I started reading Rudin's Principles of Mathematical Analysis for some Lebesgue Theory. Rudin introduces both rings and $\sigma$-rings, but I don't see the difference between them.

Assuming I'm not misunderstanding the definition, a ring is a family $\mathscr{R}$ of sets that is closed under set difference and unions. On the other hand, a $\sigma$- ring is a ring with the property that $\bigcup_{n=1}^{\infty}A_n\in\mathscr{R}$, where $A_i\in\mathscr{R}$. But isn't this just saying that a $\sigma$-ring is a ring that is closed under union, which we already know from the definition of a ring?

If someone could help clear this doubt, it would be appreciated. Thanks.

  • 3
    $\begingroup$ A ring is closed under finite unions, a $\sigma$-ring closed under countable unions. $\endgroup$
    – Hayden
    Mar 6, 2015 at 15:16

1 Answer 1


Rudin defines a family $\mathscr{R}$ of sets to be a ring if for any two $A,B\in \mathscr{R}$, we have $A\setminus B$ and $A\cup B$ in $\mathscr{R}$.

On the other hand, a $\sigma$-ring is a family $\mathscr{R}$ of sets that is a ring such that we additionally have $\bigcup_{n=0}^\infty{A_n}$ for any sequence $(A_n)_{n\in \mathbb{N}}$ of elements of $\mathscr{R}$.

As an example of a ring that is not a $\sigma$-ring, take $\mathscr{R}$ to be the set of all finite subsets of $\mathbb{N}$. This is a ring, since given $A,B$ finite, we have $A\cup B$ and $A\setminus B$ finite. However, it is not a $\sigma$-ring: Take $A_n=\{n\}$, so that $\bigcup_{n=0}^\infty{A_n}=\mathbb{N}$, which is certainly not finite, and hence does not belong to $\mathscr{R}$. This shows that $\mathscr{R}$ cannot be a $\sigma$-ring.

  • $\begingroup$ In this example $\mathscr{R}$ is a $\sigma$-algebra, right? $\endgroup$
    – Xichu
    Mar 5, 2021 at 16:04
  • $\begingroup$ It is not; if it were a $\sigma$-algebra it would also be a $\sigma$-ring, and $\mathscr{R}$ was an example of a ring of sets which is not a $\sigma$-ring. $\endgroup$
    – Hayden
    Mar 5, 2021 at 16:33
  • $\begingroup$ But it is an algebra, since it contains the whole set $\mathbb{N}$, isn't it? $\endgroup$
    – Xichu
    Mar 5, 2021 at 18:49
  • $\begingroup$ It's the set of all finite subsets of $\mathbb{N}$. $\endgroup$
    – Hayden
    Mar 5, 2021 at 19:12

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .