# When does a set $\mathcal{E}$ generates the same ring and $\sigma$-algebra?

Let $$\mathcal{E}\subset \Omega$$ and let $$\rho(\mathcal{E})$$ and $$\sigma(\mathcal{E})$$ be respectively the ring and $$\sigma$$-algebra on $$\Omega$$ generated by $$\mathcal{E}$$. The book am reading asks under what conditions on $$\mathcal{E}$$ do we have $$\rho(\mathcal{E})=\sigma(\mathcal{E})$$.

If $$\mathcal{E}$$ is finite and $$\cup\mathcal{E}=\Omega$$, then $$\rho(\mathcal{E})$$ is actually the algebra generated by $$\mathcal{E}$$, and I was able to show that the algebra and $$\sigma$$-algebra generated by a finite set coincide.

Can I say something for general $$\mathcal{E}$$?

Thanks a lot for your help.

EDIT: After further thought I believe that, in general, $$\rho(\varepsilon)=\sigma(\varepsilon)$$ if and only if $$\Omega$$ is a finite union of sets from $$\varepsilon$$ and $$\rho(\varepsilon)$$ is closed under countable unions. I posted a proof attempt below. Any feedback is greatly appreciated.

The $$ring$$ generated by $$\varepsilon$$, along with all the compliments of the sets belonging to it (w.r.t to the underlying set) will be equal to the $$algebra$$ generated by $$\varepsilon$$. The algebra along with all the countable unions of the sets belonging to it will be the $$\sigma-algebra$$ generated by $$\varepsilon$$

• Ok but under what conditions will $\rho(\mathcal{E})=\sigma( \mathcal{E})$? – Alphie Jun 28 '20 at 14:57
• Suppose $\varepsilon$ contains the underlying set, or contains sets whose finite union is the underlying set, then clearly $ring$ is equal to $algebra$ generated by it. For that algebra to be sigma algebra, the ring should be a $\sigma$ ring. – Bhaswat Jun 28 '20 at 15:06
• So in general if $\rho(\mathcal{E})$ is only a ring (not a $\sigma$-ring) then we need $\mathcal{E}$ to be finite for the equality $\rho(\mathcal{E})=\sigma( \mathcal{E})$ to hold? – Alphie Jun 28 '20 at 15:29
• Depends on what exactly the base set is. If you consider the sigma algebra and ring generated by the power set of an infinite set, they are still the same. But consider this example. Your $\varepsilon$ is the collection of all the one element subsets of the natural nos. So in the ring generated by it, you will get all the finite subsets. But in the sigma algebra, you essentially get all the subsets. Even if you ensure like the finite case, that the union of the base set is the whole set, even then the ring might not be equal to the sigma algebra. Consider all the closed intervals on reals. – Bhaswat Jun 28 '20 at 15:52
• The sigma algebra generated by it is the Borel sigma algebra, which includes all the open intervals and semi open intervals. But the ring generated by the same, won't include it all. Verify ! – Bhaswat Jun 28 '20 at 15:54

Proposition. In general, $$\rho(\varepsilon)=\sigma(\varepsilon)$$ if and only if $$\Omega$$ is a finite union of elements from $$\varepsilon$$ and $$\rho(\varepsilon)$$ is closed under countable unions.

Proof. Suppose $$\rho(\varepsilon)=\sigma(\varepsilon)$$. Then $$\Omega \in \rho(\varepsilon)$$. I will use a more explicit definition of $$\rho(\varepsilon)$$. Define an increasing sequence $$(\mathcal{F}_n)\subset \mathcal{P}(\Omega)$$ inductively as follows: let $$\mathcal{F}_1=\varepsilon \cup \{\emptyset\}$$ and let $$\mathcal{F}_{n+1}=\mathcal{F}_{n} \cup\{A\setminus B : A,B \in \mathcal{F}_{n}\} \cup \{A\cup B : A,B \in \mathcal{F}_{n}\}$$ It is not too difficult to show that $$\rho(\varepsilon)=\bigcup_{n\in\mathbb{N}}\mathcal{F}_{n}$$. Now, let $$n_1$$ be the smallest integer such that $$\Omega \in \mathcal{F}_{n_1}$$. Then either $$\Omega=A_1\setminus B_1\subset A_1$$ or $$\Omega=A_1\cup B_1$$ for some $$A_1,B_1 \in \mathcal{F}_{n_1-1}$$. Inductively, if $$n_k$$ is the smallest integer such that $$C_{k-1} \in \mathcal{F}_{n_k}$$ where $$n_1>n_2>\dots>n_k$$, then $$C_{k-1}=A_k\setminus B_k\subset A_k$$ or $$C_{k-1}=A_k\cup B_k$$ for some $$A_k,B_k \in \mathcal{F}_{n_k-1}$$. Hence after at most $$n-1$$ steps we get that $$\Omega$$ is a subset of a finite union of sets from $$\varepsilon \subset \mathcal{P}(\Omega)$$, and so $$\Omega$$ is actually the union of those sets. The fact that $$\rho(\varepsilon)=\sigma(\varepsilon)$$ is closed under countable unions is immediate.

Conversely suppose $$\Omega$$ is a finite union of elements from $$\varepsilon$$ and $$\rho(\varepsilon)$$ is closed under countable unions. Then $$\Omega \in \rho(\varepsilon)$$ and so we actually have $$\rho(\varepsilon)=\mathcal{A}(\varepsilon)$$, the algebra generated by $$\varepsilon$$. But $$\rho(\varepsilon)$$ is closed under countable unions, so in fact $$\mathcal{A}(\varepsilon)=\sigma(\varepsilon)$$.

• @Bhaswat Could I have your feedback on this proof svp? – Alphie Jul 7 '20 at 20:18