TITLE: Metric induced by measure $d(A,B) = \mu( A \triangle B)$. Ideal?
QUESTION [1 upvotes]: Let $d(A,B) = \mu( A \triangle B)$ for content $\mu$ and a ring $R$. 
I proved that 
$(1)\quad A \sim B \Longleftrightarrow  d(A,B) = 0 $ is a equivalence relation
and
$(2) \quad \mu(A) = \mu(A \cap B) = \mu(B)$.
Now I’m struggling with answer 
$(3) \quad$ Is $ [\emptyset ]$ an Ideal in $R$ or at least a Ring?
Any help here is appreciated; thanks.

REPLY [0 votes]: To clarify: if $R$ is a ring of sets, then $(R, \Delta, \cap)$ is also a ring (commutative with unity) in the algebraic sense. That's why we can take a subset $I \subseteq R$ and ask whether it is a (ring theoretic) ideal.
For $I$ to be an ideal, it must satisfy the following two properties:
$$\begin{align*}
& \bullet (\forall a, b \in I) \, a+b \in I &&&& (\text{where } + = \Delta) \\[1ex]
& \bullet (\forall r \in R)(\forall a \in I) \, r \cdot a \in I &&&& (\text{where } \cdot = \cap)
\end{align*}$$
We will now check if these hold for $I = [\varnothing] \subseteq R$.

Let $A, B \in [\varnothing]$. By definition it means that $\mu(A) = \mu(B) = 0$. Then of course $\mu(A \cup B) = 0$, but $A \Delta B \subseteq A \cup B$, hence $\mu(A \Delta B) = 0$, which implies $A \Delta B \in [\varnothing]$.
Let $A \in [\varnothing]$ and $S \in R$. By definition $\mu(A) = 0$. Since $R \cap A \subseteq A$, we also have $\mu(R \cap A) = 0$, hence $R \cap A \in [\varnothing]$.

This concludes the proof that $I = [\varnothing]$ is an ideal in $R$.
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