# discontinuity of characteristic function

For a subset $A$ of $\mathbb{R}^{n}$, the set of the discontinuity () points of the characteristic function    $\chi_{A}$ is the boundary (http://planetmath.org/BoundaryFrontier) of $A$.

Proof.  Let $a$ be a discontinuity point of $\chi_{A}$.  Then any neighborhood  (http://planetmath.org/Neighborhood) of $a$ contains the points $b$ and $c$ such that  $\chi_{A}(b)=1$  and  $\chi_{A}(c)=0$.  Thus  $b\in A$  and  $c\notin A$,  whence $a$ is a boundary point of $A$.

If, on the contrary, $a$ is a boundary point of $A$ and $U(a)$ an arbitrary neighborhood of $a$, it follows that $U(a)$ contains both points belonging to $A$ and points not belonging to $A$.  So we have in $U(a)$ the points $b$ and $c$ such that  $\chi_{A}(b)=1$  and  $\chi_{A}(c)=0$.  This means that $\chi_{A}$ cannot be continuous at the point $a$ (N.B. that one does not need to know the value $\chi_{A}(a)$).

Title discontinuity of characteristic function DiscontinuityOfCharacteristicFunction 2015-02-03 21:23:33 2015-02-03 21:23:33 pahio (2872) pahio (2872) 3 pahio (2872) Theorem msc 03-00 msc 26-00 msc 26A09