# Why modern mathematics prefer $\sigma$-algebra to $\sigma$-ring in measure theory?

Actually, i posted the exact same question before (about a year ago), but now i lost my past account so i couldn't find the past post..

So i googled this, but i couldn't find a satisfying post.

I'll illustrate two different situations below

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First definition

Let $X$ be a set and $\sum \subset P(X)$. Then, $\sum$ is a sigma algebra on $X$ iff (1)it is closed under countable union, (2)it is closed under complement and (3)$X\in \sum$

If we start measure theory via this definition, just like topology, for a given sigma algebra, we can immediately know that on which set this sigma algebra is defined.

Thus, this definition makes it possible to view a sigma-algebra as a measurable space.

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Second definition

Let $X$ be a set and $\sum \subset P(X)$. Then $\sum$ is a sigma ring on $X$ Iff (i) it is closed under countable union, (ii)$\forall A,B\in \sum, A-B\in \sum$.

If we start measure theory in this way, when we want to talk about a measure space, a set and a sigma-ring on this set should be given together. This is the only disadvantage of sigma ring in my opinion.

Let $M=\bigcup \sum$.

Then, $\sum$ is a sigma algebra on $M$ iff $M\in \sum$.

Thus, the definition of sigma ring is strictly stronger than that of sigma algebra.

I cannot understand why mathematicians got rid of this generalized definition and prefer a weaker one.

I remember that someone answered me that a lot of interesting spaces are integrable themselves. Which means, lot of interesting $\sigma$-whatsoever contains the whole space, so they are $\sigma$-algebras.

But, isn't there any interesting $\sigma$-ring which is not $\sigma$-algebra?

I really hate to go back and define something again more generally so that i have to prove every single theorem depending on property of the older definition.

For example, i started analysis with Rudin-PMA and he defined Topology as topology induced by metric space in usual sense. It was painful to me to distinguish theorems which hold only in metric space and which hold in topological space, when i learned general-topology later.

It's hard to confirm myself why mathematicians prefer this weak definition taking this risk..

• – Rahul Jan 3 '14 at 6:53
• @Rahul I saw that post too,but answers there do not answer my question. – Mathems Jan 3 '14 at 6:58
• Any easy example of a $\sigma$-ring which is not a $\sigma$-algebra would be the collection of all countable subsets of an uncountable set. Another one would be the collection of all meagre subsets of some nonmeagre topological space. However, these are both examples of $\sigma$-rings $\Sigma \subset \mathcal{P}(X)$ where $S \subset T \in \Sigma$ implies $S \in \Sigma$. – Mike F Jan 3 '14 at 7:38
• @Mike Is that space interesting in the context of measure theory? – Mathems Jan 3 '14 at 7:42
• @Mathems: Sorry, but I am not sure. – Mike F Jan 3 '14 at 7:52

This is perhaps more a comment than an answer, but it's a bit long for a comment:

I think the difference between the two theories is too small to worry about.

Let me elaborate a bit on that. First, given a σ-ring $\mathcal{R}$ on a set $X$, if it is not already a σ-algebra you can create a σ-algebra $\mathcal{F}=\{A\cup(X\setminus B)\colon A,B\in\mathcal{R}\}$. And a measure on $\mathcal{R}$ can be extended to a measure on $\mathcal{F}$ by setting it to $\infty$ on $\mathcal{F}\setminus\mathcal{R}$.

Conversely, given a σ-algebra $\mathcal{F}$ with a measure on it, the σ-finite members of $\mathcal{F}$ will form a σ-ring.

We don't quite get a simple one-to-one correspondence between measures on σ-rings and measures on σ-fields in this way, but I submit that the above makes translating back and forth between the two theories quite easy, if not trivial. I would, of course, be intrigued to see a counterexample to this conjecture.

• Just for the sake of having a simple example at hand: Let $X$ be set. For each different point $x \in X$, $\mathcal{F}_x = \{ S \subset X : x \notin S\}$ is a different $\sigma$-ring on $X$. However, each of them generates the same $\sigma$-algebra. Namely, the powerset of $X$. – Mike F Jan 3 '14 at 8:05
• @Mike Indeed; this is the sort of example I had in mind when I wrote the last paragraph. However, when you add measures into the mix, this example comes out differently. A measure on $\mathcal{F}_x$ extends to one on $\mathscr{P}(X)$ for which $\{x\}$ is an infinite atom, and if we restrict this measure to the σ-finite sets, we're back in a sub-σ-ring of $\mathcal{F}_x$ again. – Harald Hanche-Olsen Jan 3 '14 at 8:30
• "I would, of course, be intrigued to see a counterexample to this conjecture." Let $X$ be a locally compact Hausdorff space. Let $\mathcal R$ be the $\sigma$-ring generated by all compact subsets of $X$. Let $\mathcal A$ be the $\sigma$-algebra defined as follows. $M \in \mathcal A$ if and only if $M \cap K \in \mathcal R$ for all compact sets $K$. How do you interpret the relation between $\mathcal R$ and $\mathcal A$ by your conjecture? Note that $\mathcal B\subset \mathcal A$, where $\mathcal B$ is the $\sigma$-algebra generated by all open subsets of $X$. – user254385 Jul 16 '15 at 0:15

Measure Theory using $$\sigma$$-rings will lead to more complex notion of measurable function, with some non-intuitive results.

Let $$\Omega$$ be a set and let $$\Sigma$$ be a $$\sigma$$-algebra. Let $$f$$ be a function from $$\Omega$$ to $$\mathbb{R}$$. We say that $$f$$ is measurable if for every Borel set $$B$$ in $$\mathbb{R}$$, $$f^{-1}(B)\in \Sigma$$.

Now, suppose that $$\Sigma$$ is a $$\sigma$$-ring and we try to use the same definition. Then, since $$\Omega=f^{-1}(\mathbb{R})$$, either $$\Sigma$$ is a $$\sigma$$-algebra or there will be no measurable function from $$(\Omega,\Sigma)$$.

So, when working with $$\sigma$$-rings, we need a slightly different definition (as we find in Halmos' book). We say that $$f$$ is measurable if for every Borel set $$B$$ in $$\mathbb{R}$$, $$[f\neq 0]\cap f^{-1}(B)\in \Sigma$$.

This second definition allows the existence of measurable functions even if $$\Sigma$$ is just a $$\sigma$$-ring and not a $$\sigma$$-algebra. However, it leads to a few non-intuitive results. For instance: assume $$\Sigma$$ is just a $$\sigma$$-ring and not a $$\sigma$$-algebra. Then any non-zero constant function is NOT mensurable. As a consequence, if $$f$$ is measurable, then it is easy to prove, for instance, that $$f+1$$ is NOT measurable.

So, the theory of measurable and integrable functions is more naturally developed by using $$\sigma$$-algebras, instead of just $$\sigma$$-rings.

• This is a very nice answer. – Kevin Carlson Oct 31 '14 at 19:19
• More specific to understand – Bear and bunny Sep 4 '15 at 22:12