TITLE: Function fails to be of bounded variation
QUESTION [6 upvotes]: Let $f$ fail to be of bounded variation on [0,1]. Show that there is a point $x_0$ in [0,1] such that $f$ fails to be of bounded variation on each nondegenerate closed subinterval of [0,1] that contains $x_0$. I'm trying proving this directly, but I think maybe a proof by contradiction could work here. Couldn't we just assume that $f$ is of bounded variation on every closed, bounded subset of [0,1] to begin the proof by contradiction? Perhaps then we could split up the interval [0,1] so that we end up with a situation where $f$ is of bounded variation on a subinterval that contains $x_0$? I'd appreciate some help here.

REPLY [4 votes]: As Didier mentioned, the right statement probably is: 

Show that there is a point x0 in [0,1]
  such that f fails to be of bounded
  variation on each nondegenerate closed
  subinterval of [0,1] which is a
  neighborhood of x0 in [0,1].

Here is an alternative proof:
We use the standard notation $V_a^b(f))$ to denote the variation of $f$ on $[a,b]$.
Let $A : =\{ x \in [0,1] | V_0^x(f) < \infty   \}$.
Then $0 \in A , 1 \notin A$ and it is very easy to see that $x_0=\sup A$ works.