Measure of the symmetric difference of two closed regions Given two bounded closed regions $A,B\subset\mathbb{R}^n$, let $A\bigtriangleup B$ denotes the symmetric difference of $A$ and $B$ (i.e.$\left (A-B\right)\cup\left(B-A\right)$) and $m\left(S\right)$ be the Lebesgue measure of $S$,
does $m\left(A\bigtriangleup B\right)=0$ imply that $A=B$?
In other words, can $m\left(X\bigtriangleup Y\right)$ define a metric on the set of all closed regions in $\mathbb{R}^n$?
 A: We show that $A \neq B$ implies $m(A \Delta B) >0$.
If $A \neq B$, then w.l.o.g. $x \in A \setminus B$. Since $B$ is closed, this implies that there exists $\varepsilon >0$ with $B_\varepsilon(x) \cap B = \emptyset$. Since $A$ is a region $B_\varepsilon(x)$ contains an inner point of $A$, say $y \in A$. Thus for some $\delta >0$ we have $B_\delta(y) \subset B_\varepsilon(x) \cap A$, but $B_\delta(y)$ and $B$ are disjoint. Hence $B_\delta(y) \subset A \Delta B$. Thus $m(A \Delta B) >0$.
Of course, this map is symmetric by definition. We are left to show that the triangle-inequality is satisfied.
Let $A,B,C$ be bounded closed regions. Then $x \in A \Delta C$ implies w.l.o.g $x \in A$ but $x \notin C$. If $x \notin B$, then $x \in A \Delta B$, otherwise $x \in B \Delta C$, because $x \in B$. Thus $A \Delta C \subset A \Delta B \cup B \Delta C$ and
$$m(A \Delta C) \leq m(A \Delta B) + m(B \Delta C).$$
A: If by $A-B$ you mean the set $\{ x : x \in A \text{ and } x \notin B \}$ then its not true : 
Consider two different points $ \alpha \neq \beta $ and take $A = \mathbb{R}^n -\{\alpha\}$ and $B = \mathbb{R}^n - \{\beta\}$.
If you consider the equivalence relation $ A \sim B $ iff $ m(A \bigtriangleup B)=0 $ you may have a metric.
