try
For the characterisation: $\mathcal{A}=\{A\subset \Omega\setminus\{\omega_0\}:A \text{ countable or } A^c \text{ countable}\}$
However, to prove completeness, i.e. $N=\{A\subset \Omega:\delta_{\omega_0}(A)=0 \}\subset \mathcal{A}$, since we also have $N=\{A:A\subset \Omega\setminus\{\omega_0\} \}$. This seems to contradict the text of the exercise...
 A: I'm not sure your definition of completeness is correct.
You need to show that if


*

*$A \in \mathcal{A}$,

*$\delta_{\omega_0}(A) = 0$,

*$B \subseteq A$,


then $B \in \mathcal{A}$.
This is different than showing that any subset of $\Omega \setminus \{\omega_0\}$ is in $\mathcal{A}$ (in particular because $\Omega \setminus \{\omega_0\} \notin \mathcal{A}$).
Proving the above should be simple, given the [correct] characterization of $\mathcal{A}$; see below.

I should remark that your characterization of $\mathcal{A}$ is slightly incorrect. I think it should be
$$\{A : A \subseteq \Omega \setminus \{\omega_0\}, A \text{ countable}\} \cup \{A : A^c \subseteq \Omega \setminus \{\omega_0\}, A^c \text{ countable}\}$$
For justification, you need to show that this characterization


*

*is a $\sigma$-algebra, and

*any $\sigma$-algebra containing $\{\{\omega\} : \omega \in \Omega \setminus \{\omega_0\}\}$ also contains this characterization.

A: Note that this question concerns Exercise 1.3.2 from Probability Theory: A Comprehensive Course, 3rd edition, by Achim Klenke. This is a long, detailed, hopefully correct answer to the exercise in question.
Discussion
There is a perhaps natural, although totally misguided, desire to characterize $\mathcal{A}$ as $\left\{A \subset \Omega \setminus \{\omega_0\} \colon A \text{ is countable or } A^c \text{ is countable} \right\}$. Eventually you realize that this makes no sense because it cannot contain $\Omega$, so it cannot be a $\sigma$-algebra.
The connection with Exercise 1.1.4 is clearer if you define $\mathcal{B} := \{A \subset \Omega \colon A \text{ is countable}\}$, $\mathcal{C} := \{B^c \colon B \in \mathcal{B}\}$ and $\mathscr{A} := \left\{A \subset \Omega \colon A \text{ is countable or } A^c \text{ is countable} \right\}$ (the latter to reduce clutter, as in Exercise 1.1.4), and then show that
\begin{equation}
    \mathscr{A} = \mathcal{B} \cup \mathcal{C}
\end{equation}
Clearly $\mathscr{A} \supset \mathcal{B} \cup \mathcal{C}$, because we know that $\mathscr{A}$ contains all countable subsets of $\Omega$, and as a $\sigma$-algebra it contains the complements of all those sets. On the other hand, consider $A \in \mathscr{A}$. If $A$ is countable, then $A \in \mathcal{B}$. If $A$ is such that $A^c$ is countable, then $A^c \in \mathcal{B}$, and we must have $A \in \mathcal{C}$. This shows that $\mathscr{A} = \mathcal{B} \cup \mathcal{C}$.
This should make the connection between Exercise 1.1.4 and this exercise clearer, because for this exercise it is then natural to define $\mathcal{B} := \left\{ A \subset \Omega \setminus \{\omega_0\} \colon A \text{ is countable}\right\}$ and $\mathcal{C} := \{B^c \colon B \in \mathcal{B}\}$, and then we try to show that $\mathcal{A} = \mathcal{B} \cup \mathcal{C}$.
Exercise part (i)
Define $\mathcal{B} := \left\{ A \subset \Omega \setminus \{\omega_0\} \colon A \text{ is countable}\right\}$ and $\mathcal{C} := \{B^c \colon B \in \mathcal{B}\}$. We show that $\mathcal{A} = \mathcal{B} \cup \mathcal{C}$.
First, it is clear that $\mathcal{B} \cup \mathcal{C} \subset \mathcal{A}$, because $\mathcal{A}$ will contain countable unions of all elements of $\Omega \setminus \{\omega_0\}$, and all complements of those countable unions. Second, clearly $\left\{\{\omega\} \colon \omega \in \Omega \setminus \{\omega_0\}\right\} \subset \mathcal{B} \cup \mathcal{C}$. Similar to Exercise 1.1.4, we've shown that $\mathcal{A} = \sigma(\mathcal{B} \cup \mathcal{C})$, so if we can show that $\mathcal{B} \cup \mathcal{C}$ is a $\sigma$-algebra, then we're done. We check Definition 1.2.
Definition 1.2 (i)
Clearly $\emptyset \in \mathcal{B}$, so $\Omega \in \mathcal{C}$, and therefore $\Omega \in \mathcal{B} \cup \mathcal{C}$.
Definition 1.2 (ii)
Obviously $\mathcal{B} \cup \mathcal{C}$ is closed under complements by construction.
Definition 1.2 (iii)
First note that $\mathcal{B}$ is stable under countable unions, since the countable union of countable sets is countable. Furthermore, $\mathcal{C}$ is also stable under countable unions, because $\bigcup_{n = 1}^{\infty} B_n^c = \left(\bigcap_{n = 1}^{\infty} B_n\right)^c$, and $\bigcap_{n = 1}^{\infty} B_n \in \mathcal{B}$ because it's the intersection of countable sets and thereore countable, which implies that its complement is in $\mathcal{C}$.
So consider $A_1, A_2, \ldots \in \mathcal{B} \cup \mathcal{C}$, and we might as well assume that we have at least one set in this sequence from $\mathcal{B}$ and at least one from $\mathcal{C}$, because we dealt with the other cases above. We can rearrange $\bigcup_{n = 1}^{\infty} A_n$ into a union of sets from $\mathcal{B}$, unioned with a union of sets from $\mathcal{C}$. Since $\mathcal{B}$ and $\mathcal{C}$ are stable under countable unions, this reduces to showing that if $B \in \mathcal{B}$ and $C \in \mathcal{C}$, then $B \cup C \in \mathcal{B} \cup \mathcal{C}$. Note that $B \cup C \supset C$, hence $(B \cup C)^c \subset C^c \in \mathcal{B}$. Since $(B \cup C)^c$ is a subset of a countable set, it is countable, hence $(B \cup C)^c \in B$, and therefore $B \cup C \in \mathcal{C}$.
We conclude that $\mathcal{B} \cup \mathcal{C}$ is a $\sigma$-algebra, and in particular that $\mathcal{A} = \mathcal{B} \cup \mathcal{C}$.
Exercise part (ii)
To show that $(\Omega, \mathcal{A}, \delta_{\omega_0})$ is complete, we have to show that $\mathcal{A}$ contains all subsets of $\delta_{\omega_0}$-null sets. The $\delta_{\omega_0}$-null sets are those sets $N \in \mathcal{A}$ such that $\omega_0 \notin N$. Looking at our characterization of $\mathcal{A}$ from part (i) above, we see that the only sets meeting these criteria are the sets in $\mathcal{B}$, since all elements of $\mathcal{C}$ contain $\omega_0$. For any $B \in \mathcal{B}$, we know that any $B' \subset B$ is countable and does not contain $\omega_0$, therefore $B' \in \mathcal{B}$. We conclude that $(\Omega, \mathcal{A}, \delta_{\omega_0})$ is complete.
