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

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 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.