# What is the catch when introducing measure theory using $\sigma$-ring instead of $\sigma$-algebra?

I am currently using Matthew A Pons book Real Anaysis for Undergrad for introduction of measure theory

In my opinion this book is unbelievably clear for almost all the sections EXCEPT the section where measure is introduced. And the reason is because the author insists on using something called $\sigma$-ring for all the definitions, but litters the sections with ring and $\sigma$-algebra so in the problem section you will see that one question is with respect to $\sigma$-ring another is with respect to $\sigma$-algebra

And you will see some quite convoluted parts like: Also let's ignore for the time being that almost all the books written post 2000s deals with $\sigma$-algebra. This book was published in 2014.

So the main question is what is the catch when developing measure theory from the perspective of $\sigma$-ring instead of $\sigma$-algebra? I know that for a ring, the condition $A^c \in \mathcal{R}$ does not hold (not closed under complement). Can someone contrast the difference in the ways that key properties of the measure are derived (or proven) with respect to a $\sigma$-ring instead of a $\sigma$-algebra?

• What key properties? U want to prove if something is a measure on sigma algebra that it is a measure on sigma ring ? A measure has its definition - nothing is derived. Measure is a function that has 2 key properties - empty set has 0 measure and measure of disjunct union of sets is the same as sums of measures of those sets Apr 16 '16 at 23:08

There is no real catch. The concept of $$\sigma$$-ring is a slightly more general than the concept of $$\sigma$$-algebra. In fact, it is easy to see that, given a $$\sigma$$-ring $$\Sigma$$ on a set $$\Omega$$, then $$\Sigma$$ is a $$\sigma$$-algebra if and only if $$\Omega\in \Sigma$$.

So you might think: "why don't we develop Measure Theory based on $$\sigma$$-rings?" Surely it can be done, and very good books follow this track. For instance: Halmos, Measure Theory. HOWEVER, using $$\sigma$$-rings makes the definition of measurable functions more complex and with some non-intuitive results.

As I wrote to answer another question some time ago:

Measure Theory using $$\sigma$$-rings will lead to a 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 $$\mathbb{R}$$, $$f^{-1}(A)\in \Sigma$$.

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 $$\mathbb{R}$$, $$[f\neq 0]\cap f^{-1}(A)\in \Sigma$$.

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.

You may also be interested in reading my answer in Two different definitions of General Measurable Function .