Let $\Omega$ be a set and $\mathcal{E}$ be a collection of subsets of $\Omega$. Assume that we have nonempty sets $A_0 \subset A \subset \Omega$ such that $A_0 \neq A$ and $\forall B \in \mathcal{E}$ \begin{equation} A \subset B \text{ or } A \cap B = \varnothing. \end{equation} Prove that $A_0 \notin \sigma(\mathcal{E}),$ where $\sigma(\mathcal{E})$ denotes the $\sigma$-algebra generated by $\mathcal{E}$.

I figured I should try to see what happens if $A_0 \in \sigma(\mathcal{E})$ for both cases where $A \subset B \text{ or } A \cap B = \varnothing$, and then by contradiction in the basic rules of a $\sigma$-algebra find a contradiction.

(1) $\Omega \in \sigma(\mathcal{E});$

(2) $A \in \sigma(\mathcal{E}) \Rightarrow A^c \in \sigma(\mathcal{E});$

(3) $A_n \in \sigma(\mathcal{E}), n \in \mathbb{N} \Rightarrow \cup^\infty_{n = 1} A_n \in \sigma(\mathcal{E})$

Following this train of thought, we know that if $A_0 \in \sigma(\mathcal{E}) \Rightarrow A_0^c \in \sigma(\mathcal{E}).$ Yet by trying things for this I couldn't quite find a contradiction. Am I missing something crucial about generated $\sigma$-algebras, or is there something else I'm not seeing? If you want a more elaborate explanation of the things I tried, then I can post it. But I hoped that this would be enough for now.


1 Answer 1


Let $\mathcal B:=\{B\in\wp(\Omega)\mid A\subseteq B\vee A\cap B=\varnothing\}$.

Then $\mathcal B$ is a $\sigma$-algebra with $\mathcal E\subseteq\mathcal B$ so that also $\sigma(\mathcal E)\subseteq\mathcal B$.

(Observe that we can also write $\mathcal B:=\{B\in\wp(\Omega)\mid A\cap B^{\complement}=\varnothing\vee A\cap B=\varnothing\}$ which makes it immediate the $\mathcal B$ is closed under complements. Further it is evident that $\mathcal B$ is closed under countable unions.)

Now note that $A_0\notin\mathcal B$ so that also $A_0\notin\sigma(\mathcal E)$.

edit: (proof that $\mathcal B:=\{B\in\wp(\Omega)\mid A\cap B^{\complement}=\varnothing\vee A\cap B=\varnothing\}$ is a $\sigma$-algebra)

1) From $\varnothing\cap B=\varnothing$ it follows that $\varnothing\in\mathcal B$

2) If $B\in\mathcal B$ then $A\cap B^{\complement}=\varnothing\vee A\cap B=\varnothing$ or equivalently $A\cap B^{\complement}=\varnothing\vee A\cap (B^{\complement})^{\complement}=\varnothing$, telling us that $B^{\complement}\in\mathcal B$.

3) If $B_n\in\mathcal B$ for $n=1,2,3,\dots$ and $B=\bigcup_{n=1}^{\infty}B_n$ then we discern two cases:

(i) $A\cap B_n=\varnothing$ for every $n$. In this case $A\cap B=\varnothing$ so that $B\in\mathcal B$.

(ii) $A\cap B_{n_0}^{\complement}=\varnothing$ for some $n_0\in\{1,2,3,\dots\}$. Then also $A\cap B^{\complement}=\varnothing$ (because $B^{\complement}\subseteq B_{n_0}^{\complement}$) so $B\in\mathcal B$.

  • $\begingroup$ I want to say that this is correct, but I can't seem to grasp why $A_0 \notin \mathcal{A}$. $\endgroup$
    – TSpoon
    Commented Jan 24, 2018 at 14:55
  • 1
    $\begingroup$ We do not have $A\subseteq A_0$ and also we do not have $A\cap A_0=\varnothing$ so by definition: $A_0\notin\mathcal A$. $\endgroup$
    – drhab
    Commented Jan 24, 2018 at 14:59
  • $\begingroup$ Oh yeah of course, that's quite a mistake of mine. $\endgroup$
    – TSpoon
    Commented Jan 24, 2018 at 15:00
  • $\begingroup$ I think the cause of confusion might have been that I wrote $\mathcal A$ having elements $B$. I will make it $\mathcal B$ instead. $\endgroup$
    – drhab
    Commented Jan 24, 2018 at 15:02
  • $\begingroup$ Nah, I just had a brainfart. Reading set definitions can sometimes confuse me. $\endgroup$
    – TSpoon
    Commented Jan 24, 2018 at 15:09

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