# Zero-sets of BV functions

Let $f\colon [0,1]\to \mathbb{R}$ be a function of bounded variation. As $f$ is the difference of two monotone functions, the zero set $X = f^{-1}(0)$ is Borel (can we say more?).

Can we find a BV function that is strictly positive exactly on $X$?

I can do it if $X$ has countably many connected componenets but in general I don't know.

• Could you precise ? Why is $f$ is the difference of two monotone functions ? What is $X$ for you ? I know what connected components are for graphs, but what is it supposed to mean here ? – Robin Vogel Apr 20 '17 at 19:58

Theorem: a set $X\subseteq\mathbb R$ is the zero set of some function of bounded variation if and only if $\overline{X}\setminus X$ is countable.
In other words, $X$ is constructed by removing a countable set from some closed set. (I consider finite sets countable.) An example of a Borel set that does not have this property is the set of rational numbers $\mathbb Q$.
Proof of necessity: Let $f$ be a BV function such that $X=f^{-1}(0)$. If $a\in \overline{X}\setminus X$, then there is a sequence $(a_n)$ of elements of $X$ that converges to $a$. Since $f(a)\ne 0 = \lim f(a_n)$, the function $f$ is discontinuous at $a$. But a BV function has at most countably many discontinuities. $\quad\Box$
Proof of sufficiency: Recall that Every closed subset $E\subseteq \mathbb{R}^n$ is the zero point set of a smooth function. Let $g$ be such a function for $E=\overline{X}$. Inspecting the construction of $g$, one can see that $g\ge 0$ and that $\int_{\mathbb R} |g'|$ can be made finite. (Multiplying $g$ by a rapidly decaying strictly positive smooth function will do the trick).
Enumerate the elements of $\overline{X}\setminus X$ as $a_1,a_2,\dots$. Define $$f(x) = \begin{cases} g(x) + 2^{-n},\quad & x=a_n \\ g(x) \quad & \text{otherwise} \end{cases}$$ Clearly, $f\in BV$ and $f^{-1}(0)=X$. $\quad\Box$