Let $\mathscr{R}$ be a ring of sets and define set operations $\odot=\text{multiplication}$ and $\oplus=\text{addition}$ by $$E\odot F=E\cap F$$ and $$E\oplus F=E\triangle F$$
then $\mathscr{R}$ is an algebraic ring (when the element $1$ is not required to exist)
I have been submitted to a group theory course and the first time I encountered rings was on measure and integration theory course.The last exercise is from a measure theory textbook. But I have no idea what is meant to answer.
Hence the definition of ring I know.
$\mathscr{C}$ is a ring iff
i) $\emptyset\in \mathscr{C}$
ii)$A\in\mathscr{C}$ and $B\in\mathscr{C}$ so $A\cup B\in\mathscr{C}$
iii)$ A\in\mathscr{C}$ and $B\in\mathscr{C}$ so $A\setminus B\in\mathscr{C}$.
Question:
What is being asked? How should I solve the question?
Thanks in advance!
It does not make sense to say that $\mathscr{R}$ is the ring of sets, since $\mathscr{R}$ has to be a set, and the set of all sets is in fact not a set. I am guessing that you have a priori a universe $\Omega$. Then, you define $\mathscr{R}$ to be the (algebraic) ring of subsets of $\Omega$ with the operations you gave. In fact, $\mathscr{R}$ is indeed a ring with multiplicative identity $1$ in this sense, which is just $1=\Omega$.
Well, I was a bit mistaken about the ring of sets (I am not removing the first paragraph, as it probably has some mathematical value). You have a set-theoretical ring $\mathscr{R}$ of sets and you equip it with an algebraic ring structure. If the universe $\Omega$ is not in $\mathscr{R}$, then it is possible that your ring lacks the multiplicative identity. However, $\mathscr{R}$ is a boolean ring (that is, $x^2=x\odot x=x$ for every $x\in \mathscr{R}$). Note also that the additive identity is $0=\emptyset$.
You have to verify a few things.
Check that both $\oplus$ and $\odot$ are indeed binary operations of $\mathscr{R}$ (this is where you use the assumption that $\mathscr{R}$ is a set-theoretical ring of sets).
Show that $\mathscr{R}$ is an abelian group with the addition $\oplus$.
Show that the multiplication $\odot$ is associative.
Show that $\odot$ distributive over $\oplus$ on both the left and the right.
See also here for the definition of algebraic rings (ignoring the multiplicative identity requirement, as your set-theoretical ring does not assume that the universe $\Omega$ is in it). Note that the notion of set-theoretical rings is irrelevant to the notion of algebraic rings.
Here is a final remark. If the union $U$ of all sets in $\mathscr{R}$ is in $\mathscr{R}$, then $U$ is the multiplicative identity of $\mathscr{R}$. In the case that $\mathscr{R}$ is finite, $U$ is always in $\mathscr{R}$. This may fail if $\mathscr{R}$ is infinite. For example, if I take $\Omega=\mathbb{R}$ and take $\mathscr{R}$ to be the set-theoretical ring generated by singletons $\{x\}$ for $x\in \mathbb{R}$, then $\mathscr{R}$ contains only finite subsets of $\mathbb{R}$, but the union $U$ of all elements of $\mathscr{R}$ is $\mathbb{R}$, which is not in $\mathscr{R}$.