0
$\begingroup$

My question is as follows: I have two sets, $A$ and $B$. I know that $A$ is convex, $B$ is convex and that $A \subseteq B$. I have shown that the volume (as those sets are three-dimensional) of their difference is 0, i.e. $$V(B \setminus A) = V(B) - V(A) = 0$$

Is this enough to conclude that $A = B$ ? As I understand it, it is enough to conclude that $A = B$ almost everywhere (i.e. except on a set of volume 0), but I wonder if there is a way to exploit convexity to conclude.

Edit: Thanks for the answers, I have a few more then:

  • If A and B are open (resp. closed), can we conclude ?
  • Does this mean that $\mathring{A} = \mathring{B}$ and/or $\overline{A} = \overline{B}$ ?
  • 1
    Think about open vs closed intervals/cubes. – 2017-02-21
  • 0
    If one of the sets has a non empty interior, then if the symmetric difference is zero, you can conclude that the closures (or interiors) are the same. – 2017-02-21
  • 0
    @copper.hat : Thanks, I can definitely prove that $A$ and $B$ are non-empty. Do you have any reference for this property? – 2017-02-21
  • 0
    @Urukann the interiors being the same follows from the result that the boundary of a convex set is always measure zero, and then you can show that the if the two interiors (which are open convex sets) have measure zero symmetric difference then they (the interiors, not necessarily your original $A,B$) coincide. – 2017-02-21
  • 0
    @Urukann: If the interiors do not coincide, then one of the sets will contain an open set that does not intersect the other, hence the measure of their symmetric difference will be non zero. Hence the interiors coincide, and it follows from this (since they are convex) that the closures coincide. (This is under the presumption that at least one of the sets has an interior point, if this is not true then the interiors will both be empty, but the closures may differ.) – 2017-02-21

2 Answers 2

2

No, it's not enough. For instance, in $\Bbb R$, $[0,1]$ and $(0,1)$ are both convex, and their difference has measure $0$ (at least in the standard measure), but they're not equal.

For something more extreme, note that anything restricted to a plane in $\Bbb R^3$ has measure $0$, so if $A\subseteq B$ are convex subsets of the $xy$-plane, for instance, then they could be basically anything, like the whole plane and a single point.

  • 0
    Thanks! Is it enough though to conclude that the closure of $A$ is equal to the closure of $B$ ? – 2017-02-21
  • 0
    No, it's not. See my other example. – 2017-02-21
  • 0
    Thanks for the edits, this clarifies it quite nicely! – 2017-02-21
1

Consider an open ball $B(0,1)= \{(x,y,z):x^2+y^2+z^2<1\}$ of radius 1 centered on the origin, and the closed ball $\overline{B(0,1)}= \{(x,y,z):x^2+y^2+z^2\le1\}$ of radius 1 centered on the origin. In this case both sets are convex, and their difference is measure zero.