# $\sigma$-ring of subsets of $X$

I am reading Kenneth Hoffman's book on Banach spaces of analytics functions. In the first chapter Measure and integration the very first paragraph starts with this:

If $$X$$ is a set, the collection of all subsets of $$X$$ forms a ring under the following operation:

$$A+B=(A\cup B)-(A\cap B),$$

$$AB=A\cap B.$$

A $$\sigma$$-ring of subsets of $$X$$ is a subring of the ring of all subset of $$X$$ which is closed under formation of countable unions (and, a fortiori, closed under countable intersection).

I am finding it difficult to understand. Why the "closed under countable intersection" is written in bracket. Does it mean this already follows from first two conditions and closed under formation of countable unions? If not, then why it is written in bracket? If yes, how to do it?

Edit:

• Yeah. (Conventionally in ring theory, "subrings" of unital rings are required to have the multiplicative identity in them. Here, that's $X$. So if $S$ is a $\sigma$-ring, you do indeed know that $X$ is in $S$, which you implicitly use in proving that $S$ is closed under complementation.) May 20, 2021 at 4:04
• @leslietownes: $\sigma$-rings are not necessarily closed under complementation. If they are, then they form what is called $\sigma$--algebra. A ring that is closed under countable intersections is called $\delta$-ring; a ring that is closed under countable unions is called a $\sigma$-ring. The collection of Borel sets of the real line that have finite Lebesgue Measure for example, is an example of a $\delta$-ring that is not a $\sigma$-ring. May 20, 2021 at 4:06
• @pmun: it is a simple exercise to show that $\mathcal{R}$ is ring iff $\mathcal{R}$ is closed under finite union of sets and under differences of sets, that is, iff whenever $A,B\in\mathcal{R}$, $A\cup B\in\mathcal{R}$ and $A\setminus B\in\mathcal{R}$. If in addition $\mathcal{R}$ is closed under coubntable unions, then $A_1\setminus\bigcup_n(A_1\setminus A_n)\in\mathcal{R}$ whenever $A_n\in\mathcal{R}$. Notice that $A_1\setminus\bigcup_n(A_1\setminus A_n)=\bigcap_nA_n$ May 20, 2021 at 4:15
– pmun
May 20, 2021 at 4:17
• @AHusain: I am fine with $\sigma$-ring. The point here is that in measure theory and in topology, the $\sigma$ denotes countable unions, and the $\delta$ denotes countable intersections. One can also talk about a measure being $\sigma$-continuous, which means that $\mu(B_n)\xrightarrow{n\rightarrow\infty}\mu(\bigcup_nB_n)$ if $B_n\subset B_{n+1}$, and $\delta$-continuous if $\mu(B_n)\xrightarrow{n\rightarrow\infty}\mu(\bigcap_nB_n)$ if $B_{n+1}\subset B_n$. May 20, 2021 at 4:35

From $$A+B=(A\setminus B)\cup (B\setminus A)$$, you get $$A + (A\cap B)=A\setminus B$$. So a ring is closed under difference of sets. Also $$(A+B)+(A\cap B)=A\cup B$$, so a ring is closed under finite unions. Conversely, a collection of sets that is closed under differences and finite unions is a ring: for example $$A\setminus(A\setminus B)=A\cap B$$.

• A $$\sigma$$-ring is a ring that is closed under countable unions.

• A $$\delta$$-ring is a ring that is closed under countable intersections.

If $$\mathcal{R}$$ is a $$\sigma$$-ring and $$(A_n:n\in\mathbb{N})\subset\mathcal{R}$$, then $$A_1\setminus\bigcup_n(A_1\setminus A_n)\in\mathcal{R}$$ Notice that $$A_1\setminus\bigcup_n(A_1\setminus A_n)=A_1\cap\bigcap_n(A^c_1\cup A_n)=\bigcap_n(A_1\cap(A^c_1\cup A_n)=\bigcap_nA_1\cap A_n=\bigcap_nA_n$$ That is, a $$\sigma$$-ring is also a $$\delta$$-ring. The converse is not necessarily true. For example, the collection of all Borel sets in the real line that have finite Lebesgue measure is a $$\delta$$-ring, but not a $$\sigma$$-ring.

• Thankyou for a detailed answer.
– pmun
May 20, 2021 at 4:22