I am trying to construct a family $S$ of measureable subsets or $R^2$, on which the symmetric difference, defined as: $SD(A,B) = Area(A\setminus B \cup B \setminus A)$, is a metric, i.e., different elements of $S$ have a symmetric difference with positive area, and $S$ is complete under that metric.
Consider first the family of open discs centered at the origin. Here $SD$ is obviously a metric, since different discs have different radii and $SD(A,B)=\pi|r_A^2-r_B^2|>0$. To make $S$ complete under this metric, we have to add the empty set (the limit when the $r\to 0$) and the entire plane $R^2$ (the limit when $r\to \infty$).
$SD$ remains a metric when we add all open discs, since the symmetric difference of discs with a different center also has positive area.
From similar considerations, it seems obvious that I can also add to $S$ all open triangles, quadrangles or polygons with the number of sides bounded by a constant (although I don't have a formal proof for that).
Initially I thought that I can also add to $S$ all open polygons and even all simply-connected open shapes, but now I think that this will make $S$ incomplete. The example I found for this is based on an answer by Joonas Ilmavirta and an answer by sds:
Let $E_n$ be the interior of $n$-th step in the construction of a Smith–Volterra–Cantor set in $(0,1)$. Define the following sequence of sets:
$$F_n=(0,1)\times(0,0.5) \cup E_n\times(0,1)$$
$F_n$ is a comb-shaped rectilinear polygon similar to the following:
with the number of dents growing as $n$ grows. Moreover, it is open and simply-connected. But, the limit of $F_n$ when $n\to\infty$ is not open and not a polygon. The area of the limit is $0.5 + 0.5\cdot 0.5=0.75$ but the 'closest' open set is $(0,1)\times(0,0.5)$ which has area 0.5.
So, my question is now: what subset of the open simply-connected shapes can I add to $S$, such that $SD$ remains a metric on $S$ and $S$ remains complete in that metric?
Finally I think I found an answer:
SD is a metric on the set of all regular closed sets.
(A regular closed set is a set $X$ equal to the closure of its interior: $X=Cl[Int[X]]$).
Proof:
Let $X$ and $Y$ be two regular closed sets such that $Area[X\setminus Y]=Area[Y\setminus X]=0$. We prove that $X=Y$.
$X\setminus Y = X\cap Y^C$ (where $Y^C$ is the complement of $Y$).
$X\cap Y^C \supseteq Int[X]\cap Y^C$
So $Area[Int[X]\cap Y^C]=0$
These two sets are both open so their intersection is open.
The only open set with an area of 0 is the empty set (because any non-empty open set contains a ball with a positive measure). Hence:
$Int[X]\cap Y^C=\emptyset$
$Int[X]\subseteq Y$
Take the Int of both sides:
$Int[X]\subseteq Int[Y]$
Similarly:
$Int[Y]\subseteq Int[X]$
Hence:
$Int[Y] = Int[X]$
Take the Cl of both sides and use the fact that they are regular closed:
$Y = X$
QED.
CREDITS:
Shapes bounded only by lines (TonyK)
When does intersection of measure 0 implies interior-disjointness? (PhoemueX and dafinguzman)
Area of set-difference of special sets (Henno Brandsma)