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?

  • $\begingroup$ Any reaction to my answer ? $\endgroup$ – Gabriel Romon Sep 4 at 19:41
  • 1
    $\begingroup$ @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

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.

  • $\begingroup$ 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

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} $:

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

  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.