Proof $U\sim V \iff \mu(U \Delta V)=0$

Question

Let $(X,Y,\mu)$ be a measure space and $U,V\in Y$. Show that, and define the equivalence relation,

$U\sim V \iff \mu(U \Delta V)=0$

---

note: the measure is of the symmetric difference.

I know that it is a equivalence relation if

1. $U\sim U$
2. if $U\sim V$ then $V\sim U$
3. if $U\sim V$ and $V\sim W$ then $U\sim V$

but how do I show the 'iff'-statement whilst defining the equivalence relation?

Thanks!

Answer 1

My interpretation is that the question wants you to assume that you don't know what an equivalence relation formally means.

You are asked to find the meaning of an instance of an of 'equivalence' between measurable sets and prove that it holds iff $\mu(U\Delta V)=0$. Since $U\Delta V = (U/V)\cup (V/U)$, it is a union of disjoint sets, so $\mu(U\Delta V)=\mu(U/V)+\mu(V/U)$. For $\mu(U\Delta V)=0$ to hold, we must have that they are either empty sets or sets of measure zero. This also leads to $$\mu(U\cup V)=\mu(U\Delta V)+\mu(U\cap V)\implies \mu(U\cup V)=\mu(U\cap V)$$ Therefore, the sets $U$ and $V$ are 'equivalent' under $\mu$, because the sets of the elements on which they differ (i.e. $U/V$ and $V/U$) have measure zero. So

Definition. Let $(X,\mathscr{A},\mu)$ be a measure space. Let $U,V\in \mathscr{A}$. Then if $\mu(U/V)=\mu(V/U)=0$ we say $U \sim V$.

Lemma. $U\sim V$ if and only if $\mu(U\Delta V)=0$.

Proof. $(\Rightarrow)$. Since $\mu(U\Delta V)=\mu((U/V)\cup(V/U))$, by additivity $\mu(U\Delta V)=\mu(U/V)+\mu(V/U)=0$. $(\Leftarrow)$. Since $\mu(A)\geq 0,\,\forall A \in \mathscr{A}$, we have that $\mu(U/V)+\mu(V/U)=0$ implies that $\mu(U/V)=\mu(V/U)=0$.

Ultimately, you will find that $\mu(U\Delta V)=0$ describes an actual equivalence relation as you defined in the OP (but, for the purpose of the question, you are not supposed to 'know').

Answer 2

The problem is badly worded. If the relation is defined as $$U \sim V \iff \mu(U \Delta V) = 0,$$ there is no point in showing that this equivalence is true as this holds by the very definition. You are probably asked to show that the relation defined above is an equivalence relation (which I assume you know how to do since you did not ask).

Answer 3

1. $U\Delta U = (U\setminus U) \cup (U \setminus U) = \emptyset$. $\mu(U\Delta U) = \mu(\emptyset) = \mu(0)$
2. $U \Delta V = (U \setminus V)\cup (V \setminus U) = (V \setminus U)\cup (U \setminus V) = V \Delta U$. This means that if $U \sim V $ then $\mu(U\Delta V) =0$ but $\mu(U \Delta V) = \mu(V \Delta U) = 0$ which implies $V \sim U$.
3. If you assume $\mu(U \Delta V) =0$ and $\mu(V \Delta W) =0$ can you prove $\mu(U \Delta W) = 0$ ?

Bonus: notice that this equivalence relationship defines a new measure space $X/\sim$