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?

• Any reaction to my answer ? – Gabriel Romon Sep 4 at 19:41
• @GabrielRomon I want to take my time to read it and since I am busy, this might take a few more days before I can do it. But I have not forgotten! Thanks for your answer, I will respond asap. – EpsilonDelta Sep 4 at 22:56

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.

• For the proof that $\mathcal{A}$ is a $\sigma$-algebra, why the emphasis on "containing all the $\{\omega\}$ for $\omega\ne \omega_0$"? This property follows immediately if we can show that $\mathcal{A}$ is a $\sigma$-algebra. What would be missing is to show that $\Omega\in\mathcal{A}$, but this follows since $\omega_0\in\Omega$ and $\Omega^C=\emptyset$. Am I missing something? – EpsilonDelta Sep 5 at 16:44
• @EpsilonDelta In general, if $\mathcal A$ is a $\sigma$-algebra and $\mathcal F$ is a collection of sets, $\sigma(\mathcal F)\subset \mathcal A$ if and only if $\mathcal A$ is a sigma-algebra that contains $\mathcal F$. So what I wanted to show is that $\mathcal A$ is a sigma-algebra and that it contains all the $\{w\}$. Does that make more sense ? You're right about adding that $\Omega \in \mathcal A$ needs to be verified as well. – Gabriel Romon Sep 5 at 16:58
• This is not necessary. We (you) have already shown that $\mathcal{A}\subset\mathcal{C}$ and hence for having $\mathcal{C}\subset\mathcal{A}$ it is enough to show that $\mathcal{A}$ is a $\sigma$-algebra. It will follow from this statement, that the $\{\omega\}$ are in $\mathcal{A}$ follows immediately. So no need to explicitly show that $\{\omega\}$ are in $\mathcal{A}$. – EpsilonDelta Sep 6 at 8:16
• @EpsilonDelta Sorry, I don't get it. Can you explain why $\mathcal A \subset \mathcal C$ and $\mathcal A$ is a sigma-algebra directly implies that $\mathcal A$ contains all singletons except $\{w_0\}$ ? – Gabriel Romon Sep 6 at 8:30
• If we have $\mathcal{A}=\mathcal{C}$ then we know that if $\omega_0$ were in a set $A$, it must follow that $A^c$ is countable, which implies that $A$ must be uncountable. Thus, no singleton $\{\omega_0\}$ can exist. – EpsilonDelta Sep 7 at 11:41

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 $$\mathcal{A}$$:

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

• 30 seconds apart lol – Gabriel Romon Aug 30 at 17:28
• But you we're faster! And even typed out a complete solution – David H Aug 30 at 17:41
• According to your last line: If $\omega_0\in B\subset \Omega$, there exist uncountable subsets $A_1,\,A_2$ of $\Omega$ such that $A_1\cap A_2=B.$ So if $\mathcal A$ is a $\sigma$-algebra then $B\in \mathcal A$ and $\Omega \setminus B\in \mathcal A.$ So $\mathcal A$ contains every subset of $\Omega.$ This is incorrect. If $\omega_0\in A\subset \Omega$ and $A$ is uncountable then for $A\in \mathcal A$ it is also necessary that $\Omega \setminus A$ is countable. – DanielWainfleet Aug 30 at 18:03
• Thanks for pointing it out, it should be correct now. – David H Aug 30 at 21:07
• @DavidH Your characterization is a bit clearer than the one of Gabriel. Also, good quick and easy demonstration that my previous characterization was wrong! Good Job! – EpsilonDelta Sep 5 at 16:46