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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?

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  • $\begingroup$ Any reaction to my answer ? $\endgroup$ – Gabriel Romon Sep 4 at 19:41
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    $\begingroup$ @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. $\endgroup$ – EpsilonDelta Sep 4 at 22:56
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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.

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  • $\begingroup$ 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? $\endgroup$ – EpsilonDelta Sep 5 at 16:44
  • $\begingroup$ @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. $\endgroup$ – Gabriel Romon Sep 5 at 16:58
  • $\begingroup$ 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 $\mathcal{A}$. $\endgroup$ – EpsilonDelta Sep 6 at 8:16
  • $\begingroup$ @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 $\{w_0\}$ ? $\endgroup$ – Gabriel Romon Sep 6 at 8:30
  • $\begingroup$ 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. $\endgroup$ – EpsilonDelta Sep 7 at 11:41
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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} $:

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

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  • $\begingroup$ 30 seconds apart lol $\endgroup$ – Gabriel Romon Aug 30 at 17:28
  • $\begingroup$ But you we're faster! And even typed out a complete solution $\endgroup$ – David H Aug 30 at 17:41
  • $\begingroup$ 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. $\endgroup$ – DanielWainfleet Aug 30 at 18:03
  • $\begingroup$ Thanks for pointing it out, it should be correct now. $\endgroup$ – David H Aug 30 at 21:07
  • $\begingroup$ @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! $\endgroup$ – EpsilonDelta Sep 5 at 16:46

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